Arithmetick:

  • VULGAR,
  • DECIMAL,
  • INSTRUMENTAL,
  • ALGEBRAICAL.

In Four Parts: Conteining

  • I Vulgar Arithmetick, Both in whole Numbers and Fractions, in a most plain and easie me­thod.
  • II Decimal Arithmetick, With the ground and reason thereof, illustrated by divers Ex­amples.
  • III Instrumental Arithmetick, Exactly performing all Questions of what nature so­ever in a Decimal way, by Scales, with much more ease and facility then can be effected, either by Vulgar or Decimal Arithmetick, the work of Reduction being wholly avoided. Nothing in this kind having been hi­therto published by any.
  • IV Algebraical Arithmetick, Con­teining an Abridgement of the precepts of that Art, and the use thereof, illustrated by Examples and Questions of divers kinds.

Whereunto is added the construction and use of several Tables of Interest and Annuities, Weights and Measures, both of our own and other Countries.

By WILLIAM LEYBOURN.

LONDON, Printed by R. and W. Leybourn, and are to be sold by George Sawbridge at the Bible on Ludgate-hill, 1660.

Vera Effigies Gulielmi Leybourn, Philom.
[...]

TO THE READER.

HEre is presented unto thee a short Treatise of A­rithmetick, I con­fesse there are e­nough, (if not too many) alrea­dy extant, not­withstanding I have adventured to publish this, more for variety than necessity, not doubting but in the per­usal thereof thou shalt find something in this worth thy labour, and what in [Page]other Books of this kind is wanting.

The whole Treatise is divided into four Parts. The first conteins Vulgar Arithmetick in Whole Numbers and Fractions; And in every Rule there are Examples for practice added, and Questions also wrought by those sin­gle Rules. And in Division, (which is the most difficult of the four Spe­cies,) there are five varieties, so that every man may make use of that which he best understands or fancies: And in the working of the Golden Rule, &c. I have made use sometimes of one kind of Division, and sometimes of another.

The Second Part conteins Decimal Arithmetick, with the ground and rea­son thereof, Also Tables of the Mo­neys, Weights, and Measures used in England, with directions for the ma­king of those Tables, and of any other. [Page]And lastly, there are Examples wrought in Decimal Numbers, in all the most usual Rules of Arithmetick, and those Examples are incumbred with as many Fractions as can possibly happen in any Question concerning buying or selling.

The third Part is of Instrumental Arithmetick, which performeth any Question Arithmetical in a Decimal way, without the help of Tables, by which the whole work of Reduction is avoided, there being certain Scales of English Money, Weights, and Mea­sures, by me contrived, by which (by inspection onely) the Decimal Faction of either Money, Weight, or Mea­sure, may be set down as exactly, and in lesse time, than they could have been taken out of the Decimal Tables in the Second Part of this Trea­tise. I have also in this Third Part, (the better to illustrate the use, and [Page]commodiousnesse of these Scales, a fi­gure whereof is inserted between page 246 and 247,) gone through all the most usual Rules of Arithmetick, gi­ving Examples in each Rule, by which the Reader may plainly perceive what labour there is saved by using the Scales, the whole work of Reduction being taken away, and the Fraction im­mediately set down at first, without Addition, or being already set down, reduced to the known part of the Inte­ger, without Subtraction.

Ʋnto this third part there is added an Appendix conteining certain Rules of Exchanges, with Tables of the Weights and Measures of forreign Countries compared with the Weights and Measures used in London, with an Example to illustrate the use of each Table; And lastly, there are five Ta­bles calculated at 6 per Cent. Com­pound [Page]Interest, by which the true va­luation of any Lease or Annuity, or Money forborn or rebated, may be easily known, with an Example shew­ing the use of each Table.

The Fourth Part conteineth an A­bridgement of the Precepts of Alge­bra, first written in French by James de Billy, a Translation whereof came to my hands some years since, in the perunsal whereof, finding the Precepts very plain and easie, and considering that we have but very little of this kind of Arithmetick in our English Tongue, I have adventured to insert it here as a Fourth Part, thereby to make this Work the more compleat: Ʋnto which Translation there is further ad­ded diverse Questions of good conse­quence, which were not in the Origi­nall, as by comparing them togeth [...] may appear.

[Page]This Treatise thus finished, I pre­sent thee with, desiring thy friendly acceptance, and pardon for such faults as may possibly have escaped the Presse, or my self, and in so doing thou wilt encourage him, who is

A friend to all that are Ma­thematically affected, William Leybonrn.

There is now in the Presse, and ready to come forth, Euclides Elements of Geometry, all the 15 books in a most compendious form contracted and demonstrated, together with some other pie­ces never before printed in English, all published by the care and industry of my loving friend Mr. John Leeke, Reader of Mathematicks in London.

The Contents.

The First Part of Vulgar Arithmetick.
  • NUmeration pag. 1
  • Addition pag. 7
  • Addition of English money pag. 10
  • Addition of Troy weight pag. 13
  • Addition of Avoirdupois little weight pag. 16
  • Addition of Avoirdupois great weight pag. 17
  • Tables of Liquid Measures, Dry Measures, Long Measures, and Time pag. 18
  • The proof of Addition pag. 21
  • Substraction pag. 22
  • Questions performed by Addition and Substra­ction pag. 31
  • Multiplication pag. 33
  • Compendiums in Multiplication pag. 41
  • The proof of Multiplication pag. 42
  • [Page]Questions performed by Multiplication pag. 44
  • Division pag. 45
  • A second way of Division pag. 55
  • A third way of Division pag. 57
  • A fourth way of Division pag. 60
  • Questions performed by Division onely pag. 65
  • Reduction pag. 67
  • Progression pag. 70
  • Geometrical Progression pag. 75
  • The Golden Rule pag. 80
  • The Golden Rule Reverse pag. 88
  • The Golden Rule compounded of five numbers pag. 92
  • Of Fractions pag. 96
  • Numeration pag. 97
  • Multiplication pag. 100
  • Division pag. 102
  • Reduction pag. 104
  • Addition pag. 110
  • Substraction pag. 112
  • The Rule of Fellowship pag. 114
  • The Rule of Fellowship with time pag. 120
  • Alligation pag. 124
  • Position pag. 131
  • The Rule of Ceres and Virginum pag. 137
  • Extraction of the Square Root pag. 147
  • A table of Square Roots from 1 to 1000 pag. 151
  • Extraction of the Cube Root pag. 162
  • A table of Cube Roots from 1 to 1000 pag. 165
  • Some uses of the Square and Cube Roots pag. 177
The Second Part of Decimal Arithmetick.
  • [Page]TAbles of Reduction of English Coin 185
  • Of Troy weight Ibid.
  • Of Avoirdupois great weight 187
  • Of Avoirdupois little weight 188
  • Of Liquid Measures Ibid.
  • Of Dry Measures 189
  • Of Long Measures Ibid.
  • Of Time Ibid.
  • Of Dozens 190
  • The use of the Tables of Reduction 191
  • Notation of Decimals 208
  • Addition of Decimals 209
  • Subtraction of Decimals 213
  • Multiplication of Decimals 215
  • Division of Decimals 220
  • The Rule of Three in Fractions both Vulgar and De­cimal 228
  • The Rule of Three reverse in Decimals 239
  • The double Rule of Three in Decimals 240
The Third Part of Instrumental Arithmetick.
  • [Page]INstrumental Arithmetick what 245
  • A Figure of the Scales 246, 247
  • Numeration upon the Scales 247
  • Addition by the Scales 257
  • Substraction by the Scales 262
  • Multiplication by Nepeirs Bones 264
  • A Figure of the Bones 265
  • Division by the Bones 27 [...]
  • Examples in the Rule of Three direct 273
  • Examples in the Rule of Three reverse 275
  • Examples in the double Rule of Three 277
  • Examples in Barter 279
  • Examples in Fellowship 281
  • Examples in Losse and Gain 285
  • Examples in Losse and Gain upon time 280
  • AN Appendix conteining divers questions con­cerning Exchanges of the Coyns, Weights, and Measures of one Countrey, with those of another Countrey, with divers tables thereunto belonging, al­so Tables of Interest and Annuities at 6 per Cent. Compound Interest.
The fourth Part, being an Abridge­ment of the precepts of Algebra.
  • [Page]A Table of the Cossick Characters 324
  • The Alegorithm of the Cossick Numbers 325
  • Addition of simple Cossick numbers 326
  • Substraction of simple Cossick numbers 326
  • Multplication of simple Cossick numbers 326
  • Division of simple Cossick numbers 326
  • Addition of numbers composed and diminished 327
  • Subtraction 328
  • Multiplication 329
  • Division 330
  • The Alegorithm of Fractions ibid.
  • The Rule of Algebra ibid.
  • How Equations are found 331
  • How Equations are reduced 332
  • When to extract a Root 333
  • To extract the square root of numbers compounded and diminished 334
  • How to know whether a question propounded be impos­sible 336
  • The Algorithm and use of second roots 337
  • Addition of second roots ib.
  • Substraction of second roots ib.
  • Multiplication of second roots 338
  • Division of second roots ib.
  • The extraction and use of second roots ib.
  • The Algorithm and extraction of the root from surd and irrational numbers. 339
  • [Page]Reduction of simple surd roots to one and the same de­nomination 340
  • Multiplication and division of simple surd roots 341
  • How to know whether two surd roots be commensurable, or incommensurable ib.
  • Addition of simple errational roots ib.
  • Substraction of simple irrational roots 342
  • Addition and Subtraction of numbers, surd, compound, and diminished 343
  • Multiplication of numbers, surd, compound, and dini­nished ib.
  • Multiplication of universal roots 346
  • Division of universal roots 347
  • Addition and Subtraction of universal roots ib.
  • The Extraction of the root from Binomials and Apoto­mals 348
  • The use of Algebra 349
  • Questions resolved by one simple equatoon ib.
  • Questions resolved by a compound equation 352
  • Questions resolved by surd numbers 356
  • Geometrical questions resolved by Algebrae 359
  • Questions resolved by second roots 362
  • Questions resolved indefinitely 364
In Appendix.
  • QUestions in Algebra which require the Rule of thee in their operation 369
  • Examples in Algera concerning Squares 390
  • Examples relating to Cubes 391

Numeration.

NUmeration is accounted the first part of Arithmetick, and it is to know how to read a Summe of fi­gures expressed in writing; or to write down any Summe to be ex­pressed.

To the doing of which there are four things necessary.

First, to know their number, which is Nine.

Secondly, their shapes, which are 1 2 3 4 5 6 7 8 9. Of which the first toward the left hand ever signifieth One, the second Two, &c.

Thirdly, to know the value of their places.

Lastly, How their proper signification i [...] al­tered thereby.

The value of their places is thus, When two, three, or more figures stand in one Summe, tha [...] is, without any Point, Line or Comma betwixt [Page 2]them, as 321, that place next the right hand where the figure 1 standeth, is called the place of Unity, or Unities, and the figure 1 standeth in that place onely for one, and the figure 2 when it is found in that first place, stands only for two; and the like of the rest.

But in the Summe 321, above expressed, the figure 2 is in the second place, and every place conteines the value of that place before towards the right hand ten times; and therefore the fi­gure 2 doth not signifie two, but (in this second place) ten times two, that is Twenty. And so the figure 3 if it had been in that place had sig­nified ten times three, that is Thirty; but be­ing here in the third place it signifies ten times thirty, that is Three hundred. And so the whole Summe 321, is to be read, Three hundred twenty and one.

Fourthly, It is hereby seen, how their proper significations, which were Three, Two and One, are altered by being thus placed, and the Sum, which otherwise had been but Six, is Three hun­dred twenty one, as before.

In like sort, if their had been more places, as seven, the value is quite through increased ten times by being a place behinde towards the left hand; as in the Summe 1̇111̇111̇, The fi­gure 1 in the second place stands for ten times one (that is ten,) in the third for ten times ten (which is one hundred,) in the fourth for ten hundred, (which is called one thousand,) in the fifth for ten thousand, in the sixth for ten [Page 3]times ten thousand (which is one hundred thou­sand) in the last (here the seventh) place for ten hundred thousand, which is called a Milion: and so on, if there were more places, observing the same order.

Now to read this readily, make a prick over the place of Unity, another over the third from it, and over every third still towards the left hand, for so those points will be over the places of Unites, Thousands, and Milions; and so be­ginning at the last, that is, at the left hand, read one Milion, and because the three following to­wards the right signifie properly one hundred and eleven, but the prick belonging to them is in the place of thousands, call them one hundred and eleven thousand, and three remaining be­ing under the point over Unity, signifie onely one hundred and eleven; and all three points read together in one summe is, One Milion, one hundred and eleven Thousand, one hundred and eleven.

In like manner, if this number 73598624 were given to be read (according to former directions) make a prick over every third figure, beginning with the first figure towards the right hand (which is the place of Unity) and then will your number stand thus.

73̇598̇624̇

Then for the ready reading thereof (because the third prick signifieth Millions) call all the fi­gures towards the left hand, standing from that prick, Milions, which in this example are 7 and 3, [Page 4]so then this number conteins 73 Millions, 598 Thousand, 624 which in words at length we read Seventy three Millions, five hundred ninety eight thousand, six hundred twenty four.

Let thus much suffice concerning the planing of large numbers, for the ready reading of them, only take these foure Tables following, for illu­stration of what hath been hitherto delivered in words, the very sight whereof is better then a whole Chapter of information.

The first Table is thus to be read] One in the first place signifies One. One in the second place signifies Ten. One in the third place sig­nifies a hundred, &c. as in the Table.

The second Table is thus to be read] In this Table you shall find the last number thereof to consist of these figures, 357.846.903. with a point or comma betwixt every third figure, for di­stinction sake, and also every three figures in their order are connected together with this

[figure]

brace, which denominate the places of Millions, Thousands, Hundreds, so that the last number of this Table will evidently appear to be 357 Millions, 846 Thousands 903.

The third Table is only certain rowes of fi­gures set together and orderly disposed, having the signification or reading of the same numbers in words at length to them annexed, and is only inserted for the better satisfaction of such as shall doubt whether they perfectly understand what hath been before taught.

The fourth Table is much like the second, on­ly it consisteth but of one number and extends [Page 5]three places farther then the greatest number in the second Table doth: viz. to twelve places which figures are thus to be read. 736 Millions of Millions, 842 Millions, 708 Thousand, 645.

(1 Table)
One in thefirstplace signifies1 one
One in thesecondplace signifies10 ten
One in thethirdplace signifies100 a hundred
One in thefourthplace signifies1000 a thousand
One in thefifthplace signifies10000 ten thousand
One in thesixthplace signifies100000 a hundred thousand
One in theseventhplace signifies1000000 a Million
One in theeigthplace signifies10000000 ten Millions
One in theninthplace signifies100000000 a hundred Milion
(2 Table)
8
54
762
3.483
97.621
243.794
8.749.807
57.316.248
357.846.903
Millions thousands hundreds

[Page 6]

(3 Table)
number of places18 (eight,
number of places254 (fifty four,
number of places3762 (seven hundred sixty two,
number of places43483 (three thousand, four hundred eighty three
number of places597621 (ninety seven thousand, six hundred twenty one
number of places6243794 (two hundred 43 thousand 7 hundred 94
number of places78749807 (eight milions, 749 thous. 8 hundred and 7
number of places857316248 (fifty seven mil. 3 hund. and 16 thous. 248
number of places935784698 (three hundred fifty seven milions, eight (hundred 46 thous. 9 hund. and 3.
(4 Table)
7hundredmilions of milions
3tenmilions of milions
6onemilions of milions
8hundredmilions
4tenmilions
2onemilions
7hundredthousands
0tenthousands
8onethousands
6hundredhundreds
4tenhundreds
5onehundreds

Addition.

ADdition is the collecting or gathering to­gether of two or more sums, either of one or of diverse denominations, into one sum, which is called the (Aggregate) (Total) or Grosse sum.)

In addition of numbers of one denomination, the order is to set the numbers to be added one directly under the other, that is to say Unites un­der Unites, Tens under Tens, Hundreds under Hundreds, &c.

The Rule.

Having placed your numbers to be added in due or­der, one under another, draw a line under them, and begin at the lowermost figure towards your right hand, and adde that to the next figure above, and the summe of them to the next figure above that, proceeding in this or­der till you have added the whole line together, which when you have done, consider how many tens are contain­ed in that line, and for every ten keep one Ʋnite in your mind, to be added to the next row, but if there be any odde digits, you must set them beneath the stroake, just under the line you added together, having thus [Page 8]finished the addition of one line proceed to the next, and from thence to the third, and so forward, be there never so many. The Examples following will make this plain,

Example 1.

Thousands Hundreds Tens Unites
7832
5609
376
8547
22364

Let the numbers given to be added together be 7832, 5609, 376, 8547, having thus placed them in order one under another as in the Margine is done, draw a line under them, then be­gin your addition, at the lowermost figure towards your right hand, say­ing, 7 and 6 is 13, and 9 is 22, and 2 is 24 now because in 24, there is two tens, and 4 remaining, I place the 4 under the line, and carry the two zens to the next row, saying, 2 which I carryed and 4 makes 6, and 7 makes 13, and 3 makes 16, in which row there is but one ten conteined and 6 remaining, which 6 I set under the line, and carry the ten to the next row of hundreds, say­ing, 1 that I carryed and 5 makes 6, and 3 makes 9, and 6 makes 15, and 8 makes 23, in which 23, ten is conteined two times, and 3 re­maining, the three I place under the line, and carry the two tens to the next row, saying 2 which I carryed and 8 makes 10, and 5 makes 15, and 7 makes 22, in which ten is contein­ed two times, and 2 remaining, which 2 I set under the line, and because there is never ano­ther [Page 9]row to be added (to which I should carry the two tens) I therefore set it down also under the line towards the left hand, as you see done in the margine, so the total or grosse sum of these numbers being added together is 22364.

Example, 2.

Apple trees137
Pear trees76
Cherry trees107
Plum trees36
Trees in all355

A man hath in his Orchard 136 Apple trees, 76 Pear trees, 107 Cherry trees, and 36 Plum trees, and he desires readily to know how many trees he hath in all, place your uum­bers one under another as in the margine, and then begin to adde them together, at your right hand, saying, 6 and 7, make 13, and 6 make 19, and 6 make 25, place 5 under the line, and carry 2 to the next row, saying 2 and 3 is 5, and 7 is 12, and 3 is 15, place 5 under the line, and carry 1 to the next row, saying 1 and 1 is 2, and 1 is 3 which 3 I set under the line, and because there was no tens conteined in that line therefore the totall is 355, and so many Trees are in the Orchard.

Other Examples for Practice. [...]

Addition of numbers of diverse Denominations. 1 Addition of English Money

The most usual Coyns used in England are Pounds, Shillings, Pence, and Farthings, of which Coyns

  • 4 Farthings make 1 Peny thus charactred d.
  • 12 Pence make 1 Shilling thus charactred s.
  • 20 Shillings make 1 Pound. thus charactred li.

for a farthing we use q.

THE RƲLE.

In the addition of numbers of divers deno­minations this order is to be observed, viz. Place all numbers of the same denomination one direct­ly under another, as Pounds under pounds, shillings under shillings, pence under pence, and farthings un­der farthings. Then draw a line under them, and begin your Addition with the least denomination first, observing how many times the next greater denomina­tion is contained in the least, and for every time carry one unite to the next denomination, as before you did the tens, setting down the remainder if any be, then adding the next denomination together, take notice how many times the next greater denomination is contained in that lesser, carrying for every time one to the next, [Page 11]thus proceeding till you have gone over all the denominations, be they never so many.

Example 1.

li.s.d.q.
37169.3.
2109.8.1
131292
721932

Let the numbers to be added toge­ther be 37 li. 16 s. 9 d. 3 q. 21 li. 9 s. 8 d. 1 q. 13 li. 12 s. 9 d. 2 q. Place the numbers as in the margine, draw a line under them, and begin with the least de­nomination (which in this example is farthings,) first, saying, 2 q. and 1 q. is 3 q. and 3 q. is 6 q. which is one peny and 2 q. remaining, which 2 q. I place under the line, and carry the one peny to the next row which is the place of pence, saying, one peny and 9 d. is 10 d. and 8 d. is 18 d. which is 1 s. and 6 d. (Now against the 8 I make a prick with my pen for my better remembrance to signifie that there is one shilling to be carried to the place of shillings,) then go on and say 6 d. and 9 d. is 15 d. which is 1 s. and 3 d. therefore against 9. I make a prick with my pen, and (because that is the last num­ber) I set down the odde 3 d. under the place of pence, and (because I finde two pricks in the line of pence, therefore) I carry 2 s. to the place of shillings saying 2 s. which I carryed, and 12 s. is 14 s. and 9 s. is 23 s. which is one pound, and 3 s. remaining, I make a prick against 9, and go­ing on, say 3 s. and 16 s. is 19 s. which (being there is no more numbers to be added, and being also lesse then 20 s.) I set under the line, and finding one prick in the line of shillings, I therefore car­ry [Page 12]one to the place of pounds, saying, one which I carryed and 3 is 4, and 1 is 5, and 7 is 12. set down the 2 under the line (as in addition of num­bers of one denomination) and carry one to the next row, saying one that I carried and 1 is 2, and 2 is 4, and 3 is 7, which being the last I set down, and so the total or grosse sum is 72 li. 19 s. 3 d. 2 q.

Example 2

li.s.d.
2916.8.
3217.9.
811311
144084

Let the numbers to be added be 29 li. 16 s. 8 d. 32 li. 17 s. 9 d. 81 li. 13 s. 11 d. and let it be required to find the totall or grosse sum. Here in this Example the least denomination is pence, therefore I begin with them, and say, 11 d. and 9 d. is 20 d. which is 1 s. and 8 d. make a prick a­gainst the 9 and say 8 d. and 8 d. is 16 d. that is 1 s. and 4 d. make a prick against the 8 and set down the odde 4 d. then (because there are two pricks in the line of pence) you must car­ry 2 s. to the place of shillings, saying 2 s. which I carry and 13 s. is 15 s. and 17 s. is 32 s. which is 1 l. 12 s. make a prick agianst 17, and say 12 s. and 16 s. is 28 s. make a prick a­gainst 16, and (because there is no more num­bers to be added) set down the odde 8 s. under shillings, and (being there is two pricks in the line of shillings) carry 2 to the place of pounds, saying 2 and 1 is 3, and 2 is 5, and 9 is 14, set down 4 and carry 1 to the next line, and say 1 and 8 is 9, and 3 is 12, and 2 is 14, which (be­cause [Page 13]it is the last) you must set down, so is the totall or grosse sum 144 li. 8 s. 4 d.

Other Examples for Practice.
li.s.d.q.
2918.73
6311.2.1.
129402
37101
226183

li.s.d.
3628.
2902
3116.9.
625
10320

2 Addition of Troy Weight.

Troy weight is a Weight used in England, by the which is weighed, Bread, Gold, Silver, Pearl, &c. the most usual denominations of which weight are, Pounds, Ounces Peny weights, and grains, of which

  • 24 Grains make 1 Peny weig. thus charactred pw.
  • 20 Peny weight make 1 Ounce thus charactred ou.
  • 12 Ounces make 1 Pound thus charactred lib

for a grain we write gr.

The Addition of Troy weight (and conse­quently of any other weight or measure what­soever either Domestique or Forreign) differeth nothing at all from the addition of Coine last [Page 14]taught, if the affinity of one denomination to an other be first known, for whereas in money be­cause 1 [...] d. make 1 s. you therefore observe how many twelves there are in the addition of your pence, and for every 12 you adde one shilling to the place of shillings, so in the addition of Troy weight, knowing that 24 gr. make one pe­ny weight, you must therefore in the addition of Grains of Troy weight observe how many times 24 you finde in your line of Grains, and for every 24, carry one to the place of peny weights, likewise in the addition of peny weights Troy, you must consider how many times 20 is con­tained in your line, and for every 20 carry one to the place of ounces, (because 20 peny weights make an ounce.) Also in the addition of Ounces Troy, you must observe how many times 12 you finde in your line of ounces, and for every 12 carry one to the place of pounds then lastly adde your pounds together, as numbers of one denomination.

Example.

li.ou.pw.gr.
711.1319
60716.19.
3070906
18021920

Let the numbers to be added to­gether be 7 li. 11 ou. 13 pw. 19 gr. 6 li 7 ou. 16 pw. 19 gr. 3 li. 7 ou. 9 pw. 6 gr. Place your numbers as in ad­dition of money, each under other according to their respective de­nominations, as in the margine then draw a line under them and begin your Addition with the least denomination first, viz. grains, [Page 15]saying 6 gr. and 19 gr. is 25 gr. which is one pe­ny weight and one grain, make a prick against 19, and carry the odde grain to the number a­bove, saying 1 gr. and 19 gr. is 20 gr. which (because it is lesse then one peny weight) I set under the line, then finding one prick in the line of grains, I (therefore) carry one to the place of peny weights, saying 1 and 9 is 10, and 16 is 26, which is one ounce, and 6 pw. make a prick against 16, and say 6 and 13 is 19, which (being lesse then an ounce) set under the line, then for the one prick, carry 1 to the place of ounces, saying 1 and 7 is 8, and 7 is 15, which is one pound and 3 ou. make a prick at 7, and say 3 and 11 is 14, which is one pound and 2 ounces, set down the 2 ounces, and for the two pricks carry 2 pounds to the place of pounds, saying 2 and 3 is 5, and 6 is 11, and 7 is 18, which set under the place of pounds, so is your addition ended, and the sum is 18 li. 2 ou. 19 pw. 20 gr.

Other Examples for Practice.
i.ou.pw.gr.
329.1216
1711.69.
348.15.10
81047
9431818

li.ou.pw.gr.
01017.11
06.05
00168.
05219
1101619

3 Addition of Avoirdupois little Weight.

There is another kind of weight most com­monly used in England called Avoirdupois little weight, by which is weighed all sorts of wares or merchandise Garbable, as Suger, Pepper, Cloves, Mace, &c. This weight is commonly di­vided into these denominations, Pounds, Ounces, and Drams, of which

  • 16 Drams make 1 Ounce thus charact. cu.
  • 16 Ounces make 1 Pound thus charact. li.

for a dram we write dr.

In the addition of Avoirdupois weight, you must observe the very same method and order as in Money and Troy weight, having due re­spect to the quantity of the denomination, as in the addition of drams to make a prick at every 16, setting down the remainder and for every prick carrying a unite to the next place. The pre­ceding rules being so copious in this particular I shall forbear to make any verball illustration, but only give you some examples ready wrought, to­gether with the most usual parts into which the Weights and Measures now used in England are divided into: which to the ingenious will be of most validity.

[Page 17]

Examples of Addition of Avoirdupois little weight.
li.ou.dr.
1211.09
760512.
3210.00
9107.13.
321307
2460009

li.ou.dr.
0613.07.
0509.12
060309.
100000
050709
340205

4 Addition of Avoirdupois great weight.

There is a weight commonly used in England, by which is weighed all commodities that are sold by the hundred, as Corents, Wool, Flesh, Butter, Cheese, and the like, the which hun­dred weight containeth 112 pounds, and the hundred weight is divided into quarters, pounds, and ounces, so that

  • 16 Ounces makes 1 Pound thus chaeactred li.
  • 28 Pounds makes 1 quarter of a C. thus chaeactred qr.
  • 4 quarters makes 1 Hundred weig. thus chaeactred C.

for an Ounce we write oz.

[Page 18]

Examples of Addition of Avoirdupois great Weight.
C.qr.li.ou.
3703.2112.
09010603
330220.00
10000000
1203.0703
103022702

C.qr.li.ou.
0501.0007
03021806.
00010608
1103.0400
06011005
17011110

I might further proceed to shew you Exam­ples of addition of common English measures, viz. of long measures, Liquid measures, and dry measures, as also of Time, Motion, &c. but the preceding Examples being of sufficient ex­tent, I shall forbear to trouble either my selfe or the Reader with that which I conceive superflu­ous: Only, before I leave Addition, I will give you a brief view of the most usual measures in England, which take as followeth. And

1 Of Liquid Measures.

Liquid measures are those in which all sorts of Liquid substances are measred, of which (according to the Statute of 12 Hen. 7. chap. 5.) a Pinte is the least, from which the greater Li­quid measures are deduced, according as is ex­pressed in the Table following.

  • [Page 19]2 Pintes make 1 Quart
  • 2 Quarts make 1 Pottle
  • 2 Pottles make 1 Gallon
  • 8 Gallons make 1 Firkin of Ale, Sope, or
  • 9 Gallons make 1 Firkin of Beer (Herings
  • 10 ½ Gallons make 1 Firkin of Salmon or Eels
  • 2 Firkins make 1 Kilderkin
  • 2 Kilderkins make 1 Barrel
  • 42 Gallons make 1 Tierce of Wine
  • 63 Gallons make 1 Hogshed
  • 2 Hogsheads make 1 Pipe or But
  • 2 Pipes or Buts make 1 Tun of Wine.

2 Of Dry Measures.

Dry Measures are these in which all kind of dry substances are measured, as Corn, Salt, Cole, Sand, &c. of which a pinte is the least.

  • 2 Pintes make 1 quart
  • 2 quarts make 1 Pottle
  • 2 Pottles make 1 Gallon
  • 2 Gallons. make 1 Peck
  • 4 Pecks make 1 Bushel Land measure
  • 5 Pecks make 1 Bushel Water measure
  • 8 Bushels make 1 quarter
  • 4 quarters make 1 Chaldron
  • 5 quarters make 1 Wey.

3 Of Long Measures.

Long Measure is that by which is measured Cloth, Land, Board, Glasse, Pavement, Tapestry, &c [...] [Page 20]of which measures (according to the Statute of 33 Ed. 1. and 25 El.) a Barley corn is the least. So that

  • 3 Barley Corns make 1 Inch
  • 12 Inches make 1 Foot
  • 3 Foot make 1 Yard
  • 3 Foot 9 inches make 1 Ell
  • 6 Foot make 1 Fadome
  • 5 ½ Yards, or 16 ½ Foot make 1 Pole or Perch
  • 40 Perches make 1 Furlong
  • 8 Furlongs make 1 English mile.

4 Of Time.

Time consisteth of Years, Moneths, Weeks, Dayes, Houres, and Minutes. So that

  • 60 Minutes make 1 Hour
  • 24 Hours make 1 Day natural
  • 7 Dayes make 1 Week
  • 4 Weeks make 1 Moneth of 28 dayes
  • 13 Moneths, one day 6 houres make 1 Year.

5 Of Apothecaries Weights.

The weights used by Apothecaries are Grains, Scruples, Drams, and Ounces, of which [Page 21]

  • 20 Grains make 1 Scruple thus charact. ℥
  • 3 Scruples make 1 Dram thus charact. ʒ
  • 8 Drams make 1 Ounce thus charact. ℈
  • 12 Ounces make 1 Pound thus charact. li.

By help of these Tables, and the rules and cau­tions before expressed, any man may make addi­tion of any of the abovesaid measures one with another, and therefore I shall forbear to illust­rate them by Examples, but leave them to eve­ry mans own practice, and thus I conclude Addition.

The Proof of Addition.

Having placed your numbers in order, and added them together, and set the Total under the line, cut off the upper number by drawing a line with your pen betwixt that and the others, then adde all the numbers together except the uppermost, and set the Total of them under the Total before found, then adde this last Total, and the first number which you cut off with your pen together, and if the sum of those two num­bers be equal with your Total sum first found, then is your work right, otherwise not.

Example. In the first example of whole num­bers, the sums to be added were 7833, 5609, 376, and 8547, these numbers placed in due or­der and added together, the total or grosse sum of them is 22364, now to prove whether this [Page 22]Total be true or not, I cut off the uppermost number, (to wit 7832,) with a dash of the pen, and I adde the other three numbers together, name­ly, 5609, 376, and 8547, and the Total of them is 14732, which number be­ing added to 7832 (the number cut off) the sum of them is 22364 exact­ly agreeing with the To­tal first found, cleerly evi­denceth that the addition was truly performed; but if they had disagreed then the work had been erroneous. The like course must be taken for the proof of those sums which have different denominations as in Mo­ney and Weight, as by the examples following will appear.

Other Examples proved.

1 Example of Money.
li.s.d.q.
371693
2198.1
131292
1 Total 721932
2 Total 35253
Proof 721932

li.ou.pw.gr.
3291216
171169
3481510
81047
9431818
61662
9431818

[Page 23]There are are other wayes to prove Addition, by casting away of all the nines in numbers of one denomination, and of all the twelves, twenties, and nines, in pounds shillings and pence, &c. but this as the most certain and easie I embrace, and thus much for Addition and the proofe thereof.

Subtraction.

SUbtrrction is the taking of one or more small sums out of a greater, as 7 s. out of 12 s. or 37 li. out of 100 li. or 137 foot out of 983 foot, and the like.

As in Addition the sums to be Added may be either of one or of diverse denminati­ons, so likewise they may be in Subtraction, and the manner of placing them is the same, for you must set Unites under Unites. Tens under Tens, Hundreds under Hundreds, &c.

Example 1

number given986
number to be subtracted234
remainder752

Of Subtraction of numbers of one denomination, let it be required to subtract 234, out of 986, place the numbers one unde the o­ther as you see done in the Margine, draw a line under them and begin with the first figure to­wards your right hand, which is 4, saying take 4 from 6, and there remains 2, place 2 under the line, and go to the next figure which is 3, say­ing, take 3 from 8, and there remains 5, place 5 under the line, and go to the next figure which is 2, saying take 2 out of 9 and there re­mains 7, place 7 under the line, and your sub­traction is ended, and it is evident by the work, [Page 25]that if you take 234 out of 986 there will remain 75 [...], which you may thus prove, for if you adde the 234 to 752, you shall find the sum of that addition to be 986, which is equall to the whole sum from which 234 was subtracted,

Example. 2.

96527
2976
9355r

Let it be required to subtact 2976 out of 96527, place the numbers one under ano­ther as in the margine you fee done, then draw a line under them, and beginning with the first figure towards your left hand, say take 6 out of 7, and there remains 1, place 1 under the line and pro­deed to the next figure, saying 7 out of 2 I can not (wherefore you must alwayes adde 10 to the number above which in this example is 2, and it makes it 12,) therefore take 7 out of 12, and there remains 5, place 5 under the line, and (be­cause you added 10 to the 2 to make it 12, you must) carry a unite to the next figure, saying, one which I carryed and 9 is 10, take ten out of 5, which I cannot, therefore I must adde 10 to 5 and it makes 15, and say 10 out of 15, and there remains 5, place 5 under the line, and because you added 10 to 5 to make it 15, you must therefore carry a unite to the next figure saying, one which I carryed, and 2 is 3, take 3 out of 6 and there remains 3, place 3 under the line, and because there is no more figures to be subtracted from the number above, you must, say nothing fom 9 and there remains 9, set the 9 under the line, and your Subtraction is ended.

[Page 26]

Other Examples for Practice.
 li.reams ofPapersheep 
Lent5762bought9765from1000
Paid378sold6529take394
Rests to pay5384unsold3236remains606

Subtraction of Numbers of diverse Denominations, 1 Of English Money.

In Subtraction of numbers of diverse denominations, you must observe the same order as in Addition, namely, to place every number in due order, with re­spect to its denomination, as pounds under pounds, shil­lings under shillings, &c. the greater number alwayes uppermost, and drawing a line under them, begin with the least denomination first, subtracting it from the line above, and setting the remainder under the line as in whole numbers, but if the pence or shillings in the upper row be smaller then those in the nea­ther row, you must adde 12 d. or 20 s. to the smaller number, that so subtraction may be made, as by the ex­amples following will appear.

Example 1.

 li.s.d.
Lent2691810
Paid38128
rests23162

Let it be required to subtract 38 li. 12 s. 8 d. out of 269 l. 18 s. 10 d. Place your numbers as in the margine, then be­ginning with the least denomination first, [Page 27](which in this Example is pence) say 8 d. from 10 d., and there remains 2 d. set the 2 d. under the line, and proceed to the next denomination which is shillings, saying take 12 s. out of 18 s. and there remains 6 s. place 6 s, under the line, and go to the pounds, saying 8 out of 9 and there remains 1, place 1 under the line, and say 3 out of 6, and there remains 3, then (because there is no more figures to be subtracted) say nothing out of 2 and there remains 2, which set under the line, so is your subtraction ended, and the re­mainder is 231 li. 6 s. 2 d.

Example 2.

 li.s.p.
Lent93201007
Paid26281610
Rests6691118

Let it be required to subtract 2628 li. 16 s. 10 d. out of 9320 li. 10 s. 7 d. place the num­bers in order, and beginning with the pence, say 10 d. out of 7 d. I cannot, (therefore I must adde 12 d. (which is one shillings) to 7 d. and it makes 19 d.) but 10 d. out of 19 d. and there remains 9 d. set the 9 d. under the line, and (be­cause I added 12 d. to 7 d.) I must therefore carry one to the place of shillings, saying 1 s. which I carryed, and 16 s. is 17. then 17 s. from 10 s. I cannot take, therefore, I must adde 20 s. (which is one pound) to 10 s. and it makes 30 s. and 17 s. out of 30 s. and there remains 13, set 13 under the line, and carry one to the place of pounds, saying one which I carried and 8 is 9, take 9 of o I cannot, but 9 out of 10, and there [Page 28]remains 1, set 1 under the line, and carry a unite to the next place, saying 1 which I carried and 2 is 3, take 3 out of 2 I cannot, but 3 out of 12, and there remains 9, place 9 under the line, and carry 1 to the next place, saying 1 which I car­ry and 6 is 7, take 7 out of 3 I cannot, but 7 out of 13, and there remains 6, place 6 under the line, and carry one to the next row, saying 1 and 2 is 3, take 3 from 9 and there remains 6, place 6 under the line, so is your Subtraction ended, and the remainder is 6691 li. 13 s. 9 d.

Example 3. Suppose a man had lent to another man 1000 pound, and that the borrower had paid thereof at one time 127 li. at another time 430 li. 10 s. and at a third payment 50 li. and the creditor would know how much he hath recei­ved, and how much is owing of his debtor.

Place the numbers as here you see,

 li.s.d.
Money lent10000000
 1270000
paid at seve­ral times4301000
 500000
paid in all6071000
rests to pay3921000

first the sum of money lent, and draw a line under it, then set the sums paid at se­veral times one under another, and draw a line under them: Then adde all the sums which have been paid at several times together, which make 607 li. 10 s. which is the sum which the debtor hath paid in [Page 29]all, then subtract this 607 li. 10 s. from 1000 li. and there will remain 392 li. 10 s. and so much is still owing to the Creditor.

Other Examples for Practice.
 li.s.d.
Lent2601136
Paid9879
Rests250359

 li.s.d.
Owing in all1000000
Paid in all [...]61006
Rests to pay631006

 li.s.d.q.
Lent3625160803
Paid at several times100000000
Paid at several times336100602
Paid at several times039120902
Paid at several times100000000
Paid in all576030400
Rests to pay3049130403

The Proof of Subtraction.

The Proof of Subtraction is performed by Addition, for adding the number to be subtract­ed, to the remainder, the Sum of them must be equal to the number given, if you have truly wrought. As in the first example of numbers of one denomination,

the number given is986
the number to be subtracted is234
the remainer is752
Proof986

[Page 30]Adde the number to be subtracted 234, to the re­mainer 752, the sum of them is 986, equal to the number given.

Examples for Practice proved.
 li.s.d.
Lent621809
Paid371906
Rests241903
Proof621809
 li.s.d.
Borrowed1000000
Received361304
Due630608
Proof1000000

Other Examples in Weight and Measure.

1 Exa. in Troy weight
 li.ou.pw.gr.
Bought of Silver07111319
Sold05070305
unsold42041014
Proof07111319

2 Example in Avoirdu­pois great weight.
 C.q.li.ou.
Bought37032211
Sold13012306
Rests24012705
Proof37032211

3 Example in Avoirdu­pois little weight.
 li.ou.dr.
Bought841213
Sold260811
Rests580402
Proof841213

4 Example in Time.
 daysho.m.
From3642350
Take760922
Rests2881428
Proof3642350

Questions performed by Addition and Subtraction.

Question 1. What number is that which added to 376 shall make 1000? Subtract 376 from 1000, the remainder is 624, the number sought.

Question. 2. What number of pounds, shil­lings and pence must be added to 36 l. 17 s. 3 d. to make that sum up 100 li. subtract 36 li. 17 s. 3 d. from 100 li. the remainer is 63 li. 2 s. 9 d. which added to 36 li. 17 s. 3 d. makes 100 li.

Quest. 3. In the year of our Lord 1440, the famous art or mystery of Printing was invented, I would know how long it is since that time to this yeer of our Lord 1655. From 1655, subtract 1440, the remainer is 215. and so many years are expired since printing was invented.

Question 4. An Armie consisting of 13721 horse, and 26850 foot, in an engagement there were slain 3760 hors, & 7523 foot, the question is how many were slain in all, & how many horse and how many foot escaped. From the 1372 horse which went out, subtract the 3760 that were slain, there remains 9961, and so many horse escaped, Also from the 26850 foot which went out, subtract the 7523 which were slain, and there remains 19327, the number of foot [Page 32]which escaped, and by adding the 3760 horse which were slain, to the 7523 foot that were slain, their Total is 11283, and so many were slain in all.

Multiplication.

MUltiplication is that part of Arithme­tick which teacheth how to encrease one number by another, so that the number produced by their Multipli­cation shall contain one of the numbers multipli­ed so many times as there are unites contained in the other. Multiplication may fitly be ter­med a Compendium of Addition, for that it performeth at one operation the same which to effect by Addition would require many. For instance, if it were required to know how much 7 times 5 is, to perform this by addition, I must set seven fives, or five sevens, one under another, and adding them together, I shall finde that either of their Totalls shall contain 35, but this by multiplication is performed with far more brevity, as by examples hereafter shall appear.

Before you enter upon the practice of Multi­plication, it is necessary to remember the pro­duct produced by the multiplication of any one of the nine digits, by any other of the same, as readily to know; that 4 times 5 is 20, 6 times 7 is 42, 2 times 9 is 18, 7 times 9 is 63. 8 times 9 is 72, &c. Which this Table following, will[Page 34]plainly declare, and must be perfectly learned by heart before you attempt to multiply great numbers.

Multiplication Table.
  • 2 times 2 makes 4
  • 2 times 3 makes 6
  • 2 times 4 makes 8
  • 2 times 5 makes 10
  • 2 times 6 makes 12
  • 2 times 7 makes 14
  • 2 times 8 makes 16
  • 2 times 9 makes 18
  • 3 times 3 makes 9
  • 3 times 4 makes 12
  • 3 times 5 makes 15
  • 3 times 6 makes 18
  • 3 times 7 makes 21
  • 3 times 8 makes 24
  • 3 times 9 makes 27
  • 4 times 4 makes 16
  • 4 times 5 makes 20
  • 4 times 6 makes 24
  • 4 times 7 makes 28
  • 4 times 8 makes 32
  • 4 times 9 makes 36
  • 5 times 5 makes 25
  • 5 times 6 makes 30
  • 5 times 7 makes 35
  • 5 times 8 makes 40
  • 5 times 9 makes 45
  • 6 times 6 makes 36
  • 6 times 7 makes 42
  • 6 times 8 makes 48
  • 6 times 9 makes 54
  • 7 times 7 makes 49
  • 7 times 8 makes 56
  • 7 times 9 makes 63
  • 8 times 8 makes 64
  • 8 times 9 makes 72
  • 9 times | 9 | makes 81.

The use of the Table of Multiplication, and the manner how it is to be read.

This Table sheweth what the Summe of any two digits multiplyed one by another doth a­mount unto, and is thus to be read, 2 times 2 makes 4, 2 times 3 makes 6, 2 times 4 makes 8: Also 6 times 4 makes 24, 7 times 8 makes 56, 8 times 8 makes (or is) 64, 9 times 9 is 81, &c.

In Multiplication there are three terms com­monly used, that is to say, [Page 35]

  • The Multiplicand,
  • The Multiplier, and
  • The Product.

The Multiplicand is the number to be mul­tiplied.

The Multiplier is the number by which the Multiplicand is multiplied, and

The Product is the number which is produced by the multiplication of the Multiplicand and the Multiplier together.

Thus, if it were required to multiply 8 by 7, here 8 is the Multiplicand, 7 the Multiplier, and 56 is the Product, for 8 times 7, or 7 times 8, is 56.

In Multiplication it mattereth not which of the two numbers given is made the Multipli­cand, or which the Multiplier, for the Product produced by either will be the same, but the usual way is to make the greater number the Multiplicand; and the lesser number the Mul­tiplier.

THE RƲLE.

The numbers to be multiplied must be set one under another. viz. the Multiplicand (or greater num­ber). above, and the Multiplier, (or lesser number) below, the last figure of the Multiplier under the last figure of the Multiplicand, then draw a line under them, and (having learned the preceding Table per­fectly [Page 36]by heart) multiplie every digit of the Mul­tiplier, into every digit of the Multiplicand, setting the several products under the line, then having fi­nished your Multiplication, draw a line, and adde all the products together, and the Sum of those pro­ducts is the general product of the whole multiplicati­on, as by the following examples will oppear.

Example 1. [...] Let it be required to multiply 736 by 7. First, I write down 736 the Multiplicand, and under it 7, the Multiplyer, and under them I draw a line, then I multiply 7 into every di­git of the Multiplicand, saying 7 times 6 is 42, place 2 under the line under 7, and for the four tens keep 4 in mind, then say again, 7 times 3 is 21, and 4 which I kept in mind is 25, place 5 under the line, and keep the two tens in mind, then say again 7 times 7 is 49, and [...] which I kept in mind is 51, place 1 under the line, the 5 tens, (because there is no more figures to be multiplied) I set down under the line also, so is the work ended, and the pro­duct of this multiplication is 5152.

Example 2. [...] Let it be required to multiply 2417 by 5, place the numbers one under ano­ther, and draw a line under them as in the mar­gine, then begin your multiplication, saying 5 times 7 is 35, place 5 under the line, and keep the three tens in mind, then say again, 5 times 1 [Page 37]is 5, and 3 which I kept in minde is 8, place 8 under the line, and (because it is lesse then 10, I keep nothing in mind) then say again, 5 times is 20, place a cypher under the line, and keep the two tens in mind; lastly, say 5 times 3 is 15, and 2 which I kept in mind is 17, which 17 being the last number I place under the line, and so is my Multiplication end­ed, and the product is 17085.

¶ You may be satisfied of the truth of this work if you will take the pains to set down the Multiplicand 3417, five times one under another, and adde them together, as so many several sums, so shall you find the Total of that addition, to be 17085, exactly the same with the product of this Multiplication.

Example 3. [...] In the two fore-going examples, the Multiplyer consisted but of one digit, we are now to shew how multiplication is perform­ed when the Multi­plier consists of more then one fi­gure, therefore in this example, let it be required to mul­tiplie 5704 by 37, place your numbers and draw a line under them as you see in the margine, then begin your Multiplication in this manner, saying, 7 times 4 is 28, set 8 un­der [Page 38]the line, and keepe the two tens in mind, then say 7 times nothing is nothing, but the two tens in mind is 2, set 2 under the line, then say 7 times 7 is 49, set 9 under the line, and keep 4 in mind, then lastly, say 7 times 5 is 35, and 4 in mind is 39, which being the last number to be multiplied I set down under the line, so is the mulplication of one of the digits (namely 7) finished.

Then begin to multiply the second digit, say­ing 3 times 4 is 12, place 2 in the second line one place towards the left hand, and keep 1 in mind, then say 3 times nothing is nothing, but 1 in mind is 1, set down 1 by the 2 in the second line, thirdly, say 3 times 7 is 21, place 1 in the second line, and keep the two tens in mind; Lastly, say 3 times 5 is 15, and 2 in mind is 17, which 17, (because there is no more figures to be mul­tiplied) I place in the second line also.

Having thus done I draw a line under them, and adde these two lines together, as in com­mon Addition of numbers of one denominati­on, saying 8 is 8, place 8 under the line, then say 2 and 2 is 4, place 4 under the line, then say 1 and 9 is 10, place a cypher under the line, and carry 1 to the next place saying 1 and 1 is 2 and 9 is 11, place 1 under the line and carry 1 to the next row, saying 1 and 7 is 8, and 3 is 11, place 1 under the line, and carry 1 to the next place, saying, 1 which I carry and 1 is 2, place 2 under the line and so is your multiplication ended, and the product is 211048.

[Page 39] Example 4. [...] Let it be required to multiply 57325 by 4032, place the multiplicand and mul­tiplierone under another, & draw a line asbefore then proceed to the multiplication as formerly, saying, first, 2 times 5 is 10, set down a cy­pher and keep 1 in mind: then 2 times 2 is 4, and 1 in mind is 5, place 5 under the line then 2 times 3 is 6, set 6 under the line: then 2 times 7 is 14, set down 4, and keep 1 in mind, then 2 times 5 is 10, and 1 mind is 11, which 11 (being the last) I set down.

The multiplication of one of the digits being fi­nished, proceed to the multiplication of the next, saying 3 times 5 is 15, set down 5 in the second line a place more towards the left hand, and keep 1, then 3 times 2 is 6, and 1 kept is 7, set down 7, then 3 times 3 is 9, set down 9, then 3 times 7 is 21, set down 1 and keep 2 in mind, then 3 times 5 is 15, and 2 in mind is 17, which being the last set down also.

Two of the figures of the multiplier being finished proceed to the third, which (in this ex­ample) being a cypher, you may wholly neg­lect, and proceed to the multiplication of the fourth figure, onely remember to remove the product of the fourth figure one place more to [Page 40]the left hand, as in the example you may see, for the cypher, though it be not written down, yet it must keep its place, and the figures fol­lowing must be removed a place farther.

Then for the Multiplication of the fourth and last digit, say 4 times 5 is 20, set down a cypher (under 9) and keep 2 in mind, then 4 times 2 is 8 and 2 in mind is 10, set down a cypher and keep 1 in mind, then 4 times 3 is 12, and 1 is 13, set down 3 and keep 1, then 4 times 7 is 28, and 1 kept is 29, set down 9 and keep 2, then 4 times 5 is 20, and 2 kept is 22, which 22 (because the multiplicatiou is ended) set down also.

Having thus multiplied all the digits seve­rally, draw a line under their products, and adde them altogether as in the former exam­ple, so shall you find their general product to be 231134400.

Other Examples for Practice.

[...]

[...]

Compendiums in Multiplication.

1 If the Multiplier consist of cyphers in the last place or places, [...] you may omit the multipl­cation of them, and place the former figures of the Multiplier under the Multiplicand, thus if it were required to multiply 3257 by 2600. place the numbers as you see in the margent, then multiplying 3257 by 26, the Product will be 84682, to which if you add two cy­phers, (because there were two cyphers in the Multipli­er) it will be 8468200, which is the true product of the multiplication.

2 If it be required to multiply any number by 10, 100, 1000, 10000, &c, You have no more to do but to add so many cyphers to the multi­plicand as there are cyphers in the multiplier, thus if you were to multiply 365 by 10, the pro­duct will be 3650, or by 100, it would be 36500, or by 1000, it would be 365000, or by 10000, it would be 3650000, &c.

3 If any number given were to be multiplied by 5, you may abreviate your worke thus, adde a cypher to the Multiplicand, take halfe that number, and it shall be the product required, thus if it were required to multiply 8627 by 5, adde a cypher to the multiplicand, then it is [Page 42]86270 the halfe whereof is 43135, which is the product required.

The Proof of Multiplication.

The most certain proof of Multiplication is by Division, but because Division is not yet known, I will here shew a neer way by which Multipli­cation may be proved. Which is thus,

THE RƲLE.

First,

[figure]

take a Crosse as in the Margine, then, any sums being multiplied, you may prove the truth of your work in this manner, (1) Cast away all the nines which you can find in the multiplicand, what re­maineth set on the right side of the Crosse. (2) Cast away also the nines in the Multiplyer, and what re­mains set on the left side of the Crosse. (3) Multiply the figure on the right side of the Crosse by that on the left side of the Crosse, and out of that product cast away the nines, setting the figure remaining over the Crosse, then (4) Cast away all the nines in the product, and if the figure remaining be the same with that which standeth over the Crosse, then is your multipli­cation truly performed, otherwise not.

Example 1 [...]

[figure]

Let it be required to prove the Sum in the margine.

1 Cast away all the nines in the Multipli­cand, saying 4 and 3 is 7, and 2 is 9, which being rejected there [Page 43]remains 4, which I set on the right side of the crosse, then

2 Cast away all the nines in the Multiplier, saying 2 and 3 is 5, which (being lesse then 9) I set on the left side of the crosse, then

3 Multiply 4 by 5, saying 4 times 5 is 20 from which cast all the nines, and there remain 2, place 2 over the crosse, and

4 Cast away all the nines in the Product, say­ing, 2 and 5 is 7, and 4 is 11, cast away 9, and there remains 2, which exactly agrees with the figure over the crosse, and demonstrates that the multiplication is truly performed.

To multiply by any of the nine Digits without charging the memory.

To multiply any number by 2, Either double the number in your mind, or adde it, by setting it down twice, so 57325 produceth 115750.

To multiply any number by 3, To the num­ber given, adde the double thereof, the sum is the product, so 57325 produceth 171975.

To multiply any number by 4, Double the duplication in your mind, so 57325 produceth 229290.

To multiply any number by 5, Conceive a cypher added to the given number, and in your mind half thereof is the product, thus a cypher added to 57325, maketh it 573250, the halfe whereof is 286625.

To multiply any number by 6, Take half adding a cypher, and adde to the half, the figure standing next before, thus 57325|0 produceth 343950.

[Page 44]To multiply any number by 7, Take half and adde it to the double of the former figure, supposing a cypher added as before, so 57325 thus ordered, produceth 401275.

To multiply any number by 8, Double each former figure, and subtract it from the follow­ing, so 57325 produceth 458600.

To multiply any number by 9, Suppose the number multiplyed by 1, then subtract each former figure from the following, beginning with that next before the cypher, the remainer is the product, so 57325 produceth 515925.

Questions performed by Multiplication only.

Question 1. If a piece of land be 236 perches long, and 182 perches broad, how many square perches are contained therein? multiply 236 the length, by 182 the breadth, the product is 42952, and so many square perches are contain­ed in such a square piece of land.

Question 2. In a year there are 365 dayes na­tural, and in every day 24 houres, how many houres be there in a year? Multiply 365 the number of dayes, by 24 the number of houres, the product is 8760, and so many houres be there in a year.

Question 3. From London to Coventry it is ac­counted 76 miles, how many yards therefore is it from London to Coventry? Multiply 1760 (which are the number of yards contained in one mile) by 79, the product is 133760, and so many yards are between London and Coventry.

Division.

DIvision is the just contrary to Multipli­cation, for that turnes small denomina­tions to greater, as Multiplication turns greater to smaller. Or (in whole Num­bers, of which only we yet speak) Division is the asking how many times one Sum is contained in ano­ther? and the number which answereth to that question is called the Quotient.

And the number containing, which is to be divided, is called the Dividend.

And the number contained, or by which the Dividend is to be divided, is called the Divisor.

And as often as the Dividend contains the Divisor, so often doth the quotient containe Unity.

The wayes of performing Division are divers I will begin with that which is most used and taught, which is as followeth.

THE RƲLE.

Place the Divisor under the Dividend, so that the figures next to the left hand stand directly one under the other, if the rest of the Divisor be not greater: or [Page 46]if all the Divisor be greater then that above it, then the said Divisor must be devolved one place further toward the right hand; having so placed them, try how many times the lower figures are contained in the upper figures, and write that figure which answereth that question within a crooked line in the margine of the work, which is called the Quotient, and by it mul­tiply the first figure of the Divisor, and take the pro­duct out of the figures directly over it, beginning the Subtraction toward the left hand; then cancel that figure of the Divisor, and also, that of the Dividend which hath been already used, with a light dash of a pen, and write the remaine (when the product of the first figure multiplied by the quotient is subtracted as before) just over the figure used and cancelled; the [...] proceed to doe the like with the second, third, and fourth figure of the Divisor if there be so many; till having cancelled it all, and set the remaine orderly a­bove the Dividend, you have finished one work.

Now if the Dividend have still some figures un­touched towards the right hand, then remove the Di­visor still toward the right hand, but one place at a time, and then againe aske or try how many times the lower may be had in the upper, and write the answer in the quotient whether it be 1 or more, (onely it can­not be above 9) or nothing, then put 0 in the quotient, and multiply the Divisor by this new figure, and sub­tract the product, setting the remainer orderly above, as before, this work must be repeated by removing the Divisor still one place towards the right hand, un­til Ʋnity in the Divisor stand under Unity in the Di­vidend, and then the work is done.

Example. 1.

Let it be required to divide 4096 by 3 Place them thus [...]

Aske how many times 3 in 4 the quotient is 1, which is put within a crooked line by it self.

Then in your mind multiply the Divisor 3 by the quotient 1. And having said these words Once 3 is 3, [...] presently cancel the 3. And having added these words, Out of 4, cancel the 4. And after these words, and there remains 1, write 1 just over the 4, as you see here done.

Then remove the divisor one place towards the right hand, [...] saying, how many times 3 in 10, the answer is 3, which write in the quotient; and in your mind multiplie the Divisor 3 by the quotient 3, the product is o, wherefore say three times three is nine out of ten, and there remains one, then (ha­ving cancelled the 10 and the 3, write over them 1.

Again, remove the Divisor 3 one place more, [...] asking how ma­ny times 3 in 19? the answer being 6, write 6 in the quotient, and say 6 times 3 is 18 out of 19, and [Page 48]there remains 1, wherefore having cancelled the 3, and the 19, write 1 over 3, and remove the Divisor once more, and aske how many times 3 in 16? answer is 5, which write in the quotient, [...] then in mind say 5 times 3 is 15 out of 16, and there remains 1, and cancelling the 16 and the 3, write over 3, 1: Now because the Divisor 3 is advanced so far till it is come to stand under 6 in the Dividend, which 6 is the place of Unity there, the said Di­visor cannot be removed any more, and there­fore the division is ended, and the quotient be­ing 1365, shewes that the Divisor 3 is contained in the Dividend 4096, 1365 times; and 1 re­maining, which 1 being lesse then the Divisor 3, doth not contain it once, but one third part of once, which after the Reader hath skill in broken numbers must be joined to the quotient thus 1365⅓. The best proof of this Division is by multiplying the quotient into the Divisor, and to the product adde the remaine; then if the work be well done, the sum shall be equal to the Di­vidend.

So 1365 multiplied by 3 produceth 4095, to which adding the remain 1, the sum 4096 is e­qual to the Dividend.

Here we divided only by one figure 3, because the first Example being easie and clear, should be a faire Introduction to the second,

Note that if the Divisor had been greater then [Page 49]4, as 5, [...] the work must have begun thus So the quotient would have been 819½, which is one place lesse.

Example 2.

Let it be required to divide 1310720 by 4096, Place them thus [...]

Now the question to be asked is, how many times 4096 is there in 13107? to find an answer to this question, the Reader which hath but an in­different faculty of judging; may do it, (for the most part) by considering the first figure in the divisor, as here 4, for presently he knowes that 4 times 4 is 16, which cannot be had in 13, there­fore the first figure in the quotient must be lesse then 4.

Again, it cannot be much lesse, because the second place in the Divisor is 0.

He may therefore venture on 3, And having put 3 in the quotient, say 3 times 4 is 12, [...] which subtract out of 13 over it thus, say 2 out of 3, and there remains 1, and 1 out of 1 and there remaines 0, cancelling the 0 and the 13, and set the remaine 1 over the 4, as you see.

Then go on saying 3 times 0 is 0, out of 11, and remains still 11, againe 3 times 9 is 27, take 7 out of 10 there remains 3, to be set over the 0, [Page 50]which is over 9, and 2 with 1 borrowed (to make the 0 10, from the which the 7 was taken) is 3, and 3 out of 11, remains 8, which write over the place of 0 in the Divisor, cancelling the 9 in it, and also those figures of the Dividend 110, out of which you have taken any thing; then lastly, say 3 times 6 is 18, take 8 out of 7 I can­not, therefore (borrowing 10,) say 8 out of 17 remains 9, which write over the 7, then in the next place, the 10 borrowed is 1, and the 10 in the 18 is 1. Say therefore 1 and 1 is 2, out of 3, and the remain is 1, which write over the 3, having still cancelled the figures which you have used orderly as you go.

Now the whole Divisor being cancelled, it must be removed on one place further, and pla­ced as here.

Now aske again how many times 4096 in 8192? [...] it will be found (because of the 0 after the 4) as often as 4 in 8, that is two times, put therefore in the quotient 2, and work as before, saying 2 times 4 is 8, out of 8 remains 0, then 2 times 0 is 0 out of 1 remains 1, then 2 times 9 is 18, take 8 out of 9 remains 1, (which put over the 9,) and 1 out of 1 in the next place remains 0, then lastly, 2 times 6 is 12, that is, 2 out of 2 there remaines 0, and 1 out of 1, there also remaines 0, cancel and put the remainer over as formerly.

[Page 51]Now againe, the Divisor being all cancelled, should be removed; and aske how many times 4096 in 0000, the answer is 0, which being put in the quotient, the work is all done.

And the quotient 320, shewes that the Divisor 4096 is contained in the Dividend 1310720, three hundred and twenty times.

And whether the Divisor have 2, 3, 5, 7, or more places, the working is still like this, not dif­fering from it at all.

Except for brevity in some cases, where the Divisor toward the right hand hath one Cypher or more; for those Cyphers may be placed or­derly under the Dividend at first, and remain there till the work be done, with the rest of the Divisor, which must needs shorten the Division.

As if 2587645 were to be divided by 15000, place them thus [...]

And say how ma­ny times 15 in 25? [...] an­swer is 1, which write in the quotient, and multiply in minde the divisor by the quotient, saying once one is one, out of 2 there remains 1, then cancel the 2, and the 1 un­der it, and write the remaining 1 over it, as here is done, then say once 5 is 5 out of 5, and there remains 0, therefore cancel the two 5, and over the uppermost write 0.

[Page 52]Now remove the Divisor one place, and aske how many times 15 in 108? answer is 7, which write in the quotient, and multiply, saying 7 times 1 is 7, out of 10 remains 3, which write o­ver the 0, then say 7 times 5 is 35, take 5 out of 8 remains 3, which write over 8, and 3 out of 3 remains 0.

Now againe, remove the Divisor, and aske how many times 15 in 37? answer is 2, which write in the quotient, and say two times 15 is 30, out of 37 remaines 7, and the Division is ended, the remaine being 7645.

Proof of this.
Multiply the quotient172
by the Divisor15
 860
 172
The Product is2580

Before which put the three Cyphers, And then it is2580000
To which adde the remaine7645
The Total is2587645

Which is equal to the Dividend, and there­fore the work is right.

So if one would divide any sum by 10, 100, 1000, 10000, &c. he need but cut off the first, two first, three first, or four first figures towards the right hand, the other figures shall be the quo­tient, and those cut off, the remaine.

As if 1587645 be divided by 1000, the quotient is 1587, and the remaine 645.

[Page 53]If one would know how many pounds Ster­ling are in 95670 shillings; having placed them thus, that is, 2 under 9, [...] and 0 un­der 0, then divide by 2, which is very easie. So the quoti­ent 4783, shewes that in the number of shillings given, is contained 4783 pounds, and 10 shillings remaining.

The reason why the Divisor was 2, is because there are 20 shillings in one pound, and therefore any number of shillings is turned into pounds, by dividing by 20, that is by 2, putting the cypher under unity, onely to fill a place at last. And this way of turning any number of shillings in­to pounds may be easily effected by memory, if you suppose the last figure of your given number to be cut off with a line, or comma, and taking the half of the other figures. Thus, let the given number of shillings be 5739; Imagine the last fi­gure 9 to be seperated from the rest by a line thus 573 9, now by memory take the half of 573, by saying in your mind the half of 5 is 2, (and 1 remaining, which makes the 7 following 17,) the half of 17 is 8, (and 1 remaing) the half of 13 is 6, (and 1 remaing, which 1 is 10 s. to be added to the 9, and the whole is 286 li. 19 s.

In this way of Division, as in all others, if the remaine at last be greater then the Divisor, the quotient is not just, but too little, which may be remedied (without beginning the work a­gaine) by dividing the remainer onely by the same divisor; for thereof will arise a new quo­tient, [Page 54]which added to the former quotient, the summe will be the just quotient.

So if 7290 be divided by 27, the numbers being placed thus,

Because 27 can be had in 72 but twice, [...] put 2 in the quotient, saying 2 times 27 is 54, out of 72, and there remains 18 which write over 72, cancelling as be­fore is shewed; then removing the Divisor, say how many times 27 in 189? answer is 7, but if one should mistake, and write 5 in the quotient, and say 5 times 27 is 135, out of 189, remains 54 and write it over as before; and remove the Divisor, and say, how many times 27 in 540? answer is 20, but should not be above 9, say therefore, 9 times 27 is 243, out of 540, and the remain is 297, which being the last remain, and greater then the Divi­sor, shewes the quotient 259 is too little.

Wherefore divide the last remain 297, [...] By the Divisor 27 saying, twenty seven in 29 once, and write 1 in the new quotient; and say once 27 is 27, out of 29, remains 2 which write over 9, and remove the Divisor, and say, 27 in 27, justly once, so write 1 in the quotient, so the quotient is 11, which ad­ded to the former quotient 259, gives 270, which is the true and whole quotient.

A Second way of Division.

Although (as I have said) the former way is more used, yet this may seem plainer and more natural to some, I will therefore give one exam­ple of it.

Example.

Let it be required to divide 6477734 by 334, [...] where the first figure in the quotient is ea­sily seen 1, subtract once 334 out of 647, and write the remain under a line, then see how many times 334 in 3137, answer 9, by which multiply the divisor 334 in ano­ther paper, the pro­duct is 3006, which subtract out of the re­maine, then the new remaine is 131734.

Again, how many times 334 in 1317, answer is 4, for the third figure in the quotient; by which multiply the Divisor, the product is 1002, which take out of the later remaine, as in the margine, then will remain 31534.

Now try how many times 334 in 3153, an­swer is 9, for the fourth figure in the quotient, by which multiply the divisor (in a by paper) the [Page 56]product is 3006, (as might have been seen above in the second subtraction) which subtracted out of the later remaine, there remains now 1474.

Lastly, aske how many times 334 in 1474, answer is 4, for the last figure in the quotint, by which multiplying the divisor 334, the product is 1336, which subtracted from 1474, there final­ly remaines 138, which being lesse then the divi­sor, shewes the division is done.

Proof of this.

The several subtractions and the final remaine added together. [...]

The Total equal to the dividend 6477734

If the former (as I have said more usual) way seem difficult to beginners, because the pro­ducts of the divisor into the several figures in the the quotient are not set down but mentally made; and also because the subtraction of them begins towards the left hand: and lastly, because the remaine is still set above: yet this later way which agrees altogether with plain Subtraction before taught, I hope is so plain, that any diligent rea­der may acquire it, without a Tutor. And yet for the better satisfaction and help of the young Learner, I will adde another way or two more of Division.

A Third way of Division.

There is another kind of division which is ve­ry much used, and is in most request with those who have most occasion to divide great num­bers, the manner of working is not much unlike the way before taught, one or two examples will make it plain.

Example 1. [...] Let it be required to divide 162483 by 1321, set down your numbers as you see them placed in the margine, viz. First, set down 162483 the dividend, then on the left hand there­of set the divisor 1321 with a crooked line between them, then on the right hand there­of make another crooked line which must serve to set the figures of the quoti­ent in, so are your numbers placed in due order, then draw a line under the dividend, and make a prick under the figure 4, because so far the fi­gures of the divisor would extend if they had been placed under-neath the dividend, according as in the other examples, this prick serves onely to shew how far you have proceeded in your work, and must at every division be removed a place further, till at length you come to the last figure of the dividend: your numbers being thus placed with a line under them, you are ready for the work, which must be performed according to the directions of the following rule.

[Page 58] THE RULE.

Demand how often the divisor may be had in the di­vidend, and place that number in the quotient, then multiply the divisor by the quotient, and place the pro­duct under the line: then subtract this product from the dividend, and set the remainer under the product, then make a prick under the next figure of the dividend, and bring that figure down to the remainer, and then proceed as before.

Example, [...] Your numbers being placed as is be­fore directed, you may begin your work in this manner, first, say how many times 1321, can I have in 1624? say once, place 1 in the quoti­ent, by which 1 mul­tiply the divisor 1321, beginning at the left hand, saying, once one is 1, place 1 under the line, then once 2 is 2, set 2 under the line, then once 3 is 3, place 3 under the line, lastly, once 1 is 1, place 1 under the line, then subtract this 1321, from 1624, and there will remain 303, to this 303 bring down the next figure in the dividend, namely 8, so will that number be 3038, under which draw a line, and repeat the same work again, saying, how many times 1321 can I have in 3038, which [Page 59]may be had two times, place 2 in the quotient, by which 2 multiply the divisor 1321, saying 2 times 1 is 2, place 2 under the line, then 2 times 2 is 4, place 4 under the line, then 2 times 3 is 6, place 6 under the line, lastly, 2 times 1 is 2, place 2 under the line, and subtract this 2642 from 3038, and there will remain 396, to this 396 bring down the next figure of the dividend, which is 3, so is this number made 3963, under which draw a line, and repeat the work once a­gain, saying, how many times 1321 can I have in 3963, which may be had 3 times, by which 3 multiply the divisor 1321, saying 3 times 1 is 3, then 3 times 2 is 6, then 3 times 3 is 9, and lastly, 3 times 1 is 3, which place under the line, and subtract it from the line above, which in this example is the same number, therefore there re­mains nothing, and the work is ended, but if a­ny remainer had been, that should have been set under the line, as by the examples following will appear.

Other Examples for Practice.

In this Example where 793058, [...] is di­vided by 5624, you may perceive that the quotient is 141, and 74 remaining, so that the real quo­tient is 141 74/5624.

[Page 60] The Proof of this Division.

This kind of division is proved by Addition, for If you adde the several products arising from the multiplication of the several quotients into the divisor, and also adde thereunto the remainder (if any be) the total of this addition shall be equal to the dividend, if there be no errour in the work.

So in the example following, [...] if you adde 4325 the first product, and 30275 the second product, and 2796 the remainder, to­gether, in the same order as they now stand in the exam­ple, you shall find the Total of this Addition to be 76321, equal to the dividend, which demonstrates the work to be true.

A Fourth way of Division.

There is a fourth way of division used by some, not inferiour to any of the preceding, for that it is no burthen to the memory, and it is also proved by Addition.

The manner of placing the figures is the same [Page 61]with the third kind of division last taught; And for the performance of the work, this is

THE RƲLE.

First write down the dividend, and on the left hand thereof the divisor, with a crooked line betwixt them, and on the right hand of the dividend make another crooked line wherein to place the figures of the quoti­ent, then draw a line under the dividend, and also make a prick under that figure of the dividend under which the last figure of the divisor would fall, were it to be placed as in the first kind of division. This done, demand how often the divisor may be found in those figures of the dividend, and place that digit in the quotient, then by this digits multiply the divisor, and set the product of this Multiplication directly under the dividend, beginning at the place where you made the prick, then subtract this product from the figures of the dividend, and place the remainder over the dividend, cancelling the figures of the dividend as you proceed, so is the first figure of the quotient finish­ed, then make a prick under the next figure of the di­vidend, and demand how often the divisor may be found in the last remainder, and the other figure being added thereto, which place in the quotient, and pro­ceed in all respects as before, till you have pointed all the figures of the dividend.

Example, [...] Let it berequi­ren to divide 763258 by 2345, place your numbers as you see in the margent, and because there are 4 figures in the [Page 62]divisor, therefore make a prick under the fourth figure of the dividend, which is 2, and draw a line, then begin your division in this manner: saying,

First, [...] how many times 2345 can I have in 7632, (or how many times 2 can I have in 7) say 2 times, place 3 in the quotient, by which 3 multiply the divisor, saying 3 times 5 is 15, place 5 under the prick, then 3 times 4 is [...] and 1 is 13, place 3 under the line, then 3 times 3 is 9, and 1 is 10, place a cypher under the line, then 3 times 2 is 6, and 1 is 7, place 7 under the line, then subtract 7035 from 7632, saying 5 from 12, and there remains 7, place 7 over 2. (cancelling the 5 and the 2,) and bear one in mind, then 1 and 3 is 4, out of 13, there remains 9, place 9 over 3, and cancel 3 and [...], then 1 which I carried, from 6, and there remains 5, place 5 over 6, and cancel 0 and 6, lastly, 7 from 7 there remains 0, which you need not set down, but cancel the two sevens, then will the work stand as above, and the remainder will be 597.

Secondly, make a prick under the next figure of the dividend, namely 5, and say how many times 2345 can I have in 5975, answer two times, place 2 in the quotient, by which multi­ply the divisor, saying 2 times 5 is 10, place 0 under 5, and carry 1, then 2 times 4 is 8 and 1 is 9, [Page 63]place 9 under 7, then 2 times 3 is 6, place 6 under 9, lastly, 2 times 2 is 4, place 4 under 5, so is the product of this multiplication 4690, which you must subtract from 5975, saying 0 from 5 and there remains 5, place 5 over 5, and cancel 0 and 5, then 9 out of 17, there remains 8, place 8 over 7, and cancel 9 and 7, then 1 carryed and 6 is 7, from 9 there remains 2, place 2 over 9 and cancel 6 and 9, [...] last­ly, 4 from 5 rests 1, place 1 over 5, and cancel 4 and 5, so have you finished your second figure, and your work will stand thus, and your remainder will be 128.

Thirdly, make a prick under the next figure of your dividend (namely under 8,) and aske how many times 2345 can I have in 12858, (or how many times 2 can I have in 12,) say 5 times, place 5 in the quotient, by which multiply the divisor, saying 5 times 5 is 25, place 5 under 8 and carry 2, then 5 times 4 is 20, and 2 is 22, place 2 under 0, and carry 2, then 5 times 3 is 15, and 2 is 17, place 7 under 9, and carry 1, then 5 times 2 is 10, and 1 is 11, place 11 under 4 and 6, so is the product of this multiplication 11725 to be subtracted from 12858, saying 5 from 8 rests 3, place 3 over 8, and cancel 5 and 8, then 2 from 5 rests 3, place 3 over 5, and cancel 2 and 5, then 7 from 8 rests 1, place 1 over 8 and cancel 7 and 8, then 1 from 2 rests 1, place 1 over [Page 64]2, and cancel 1 and 2, lastly, [...] 1 from 1 rests nothing, so is your work ended, which you shall find to stand as in the margent, the re­mainder being 1133.

The Proof of this Division.

This kind of division is also proved by Additi­on, for, if you draw a line under the work, and adde all the figures between the two lines together, (in or­der as they there stand) taking in the remainder (if any be) the Total of this addition will be equal to the Di­vidend, if the work be true.

Other Examples for Practice proved.

[...]

[Page 65] [...]

Questions performed by Division only.

Question 1. If a square piece of Land contained 42952 square perches, and one of the sides there­of be 236 perches long, how long must the other side be? Divide 42952 by 236, the quotient will be 182, and so many perches long must the other side be.

Question 2. In a yeaar there are 8760 houres, and in every natural day there are 24 houres. I demand how many dayes be there in a year? Di­vide 8760 by 24 the Quotient will be 365, and so many dayes be there in a year.

Question 3. The distance from London to Co­ventry is 133760 yards, and in one mile there is [Page 66]contained 1760 yards, now I would know how many miles it is from London to Coventry; Di­vide 133760 by 1760, the quotient will be 76, and so many miles is it from London to Coventry.

These Questions performed by Division only, are the converse of those that were performed by Multiplication, which I the rather make choice of, that the reader might see how Multiplication and Division prove each other.

There are one or two more kinds of Division, something like these last, but I shall forbear ex­emplifying them; for much variety helps to make a Book rather great then fit.

¶ Here is to be noted that in the following Rules, where there is continual use of Divi­sion, I sometimes use one kind of Division, and sometimes another, for variety sake, but the practicioner may use which he is best skill'd in, for they all produce the same effect.

Reduction.

IS twofold, first, that which turns greater de­nominations into smaller, as pounds into shil­lings or pence, this is done by Multiplicati­on as followeth.

Example 1.

Let it be asked how many pence are contained in 729 li. 11 s. 7 d?

First, a shilling is contained in a pound 20 times, therefore multiply 729 by 20, or (which is the same, but shorter) by 2, and put o to the product, as in the margine, [...] this shews that in 729 l. there are 14580 shillings.

To which adde 11 s. it makes 14591 shillings.

Again, because one peny is con­teined in one shilling 12 times, multiply 14591 by 12, it produ­ceth 175092, to [Page 68]which adde the 7 pence, so the summe will be 175099, and so many pence are contained in 729 li. 11 s. 7 d.

Example 2.

Let it be asked how many pintes are contained in 4 Tun, 3 Hogsheads, and 27 Gallons?

First, 1 Tun is equal to 4 Hogsheads, there­fore 4 Tun is equal to 16 Hogsheads, to which adde the 3 Hogsheads, so there is 19 intire Hog­sheads.

Againe, because one Hogshead containes 63 Gallons, multiply 19 by 63, [...] it produceth 1197 Gallons, to which add 27, it gives 1224 Gallōs.

Lastly, because every Gallon contains 8 pintes, multiply 1224 by 8, it produceth 9792, and so many pintes are con­tained in 4 Tuns, 3 Hog­sheads and 27 Callons.

After the same sort might drie Measures be reduced, as quarters to bushels, pecks, or gallons, and likewise all weights and outlandish Coins, of which the proportion of the greater to the lesser is (before) known, or given.

Secondly, it is often requisite to turn smaller denominations to greater: this is done by Divi­sion, as followeth.

Example 1.

Let it be asked how many pounds are contain­ed in 80976 shillings?

Divide 80976 by 20, [...] the quo­tient is 4048 li. and 16 s. re­maining, which is the true an­swer.

Example 2.

Let it be asked how many pounds are in 109754 d?

Because a pound contains a shilling 20 times­and a shilling containes a peny 12 times, there fore if 109754 be divided first by 12, the quotient 9146 shillings, and 2 pence over, then if 9146 be divided by 20, the quotient is 457 pounds and six shillings remaining; so that 109754 pence is equal to 457 li. 6 s. 2 d.

Or if 109754 had been at first divided by 12 times 20, that is by 240, (which is the number of pence contained in a pound) the quotient had been 457 pounds, and 74 pence remaining, which is all one with the former; for 74 pence is equal to 6 shillings 2 pence.

More instances shall not need herein, because the thing of it self is very clear.

Progression.

IS also of two sorts, the first is of certain num­bers in Arithmetical Proportion from 1, that is, such as differ equally, as 1, 2, 3, 4, 5, 6, where the common difference is 1, (as is easily seen,) or 1, 3, 5, 7, 9, 11, where the common difference is 2, or any other, as 1, 8, 15, 22, 29, 36, where the common difference is 7, this is called Arithmetical Progression.

2 Secondly, of certaine numbers in Geometri­cal proportion from 1, that is such as increase by a common Multiplication, as 1, 2, 4, 8, 16, 32, where the common multiplier is 2, that is the first by 2, produceth the second, and the second multiplied by 2, produceth the third, and so on.

Or as 1, 3, 9, 27, 81, 243, where the common Multiplier is 3, this is called Geometrical Progres­sion.

Both the common difference (in the first) and the common Multiplication (in the later) shall for shortnesse hereafter be called the common ex­cesse.

First, now of the first sort, or Arithmetical [Page 71]Progression, the principal use of this is,

1 If the number of places, and common ex­cesse be given, to find the last number.

2 When the number of places, and the last number is given, to find the aggregate, or total summe of all the numbers.

3 When the last number, and the total summe is given, to find the number of places.

4 The Number of places, and total summe being given to find the last number.

5 The last number, and number of places gi­ven, to find the common excesse.

6 The last number, and common excesse be­ing given; to find the number of places.

I will instance in no more, few of these ever hapning to be used.

For the first of these, let there be given the number of places100
The common excesse1
To find the last number also;100

THE RƲLE.

Multiply the number of places lesse by 1, by the common excesse, and to the product adde the first number; the summe is equal to the last number.

So here multiply 99 by 1, the product is 99, (for 1 neither multiplies nor divides) to this adde the first number 1, it gives 100 for the last number.

Or let the numbers be 1, 7, 13, 19, 25, 31, [Page 72]where the common excesse is 6. and the number of places also 6.

Now (if the number of places lesse by 1, that is) 5 be multiplied by the common excesse, which is 6, the product is 30, to which adding the first number which is 1, the last number 31, is thereby composed. This is so casie that it needs no proof.

2 For the second, which is, The last number, and the number of places given, to find the total summe of all the numbers.

THE RƲLE.

Adde the first and last numbers together, and mul­tiply the summe by half the number of places, the pro­duct is equal to the aggregate or summe of all the num­bers added together.

So if to the first number 1 be added the last number 100, it gives 101, which multiplied by 50 (which is half the number of places) produ­ceth 5050, which is equal to all the hundred num­bers added together.

And hereby may that vulgar question be an­swered, which is,

If a man take up 100 stones placed a yard one from another all in a right line, by one at once, and bring them back one by one to his first standing, how many yards doth he goe backwards and forwards?

It is shewed before that he goes forward 5050 yards, and he must needs come back just as much; that is, in all 10100 yards. which is 5 miles and 3 quarters; wanting 20 yards.

[Page 73]Or secondly, suppose the numbers were 1, 9, 17, 25, 33, 41. Whereof the common excesse is 8, the first and last added gives 42, which multiplied by 3, (half the number of places) the product is 126, which is the summe of them all.

3 For the third thing, that is, by the last num­ber and the total, to find the number of places.

THE RULE.

Adde the first and last numbers, and by the summe of them divide the total, [...] the quotient will be equal to half the number of places.

This is so plain it needs no cleering.

4 For the fourth, if the total, and number of places bee given, to find the last number.

THE RULE.

Divide the total by half the number of places, the quotient is a number, from which if 1 be taken, the rest is the last number.

As let the number be 1, 3, 5, 7, 9, 11, 13, 15, or any other (in arithmetical proportion) whatso­ever. The summe of these is to be 64. And the number of places is 8, the half of it 4. Now if 64 be divided by 4, [...] the quotient is 16, from which if 1 be ta­ken, there remains 15 for the last number.

5 Now for the fist variety, If the last number [Page 74]and number of places be given, to find the com­mon excesse.

THE RƲLE.

From the last number take 1, and the remaine shall be the Dividend; then from the number of places also take 1, and make this later remaine the Divisor▪ then the quotient of this Division shall be the common excesse.

Example, Let the numbers be 1, 4, 7, 10, 13, 16, from 16 take 1. remains 15, for the dividend, then from 6, (which is the number of places) take al­so 1, remains 5 for the Divisor.

Now when 15 is divided by 5, the quotient is 3. And 3 is also the common excesse, or dif­ference between 1 and 4, or 4 and 7, &c.

6 Lastly, let the last number, and the com­mon excesse be given, to find the number of places.

THE RULE.

From the last number take 1, and divide the re­maine by the common excesse; then to the quotient adde 1, the summe is the number of places.

As let the numbers be 1, 5, 9, 13, 17, 21, 25, 29, from 29 take 1, remains 28, which divide by 4, (which is the excesse) the quotient is 7, to which adde 1, the summe is 8, which is the number of places, as the reader may easily count.

Geometrical Progression.

I shall not be so large in this as in the former, because these things are of little use to the A­rithmetician, except where a number is to be many times doubled, tripled, or the like, which cannot be so easily abridged here, as in the other, because there the last number arising of many additions of the excesse to 1, was easily found by one multiplication: but here the last number being made by many Multiplications of the ex­cesse, is therefore many times harder then the other

The varieties here shall be but two.

1 The common excesse and number of pla­ces being given, to find the last number.

2 The excesse, and last number being given, to find the total summe.

The first of these may thus be found. Let the numbers be 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, the exceesse is 2, the places 10, find out the fifth number (which is easily done, for any one may reckon so farre by heart,) that is here 16, and multiply 16 by 16, it produceth 256, which is the ninth number, lastly, multiply 256 by the excesse 2, thence ariseth 512, the number desi­red.

So if the places had been more, as 72, having found the 9 number 256, multiply it by 256, thence comes 65536 for the 17th. number, which mul­tiplied [Page 76]by the excesse 2, gives 131072 for the 18th. place, which multiplied by 131072, gives 17179869184, for the 35 place; and that multi­plied againe by the excesse 2, gives 34359738368, for the 36 place, that multiplied by 34359738368, then the product will be 1180591620717411327|424, for the 71 place, which lastly, multiplied by the excesse, gives 2361183241434822654848, for the 72 place, which is the last number of the pro­gression required to be found.

Perhaps this may seem somewhat tedious but where things cannot be performed without la­bour, the Reader must content himself with such Rules as make it lesse; for it is certaine that this way is much shorter then to have multiplied still by the excesse 71 times, which else he must have done.

All this notwithstanding, he is not bound to use the same numbers, much lesse in other que­stions where the number of places is not the same, but whereas I began from the place 9, he may begin at 8, 10, or 12, or where he pleases; so as he remembers still where he is; for this is general, if the number belonging to any place what­soever, be multiplyed by it self, the product shall be the number belonging to twice so many places want one place.

Now for the second thing, which is to find the summe of all the numbers.

THE RULE.

From the last number take the first, and divide [Page 77]the remaine by the excesse want 1, then multiply the quotient by the excesse; and to the product adde the first number, the summe of them is equal to the summe of all the numbers.

So if from the last number, or 72 place be ta­ken 1 remain is 2361183241434 22654847 which should be divided by 1, (that is the excesse want 1 for the excesse is but 2) but because 1 neither multiplies nor divides that labour is saved; Now multiply this remain by the excesse, the product is 4722366482869645309696, to which adding the first number 1, by making the fi­gure 6 next the right hand to be 7, you have the total summe of all the 72 numbers.

A Question resolved by Geometrical Progression.

A Londoner sojourning in a Countrey Mar­ket Towne, in Winter, made himselfe a new freez, Suit and Coat, on which were set 6 dozen of buttons of Silke and Silver; A Baker being in his company liked it so well he would buy it of him; the Citizen consented to let him have it, paying for the first button a single barley corn, for the second 2, for the third 4, and so on doubling to the last.

The bargaine was liked on both parts for the present, but shortly after revoked, for it could not be performed, and no man can be holden to an impossibility.

But why this could not be performed: may be judged; first, by inquiring the worth of so [Page 78]much barley in Money: And secondly, the weight of it; and how it should be removed.

1 For the first, allowing 10000 corns to a pinte (which is more then enough) then 5120000 corns make a quarter; and yet (for shortning the division) we will allow 10000000 corns to a quarter; by which dividing the whole number of corns (which is done by cutting off the first 7 figures toward the right hand,) the quotient will be 472239648286964, and so many whole quarters there are omitting the remaine, as in this case not considerable.

Now allowing barley were to be sold at 15 d. the bushel (which is cheap,) it is so many Angels: and therefore dividing by 2 it is 2361183241414|3482 pounds sterling: which is in words, Two hundred thirty six millions of millions, one hundred and eighteen thousand, three hundred twenty four millions: one hundred forty three thousand; four hundred eighty two pounds, whick I take to be too much for any Trades-man to get or keep.

And reckoning land for ever at twenty yeares purchase, if this summe of pounds be divided by 20, the quotient is the yearly rent of 118059|16207174 pounds.

And this divided again by 365 (the number of dayes in a year) the quotient 32344975918, that is above thirty two thousands of millions a day for ever. So great a vanity may be conclu­ded on for want of a little premeditation.

2 Now, secondly, for the weight of it, if we [Page 79]put 8 Bushels to weigh 2 hundred pound weight (for sure it doth weigh more) then the whole number of quarters multiplied by 2, gives the weight of all to be 944473296573928 hundred weight, and if this be divided by 20, (which is but cutting off one figure toward the right hand, and dividing the rest by 2) or which is all one, cut off one figure from the number of quar­ters, the quotient 47223664828696, is so many tuns. And therefore it will require 47223664828 ships of 1000 a piece to carry it: And conse­quently if every Nation in the World had above 10000 such ships, yet there must be above four millions of such Nations: which I suppose are not to be found in this World.

And here I will leave this, having used this long example, (which though it require more labour as all great examples do, yet the same skill will do it, as if the places had been fewer) that the Reader being throughly exercised thereby, may the easier leap over others which are shorter.

THE GOLDEN RULE, Or Rule of Three.

THis is the most useful and most easie Rule in Arithmetique, and deserves a Golden name. It is when there are three numbers given or known to find a fourth in proportion with them.

But 4 numbers are in proportion, and called Proportional, when as the first is to the third, so is the second to the fourth.

As if there were given, 3, 4, and 6, to find a fourth which may be to 4, as 6 to 3, that is double, and that fourth number is 8; And this is called Proportion direct: and the Rule whereby it is done, The Direct Rule.

There is also another proportion called Recipro­cal; which is when as the first is to the third, so is the fourth to the second: As 3, 4, 6, and 2, this is called The Reverse Rule.

In direct proportion, the product of the two middle numbers multiplied together, is ever e­qual to the product of the first and last multipli­ed together; which serves not onely for a Proofe, [Page 81]but a ground of the Rule, which Rule shall here follow: the Reverse Rule being deferred till we have done with this.

The Rule Direct.

Multiply the second terme (or number) by the third and divide the product by the first; the quotient shall be the fourth number desired.

Example. Let the three numbers given be 2, 6 3, multiply 6 by 3, the product is 18, then divide 18 by 2, the quotient is 9, which is th fourth number in proportion with 2, 6, and 3.

For as 2 to 3, so 3 times 2, which is 6, is to 3 times [...] ▪ which is 9.

And so the product 18 divided by 2, and the quotient 9, causeth that the product of 2 into 9 shall be also 18, and consequently if 2 be the first of the 4 proportional numbers, and 6 and 3 the two middlemost, then 9 is the last.

Otherwise.

Divide the second by the first, and multiply the third by the quotient, the product shall be the fourth.

So if one divide 6 by 2, the quotient is 3, by which multiply 3, the product is 9, for the fourth number, as before, otherwise still this Rule might be expressed; but where the first way is so short and clear, there many other wayes would rather trouble then help the person that should use them.

In the first way (which here we mean to use [Page 82]and no other) if the first number be 1, then the product of the second and third gives the fourth, without any Division. Or if the second, or third number be 1, then there needs no multiplicati­on, but dividing the greater of them by the first, the quotient (in whole numbers for yet we speak of them) is the fourth number, which was sought.

Note 1.

To know when to use the direct, or the Re­verse Rule, consider, if more, require more; or if lesse, require still lesse: then use the direct Rule, but if more require lesse, or lesse more, then use the Reverse Rule, this will be easily understood when we come to examples.

Note 2.

To know how to place the three numbers when they are confusedly given, remember that two of them are alwayes of one denomination, as both pounds, or both sheep, or both yards, or acres; and the other number hath another denominati­on, now know that this single number is ever the second number in order.

And one of the other two, namely, that which hath some relation to this second, is the first; and the other is the third number, whose relati­on is sought for in the fourth, whence its plaine that the second and fourth are also of the same denomination.

[Page 83]And having premised these things, let us now exemplifie the Rule in some questions.

Question 1.

If 3 yards of cloth cost 4 li. what shall 21 yards cost?

Set the numbers in order as in the example, if 3 yards cost 4 li. what 21 yards? Here you see that the first number and the third number are both of one denomination, viz. both yards, and the second number is of another denomination, namely pounds, wherefore the fourth number which is sought for, must be also pounds, there­fore multiplying (according to the Rule before given) the second number by the third, and di­viding the product by the first, the quotient shall answer the question.

First, 21 multiplied by 4, (which is the third [...]mber multiplied by the second) produceth 84, which divided by 3 the first number, the quoti­ent is 28 li. and so much shall 21 yards cost: for 28 is to 4, as 21 to 3, seeing each containes either 7 times.

And the work will stand thus, [...]

Question 2.

If 4 men eat 2 pecks of corn in a week, how many pecks shall serve 100 men?

Place your numbers as here you see, then mul­tiply 100 by 2, (that is the third number by the second, and the product is 200, which divided by 4, the quotient is 50, for the number of pecks required.

[...]

Question 3.

If 20 sheep cost 13 pound 13 shillings 4 pence, what is that for every sheep?

Turn the shillings and pounds into pence, thus, [...]

[Page 85]

Multiply 12 s. by 12 the product is156
And 13 li. by 240 (because 240 pence make one pound) the product is3120
To which adde the 4 d.4
It makes in all3280

Then the question will be, If 20 sheep cost 3280 pence, what shall one sheep cost?

[...]

By the rule before delivered, I should multi­ply the second number by the third, but in this example, the third number being 1, it doth not multiply; I therefore divide 3280 the second number, by 20 the first number, and the moti­ent 164, is the price of one sheep in pence, which divided by 12, the quotient is 13 s. and 8 d. re­maining, the price of every sheep therefore is 13 s. 8 d.

Question 4.

How many 10 inch tiles will pave a floore that contains 16 square yards?

[Page 86]First, remember there are 36 inches in one yard in length; which multiplied into 36, gives 1296, for the square inches in one square yard; multiply 1296 therefore by 16, thence comes 20736, the summe of all the 16 yards in inches.

Secondly, seeing every tile is 10 inches in length, and 10 in breadth, multiply 10 by 10, it produceth 100 for the square inches in one tile; See the manner of worke.

[...]

Then by the golden Rule.

If 100 inches require 1 tile; what shall 207; 6 inches require?

[...]

[Page 87]Here because 1 doth neither multiply nor di­vide (as hath been several times intimated) I therefore divide the third 20736, by the first 100, the quotient is 207, and 36 remaining.

So it appears that 207 is too little, and 208 too much to doe the worke: the just number being 207 36/100, we shall not trouble the reader with this till he know something of Fractions.

Question 5.

If 100 li. give 6 li. interest for a year, how much shall 750 li. give?

Multiply 750 by 6, the product is 4500, which divided by 100, the quotient is 45 li. for the thing required.

[...]

Question 6.

If 750 li. give 45 li. interest for a year, what shall 100 li. give?

[Page 88]Multiply 45 by 100, the product is 4500, which divided by 750, the quotient is 6 li. for the inte­rest of 100 li. for a year.

[...]

Many other questions might be added, but the rule is so plaine that it needs them not; and so general, that he which can resolve one, may aswel resolve any other: And for that reason; and because in all the Rules which follow, this rule will be constantly made use of, I will say no more of it here.

The Golden Rule Reverse.

Question 1.

IF 12 workmen do any piece of worke in 8 moneths: how many workmen shall do the same in 2 moneths?

[Page 89] THE RƲLE.

Multiply the first term by the second; and divide the Product by the third, the quotient is the number desired.

Here 12 is not the first number, though it be first named; but the three numbers placed in or­der, stand thus, 8, 12, 2, for the middle terme must alwayes be of the same denomination with that which is required.

Now multiply 12 by 8, the product is 96, which divided by 2, the quotient is 48, which answers the question, As in the example, [...]

For if 8 moneths require 12 men; then (a fourth part of 8) 2 moneths shall require (four times 12, that is) 48 men.

For here lesse requires more, that is lesse time, more hands; and therefore is it wrought by the Reverse Rule.

Question 2.

How many Ells of Tapestry will serve to hang a roome 3 yards high, 6 yards long; and 5 yards broad; not regarding Doors, Windowes, or Chimney, but as if there were no such?

First, multiply 6 by 3, the product is 18, which doubled (because there are 2 sides called lengths) is 36 yards for all the length.

Secondly (for the same reason) multiply 3 by twice 5, that is by 10, the product is 30 yards, for all the breadth; which added to 36, gives 66 yards equal to all the length and breadth in yards.

But now because Ells, that is, Flemish Ells (for such measure are hangings sold by) is equal to 3 quarters of a yard, that is, their Ell is to our Yard as 3 to 4. Say therefore if 4 give 66, what 3? multiply 66 by 4, it produceth 264; then divide 264 by 3, the quotient is 88. Againe, multiply 88 by 4, and divide the product (which is 352) by 3, the quotient is 117, and 1 remaining, to which the divisor 3 being applyed; the number justly answering the question is 117 Ells, and one third part of an Ell.

Note 1.

Because here we had to deal with things which had equal length and breadth, that is square yards, and square ells; therefore one multiplication and division was not sufficient to [Page 91]proportion this; but if instead of working by 4 and 3, we had done it by their squares which are 16 and 9, it might have been performed at once, thus, multiply 66 by 16, product is 1056, which divided by 9, the quotient is 117⅓, as before, but I began not with this way, for I supposed my reader ignorant of squares.

Note 2.

It might also have been done, by reducing all the terms into quarters of a yard at the first, and after the number is found, reducing them a­gaine to ells; but because it is more proper to worke thus, till fractions have been taught, I leave that, and proceed to another question.

Question 3.

If 1 close would graze 21 horses for 6 weeks; then (supposing no waste to be made) how many horses would it feed for 7 weeks?

Multiply 21 by 6 it produceth 126, which di­vided by 7, the quotient is 18. At that rate therefore it would keep 18 horses for 7 weeks.

Question 4.

If 1 close will feed 18 horses for 7 weeks, how long shall it feed 63 horses.

Multiply (according to the rule) 18 by 7, the pro­duct is 126, which divided by 63, the quotient is 2, therefore 2 weeks it shall keep them.

[Page 92]The like way serves for hay, oates, or any o­ther provision for Man or Beast; which may be of use in Garrisons, and such like cases where scar­city may be feared, to proportion either the mouthes to the meat, or meat to mouthes

I will say no more of this Rule; Neither will I treat of the double Rule of Three, as a rule by it selfe; but come to the Rule of five numbers, which is held an abridgment of the other.

The Golden Rule Compound of five Numbers.

Question 1.

IF a hundred pound weight (that is 112 pound weight) carryed 120 miles cost 14 s. how much shall three quarters of a hundred, (that is 84 pound) cost, being carryed 40 miles?

THE RULE.

Multiply the three last numbers one into another, (that is) the third by the fourth, and that product by the fifth; the last product shall be the Dividend.

Again, Multiply the two first numbers together; the product shall be the divisor. This Division being made, the quotient will be the number of shillings desired.

Example of the former Question.

[Page 93]First, place your numbers according to the te­nor of the question thus, [...]

Your numbers being placed in order, reduce the 13 s. into pence, and it is 168 d. then multiply 168 by 84, the product is 14112, which multi­plied by 40 the later product, it produceth 564480, for the dividend.

Then multiply 112 by 120 it produceth 13440, for the Divisor.

Divide 5644800, by 13440, the quotient will be 42 pence; which is 3 s. 6 d. and answers the question.

In this Rule the first number and fourth, also the second and fifth; and also the third and sixth, are of like denomination and nature.

Question. 2.

If 100 li. for 6 moneths yield 3 li. interest, what shall 625 li. yeield for 36 moneths?

Place them, 100, 6, 3, 625, 36.

Multiply the three last, as before is shewed, the later product is 67500 for the dividend; And the 2 first multiplied make 600 the Divisor, then divide 67500 by 600 (or 675 by 6, which is all one) the quotient will be 112 whole pounds; and 300 (or 3) remaining, which because it is half the divi­sor, fignifies the half of a pound; that is 10 shil­lings. So the answer to the question is 112 li. 10 s.

[...]

Which might have been given in one denomi­nation, namely 2250 shillings, if before the work the pounds had been turned into shillings, by multiplying them by 20, as hath been shewed before.

[Page 95]But since most questions, except such as are stu­died for the purpose, are apt to end in some fra­ction, I shall next treat of fractions.

Onely first, having spoken of the double Rule of Three, this may let you know, that all questi­ons which are wrought at once by the compound rule of five; may be done at twice by the single rule of three: and the doing of them so by two operations is is called the double rule.

As in our last question there are two things considerable, the difference of money; and the difference of time.

First, for the money,

Say, if 100 li. give 3 li. what 625 li.? answer 18 75/100 li.

Secondly, for the time;

Say if 6 li. give 18 75/100 li. what 36 mo. answer 112 50/100 li.

But this will be better understood anon; and then the Reader may use that which he likes best.

Of Fractions.

THe word Fraction signifies a breaking or breach of any intire thing into parts; and when a number is broken so, the parts (which must needs be every one lesse then the whole; and the whole is accounted but One, or Ʋnity) being lesse then Unity, are called fractions (that is fragments, or pieces) of Unity. Now the Unite, or intire number which is to be broken, may be any thing, as one pound, in re­spect of which, shillings and pence, and far­things are fractions; or one shilling, in respect of which, pence, and farthings are fractions; or one peny, in respect of which, farthings are fra­ctions; and the like of weights and measures, or any other thing to be broken into parts.

In Fractions, we shall treat first of Numeration, then of Multiplication and Division, then of Redu­ction; then lastly, of Addition and Subtraction.

The reason of this Order will soon be seen; for Multiplication and Division are here much easi­er then Addition, &c. and therefore ought to be learned before them.

Numeration.

NUmeration is nothing else but the way of writing Fractions; and that this may be done, we must consider that any Unite, or Number representing an Unite, may be broken into two parts equal; and then each of the parts is called one second, or halfe; or it may be parted in­to three equal parts, and then each part is called one third; and two of them are called two thirds; and the like may be understood if it were parted into 4, 5, 6, 7, 8, 9, 20, 50, or 100, or how many soever.

Now to write these; doe thus,

  • Write one half Thus ½
  • Write one third Thus ⅓
  • Write one fourth Thus ¼
  • Write one fifth Thus ⅕
  • Write one sixth Thus ⅙
  • Write one seventh Thus 1/7
  • Write one eighth Thus ⅛
  • Write one nine Thus 1/9
  • Write one tenth Thus 1/10

[Page 98]In every one of these 10 fractions, the Num­ber below the line is called the Denominator, and it shews into how many parts the Unite is broken.

The Number above the line shews how ma­ny of those parts are taken, or contained in the fraction, and is therefore called the Numerator: So in the fraction ⅗: the Denominator [...] shewes the Unite to be broken into 5 parts: and the Numerator 3 signifies 3 of such parts to be con­tained in the fraction; which fraction therefore is called three fifths.

And here it is plaine, that as the Numerator is in proportion to the Denominator: so is the fraction to 1, for 3/3 or 5/5: or any the like, is equal to 1.

And therefore all fractions are quotients of lesser numbers divided by greater, as 4/7 signi­fies 4 to be divided by 7, and as the dividend 4, is to the divisor 7: so is the quotient 4/7 to Unity.

And therefore this line of separation which is drawn between the dividend and divisor, doth properly signifie division.

Hitherto we have spoken onely of such fra­ctions as are lesse then 1, and those are called Proper Fractions: but there are also 2½, 3¾, 5 1/7, 6⅗, and the like mixed Numbers; which so written signifie two and an half, 3 and 3 quarters, five and a seventh, 6 and 3 fifths. These by multi­plying the whole Numbers, by the Denomina­tor, and to the product adding the Numerators [Page 99]respectively are turned to 5/2, 15/4, 36/7, 33/5, which are called Improper fractions, because every one of them contains more then Unity.

These, neverthelesse may be multiplyed, divi­ded, added, or subtracted in the same way as are proper fractions. And this shall serve for Numeration of Fractions.

Multiplication.

THE RULE.

MƲltiply all the Numerators together, the last product shall be the Numerator of the product required: Likewise multiply all the Denomina­tors together, the last product shall be the Denominator of the product sought.

Example 1.

If ⅗ be multiplyed by 4/9, multiply the Nume­rator 3 by the Denominator 4, the product is 12, for the Numerator of the new product. Also multiplying the Denominator 5, by the Denomi­nator 9, they produce 45, for the Denominator of the desired product, so that product which was required, is 12/45.

Example 2.

If ½, ¾, ⅘, 5/9, and 3/11 were to be multiplyed all together, begin with the Numerators, saying, once 3 is 3, and 3 times 4 is 12, and 12 times 5 is 60, and 60 times 5 is 180, for the Numerator: Then multiply the Denominators: saying 2 times 4 is 8, and 8 times 5 is 40, and 40 times 9 is 360, and 360 times 11 is 3960, for the new De­nominator. So that the product of all these is 180/3960, that is equal to 1/22, as shall be seen hereafter in Reduction.

[Page 101]And thus it appears that proper fractions be­ing lesse then one, are still made lesse by multi­plying; as here the product 1/22 is much lesse then 3/11, which is the least Multiplyer, and the reasons hereof is plain, for seeing Multiplication is but the taking of a Number, a certaine num­ber of times, if that number of times be more then 1, then the Number to be taken is increa­sed by being taken more then once; but if the Number of times be 1, it is not increased, nor di­minished, but is still the same; Lastly, If that number of times be lesse then 1, as ½, the num­ber not being taken once, but half of once, pro­duceth a number lesse by half; that is the half of the number to be taken; and the like reason is of all others.

Example 3.

Multiply the mixt Numbers 3½, 4⅓, and 5¾: First, (as hath been shewn already) turn them to improper fractions: thus, first, say 2 times 3 is 6, and 1 is 7. So the first is 7/2. Secondly, 3 times 4 is 12, and 1 is 13: so the second is 13/3. Lastly, 4 times 5 is 20, and 3 is 23: so the last is 23/4 Now the fractions to be multiplyed are 7/2, 13/3, and 23/4; First, for a new Numerator, say, 7 times 13 is 91, and 91 times 23 is 2093, for a new Numerator.

Then say, 2 times 3 is 6, and 6 times 4 is 24. So the new Denominator is 23.

And the product of all these fractions is 2093/24, that is, if real division be made, 87 5/24.

Division.

DIvision of one fraction by another, is but crosse multiplication of them; that is, the Numerator of the one, by the Denomina­tor of the other, and hereby the proportion of one fraction to another is seen.

Example 1.

Divide ¾ by 6/8,

[figure]

to do it, set them thus: and multiply as the crosse leads, saying, 3 times 8 is 24, for a new Numerator, and 6 times 4 is also 24, for a new Denominator; so the quotient is 24/24, that is 1, which shewes the fractions to be equal one to another.

Example 2.

Divide ⅗ 4/9.

[figure]

First, set them thus: And say, 3 times 9 is 27, for a Numerator, and 5 times 4 is 20, for the Denominator: so the quotient is 27/20, and so many times is 4/9, con­tained in 3/5, that is, as 27 is to 20, so is 3/5 to 4/9, and so is 27/20 to 1.

In division it is to be remembred that the Nu­merator [Page 103]of the quotient ever ariseth of the Nu­merator of the Dividend; And the Denominator of the quotient comes of the Denominator of the Dividend, each being crosse multiplied, as before.

If a Fraction be to be divided by a whole num­ber, multiply the Denominator by that num­ber, the product gives the new Denominator, and the Numerator remaines the same. So if ¼ be divided by 9, say 9 times 4 is 36. So the quotient is 1/36.

Or if ¼ were to be multiplyed by 9, the pro­duct (by multiplying the Numerator by 9), will be 9/4: that is, 2¼.

Example 3.

Divide 320/8 by 45/9, thus:

[figure]

say 320 times 9 is 2880, for a Numerator: And 8 times 45 is 360 for a Denomina­tor. So the quotient is 2880/360, or 8/1.

For 320/8 is equal to 40, and 45/9 equal to 5, but 40 contains 5 eight times.

And so in the second example, it may be pro­ved, that as 27 to 20, so is ⅗ to 4/9. For first multi­ply the two middlemost, then 20 times ⅗ is 60/5, that is 12.

Secondly, multiply the first and last, and then 27 times 4/9 is 108/9: that is also 12.

Wherefore by that which hath been said in the Golden Rule, the four Numbers 27, 20, 3/5, 4/2, are proportional.

Reduction.

REductioon of Fractions is threefold.

1 To reduce one fraction (which is not already in the least) to a lesser Denomi­nation.

2 To reduce many fractions of divers deno­minations, to one denomination.

3 To reduce any fraction from one denomina­tion (as neer as may be) to any other denomi­nation desired.

For the first of these, divide both the Nume­rator, and the Denominator, by the greatest com­mon divisor that you can think of; the two quo­tients being placed respectively in a fraction, that fraction shall be equal to the former fraction, and in lesser termes.

So (in the 3 Example of Division) to reduce 2880/360 to 8/1, divide 2880 by 360, quotient is 8, then divide 360 by 360, the quotient is 1, and the new fraction 8/1 is equal to the former fraction, 2880/360, and in lesse termes; as you may see. But to find the greatest common divisor.

The Rule is,

Divide the greater terme by the lesser (I mean by [Page 105]termes, the Numerator and Denominator) and by the remainer (if any be) divide the divisor, and if any thing still remains, by that divide the last divi­sor, continuing this course till nothing remain greater then Ʋnity; that divisor which is last of all is the greatest common measure of both termes, by which both being divided, and the quotients placed like a fra­ction, that fraction shall be equal to the former fraction, and in the least termes.

Example.

Reduce 148/16 to the least termes; first divide 148 by 16, the quotient is 9, and 4 remaines: a­gaine, divide 16 by 4, the quotient is 4, and no­thing remaines; wherefore taking 4, (the last di­visor) for the greatest common divisor, by it di­vide 148, the quotient is 37, and by it divide 16, the quotient is 4. These two last quotients pla­ced orderly in a fraction, make 37/4, which is e­qual to 148/16, and in the least terms, for no num­ber greater then 1, will divide evenly both 37 and 4.

Other wayes there are of lessening fractions, as dividing the termes (if they be even Num­bers) by 2, and the quotients (if even) again by 2, or else by 3, o [...] any other Number that will divide them both evenly, that is, leave nothing remaining, but the former Rule being general and easie, shall serve for all.

Now secondly, to reduce many Denominati­ons to one common denominator; let the fra­ctions be ½, ¾, ⅘, ⅞, 9/10, to be reduced all to the de­nomination 3200.

[Page 106] THE RULE.

Multiply all the Denominators together: saying 2 times 4 is 8, and 8 times 5 is 40, and 40 times 8 is 320, and 320 times 10 is 3200, this last product 3200, shall be the common Denominator. Then to get Numerators for every one of them. Multiply eve­ry particular Numerator into all the Denominators except his own: As first, for the first, say 1 time 4 is 4, and 4 times 5 is 20, and 20 times 8 is 160, and 160 times 10 is 1600.

For the first Numerator; so the first fraction reduced is 1600/3200. Then for the second Numera­tor: say, 3 times 2 is 6, and 6 times 5 is 30, and 30 times 8 is 240, and 240 times 10 is 2400. So the second fraction reduced, is 2400/3200. After the same manner may the other three be reduced to 2560/3200 for the third: 2800/3200 for the fourth: and 2880/3200 for the last; these are severally equal to the o­ther, the first to the first, &c. as may be proved; thus,

Let the Unity be a pound Sterling, then

 s. 
The ½ of it is10 
and ¾ is15 
and ⅘ is16d.
and ⅞ is176
and 9/10 is18 
 s.d.
In all766

That is 3 whole Unites, and 16 s. 6 d. over, Turn 16 s. 6 d. all to six pences, it is 33, and [Page 107]because 6 d. is the fortieth part of a pounds, there­fore all the fractions are equal to 3 33/40.

Now adde the new Fractions (which being all of one denomination) may be added like whole Numbers: thus, [...]

Which divided by the Denominator 3200, the quotient is 3 2640/3200. Now 2640/3200, reduced to the least termes, as hath been shewed how it may, will be 33/40, so the sum of these also is 3 33/40. which is equal to the sum of the fractions given to be reduced, and therefore they are equal in sum, and might be thus proved equal severally, that is, the first of them propounded, to the first re­duced. Divide the Numerator 1600 by the Nu­merator 1, quotient is 1600. Also divide the Denominator 3200, by the Denominator 2, the quotient is also 1600, and so may any of the rest be proved equal by the equality of quotients. But I leave it as plaine enough already.

Thirdly, Any fraction being given, to change the denomination to any other more requisite, retaining still (as neer as may be) the same value.

THE RƲLE.

Multiply the Numerator given, by the Denomina­tor [Page 108]required, and divide the product by the Denomina­tor given; the quotient shall be the Numerator required.

Example.

Let the Fraction given be 7/13 of a pound Ster­ling, what is that in twentieth parts or shillings? Multiply 7 by 20, the product is 140, which di­vided by 13, the quotient 10 10/13, that is, 10 s. and 10/13 of a shilling: which may be brought to pence thus, multiply 10 by 12, product is 120, which divided by 13, quotient 9 3/13 d. And againe, multiply 3 by 4, the product is 12, which divi­ded by 13, quotient is 12/13 of a farthing, so seven thirteenths of a pound is 10 s. 9 d. and almost a farthing.

But he which is resolved to have it in the smal­lest coine, may do it at first work; for seeing a farthing is the 960 part of a pound, multiply 7 by 960, they produce 6720, which divided by 13, the quotient is 516 farthings, and 12/13 of a far­thing: these farthings may be turned to shillings, dividing by 48, or to pence by 4.

This Rule though it be brief and plain; is of great use in Arithmetick; either for turning Natural and surd Fractions into Decimals; or any other desired Denomination, with such fa­cility and speed as may be wished.

Fractions of Fractions.

In reduction of Fractions, some make another, or more parts, as fractions of fractions for one: that is, when there is a part of a fraction, or a part [Page 109]of a part of a fraction, &c. to be valued in one fraction.

THE RULE.

Multiply all the Numerators together, the last pro­duct shall be the Numerator desired: then multiply all the Denominators together, and this last product shall be the Denominator sought.

Example.

Let the Fractions of Fractions propounded, be ⅘ of ¾ of ½, for so they are usually written; and let the Numerator be multiplyed: saying, 4 times 3 is 12, and 12 times 1 is 12, the Numerator therefore required is 12: then for the Denomi­nator, say, 5 times 4 is 20, and 20 times 2 is 40, for the Denominator required; and 12/40 is equal to ⅘ of ¾ of ½.

Proof.

Let the Unite be 40 s. one fifth of 40 is 8, and therefore ⅘ is 32, of which one fourth is 8, and ¾ is 24, of which one half is 12, and therefore 12/40 is the just sum of all the Fractions, This needs no further exemplifying.

Addition.

TO adde many fractions into one sum, consi­der whether they be of one Denomination or divers; if of one, then adde all the Numera­tors together into one summe, that summe is the new Numerator: and the Denominator, in this case is not altered.

Example.

Let the fractions to be added be 2/4, 4/4, 5/4, ¼, Adde the Numerators: saying 2 and 4 is 6, and 6 and 5 is 11, and 11 and 1 is 12. So the summe of them all is 12/4, that is 3 Unites.

As, let the Unite be 20 s, one fourth is 5 s. and 2/4 is 10 s. and 4/4 is 20 s. which added to 10 s. is 30 s. then 5/4 is 25 s. which added to thirty shil­ling gives 55 s. And lastly, ¼ is 5 s. which added to 55 s. makes 60 s. that is 3 times 20 s. that is 3 l. or 3 Unites.

But if the fractions to be added, be of divers denominations, as let them be 3/2, ¾, ⅘, ⅞, then (by the reduction afore-going) they must be tur­ned all into one denomination, and then they [Page 111]will be 320/480, 360/480, 384/480, and 420/480, and may be ad­ded like those before: thus, [...]

So the summe of all is 1484/480, or 371/120, that is, 3 11/120, which if it be money, and the Unite 1 l. it is then 3 l. 1 s. and 10 d. as may be tryed thus: First, ⅔ of a pound, is 13 s. and 4 d. and ¾ is 15 s. and ⅘ is 16 s. Lastly, ⅞ is 17 s. 6 d. These all ad­ded together, the summe if 3 l. 1 s. 10 d.

Subtraction.

IN Subtraction of one fraction from another, if they be both of one denomination; it is done by taking the Numerator of one from the Numerator of the other, the remaine is the new Numerator, and the Denominator the same as before.

So if 2/9 be subtracted from 8/9. the remaine is, 6/9 the like of all others.

But if they be not of one Denomination, they must first be reduced to be so; then that which is said before is sufficient.

The golden Rule in fractions is the same as in whole Numbers, I will give you but one in­stance.

If ¾ of a yard of tape cost ½ of a peny, what shall one inch, that is, 1/36 of a yard cost?

Multiply the second by the third, the product is 1/72, which divided by ¾, the quotient is 4/216 of a peny, for the price of 1/36 of a yard.

Otherwise,

Seeing ¾ of a yard may be turned to 27 inches; [Page 113]Say, if 27 cost ½, what 1? divide ½ by 27, it makes 1/54 for the answer: which is equal to 4/216, and in the least termes.

And wheresoever this may be done, to have the first and third Numbers fractions of one de­nomination, the best way is to work with their Numerators, not regarding their denominator at all: as if ⅔ cost ¾, what 7/3? instead thereof write, if 2 cost ¾, what 7? multiply ¾ by 7, it produceth 21/4, which divided by 2, the quotient is 21/8, and that is the answer in the least terms.

And all this while it should have been noted that the fractions are ever written in a smaller figure then the whole Numbers.

The Rule of Fellowship.

THis Rule is usefull for Merchants, and all such as trade in Companies, with a joint stock; and must share a proportional part of the gaines, or losse; every one accord­ing to his stock which he laid in.

The Rule is two-fold, with equal time; or with unequal time.

That which is with equal time, is commonly called, The Rule of Fellowship without time.

Of this we will first speak.

THE RULE.

As the whole joint stock, is to all the gain or losse; So is each mans particular stock, to his part of the gaine, or losse.

Example 1.

Two Merchants A. and B. buy 700 l. a yeare Land for ever, (when money is at 8, per cent.) for 14000 l. of which A. paid 8000 l. and B. 6000 l. after 5 yeares (money being fallen to 6 per Cent.) [Page 115]they sell it for 18700 l. so there is gained 4700 l. how much of this must A. have?

First for A.

Say, if 14000 gaine 4700, what 8000? answer, 2685 10000/14000.

Then for B.

If 14000, gaine 4700, what 6000? answer 2014 4000/14000. As by the following operation doth appeare.

(I) For A.

[...]

[Page 116](II) For B.

[...]

Here note, that this work might have been much abreviated, if from each of the three num­bers you had cut off two Cyphers towards the right hand, as hath been formerly shewed in the Compendium of Multiplication and Division.

Now for the proof hereof,

If you adde2685 10000/14000.
which is the sum that A. gained; 
 To—2014 4000/14000.
 The summe of them is 4700.

Which is equal to the total gain.

And according to the proportion of these two Numbers: that is, as 8 to 6, or 4 to 3. So they ought to have parted the yearly rent also, all the time they received it: that is, A. ought to have 400 l. yearly; and B. 300 l.

Example 2.

A. B. and C. joyne their moneys to make a stock of 25000 l. of which A. laid in 10000 l. B. 8000 l. and C. put in 7000 l. with this (after a cer­tain time in trading) they gained 7500 l. how must this be parted?

First for A.

Say, if 25000 gaine 7500, what 10000?

Or shorter, if 25 get ½, what 10? Multiply 7½ by 10, it produceth 75, which divided by 25, the quotient is [...], that is, (restoring the 3 Cy­phers) 3000 l. for A.

Then for B.

Say, if 25000 gaine 7500, what 8000?

Or shorter, if 250 get 75, what 80.

Multiply and divide as the Golden Rule requires, and to the quotient, restore the 2 Cyphers, then it will be 2400 l. for B.

Lastly, for C.

Say, if 250 give 75, what 70? answer 21, to which put the 2 Cyphers, it makes 2100 for C.

And these three 3000, 2400, and 2100, being added together, make 7500. And have that pro­portion as the particular stocks had: and there­fore the work is right.

[Page 118](I) for A.

If 25 gaine 7 ½, what 10 s.

[figure]

[...]

(II) for B.

[...]

[Page 119](III) for C.

[...]

And if instead of gaining 7500 l. whereby e­very one is supposed to have his stock, and a part of the gaines; they had lost 7500 l. then their particular stocks had not been due to them, but so much as would be left after their proportional parts of the losse were abated.

Example 3.

A. B and C. with a joynt stock of 25000 l. gaine 7500: of which A. gets 3000. B. 2400. C. 2100, what was their stock?

This is but the Converse of the former, there­fore say, if 5700 require 25000, what doth 3000 require? 10000 for A. and so work for the other two.

Many examples are of little use, (except to load the Readers memory) where the Rule is so short and plain, I will therefore adde no more to this part of the Rule, but immediately come to the Rule of Fellowship with time.

The Rule of Fellowship with time.

THis Rule is to be used when the times of the continuance of the particular stocks are un­equal, and differ; so that here the difference of time, and also the difference of stock being both to be considered; it can be done no better way then by taking the Power of them both to be the particular stock; and all those Powers added, to be the whole stock, that which I call the Power, is the product of the money of every one multi­plyed by his time; And then

THE RƲLE.

As the summe of those Products, is to the whole gain; so is each particular product, to its part of the gain.

Question 1.

Three Merchants A. B. C. make a stock of 10000 l. of which A layes in 4000 for 3 moneths, B. 3000 l. for 6 moneths; and C. 3000 l. for 8 moneths, with this they gain 2000 l. what is each mans share?

[Page 121]First, for A. multiply 4000 by 3, it makes 12000, let that be accounted his particular stock.

Secondly, for B. multiply 30000 by 6, it makes 18000.

Lastly for C. multiply 3000 by 8, it produceth 24000, for his stock, adde these, they make 54000 l. for the general stock; then say,

For A.

If 54000 give 2000, what 12000? answer, 666 24000/54000.

Then for B.

If 54000 give 2000, what 18000? answer 444 36000/54000.

Lastly, for C.

If 54000 give 2000, what 24000? answer, 888 48000/54000.

The three Fractions may be reduced (by divi­ding each Numerator, and Denominator by 6000) and then the three shares will be 444 4/9, 66 6/9, and 888 8/9, which altogether make 2000, as they ought.

Question 2.

Three Farmers, A. B. and C. lay out 1000 l. to stock their grounds with Cattel, of which A. put in 200 l. for 6 yeares; B. had 300 l. going for 4 yeares; and C. 500 l. for 2 years; at the end (by unseasonable times) there was lost 200 l. which made the remain of their stock but 800 l. what had each man left?

Multiply 200 by 6, it gives 1200: Likewise, 300 by 4, it gives 1200. Lastly, 500 by 2, the pro­duct is 1000: all these are 3400 for the joynt stock.

[Page 122]Then first for A.

Say, if 3400 lose 200, what 1200? answer, 70 2000/3400 for A, to which B is equal, because the power of his stock is so.

Therefore for C.

Say, if 3400 lose 200, what 1000? answer, 58 2800/3400. So the 3 shares are 70 20/34, 70 20/34, and 28 2 [...]/34, equal to 200.

Now because A put in 200 l. and lost 70 20/34 sub­tract the losse from the stock, remains 129 14/34.

And so doing for B, his remain will be 289 14/34.

And for C, his remain is 441 6/34. Now these three remains, 129 14/34, 229 14/34, and 441 6/34, make up 800 l. which was the whole remain.

Quest. 3.

A. rents a close for a year, to pay 80 l. he puts into it 200 sheep: 2 months after B. puts 40 sheep in; and 5 moneths after that C. puts in 100 sheep; how much must every one pay of the rent?

Multiply 200 by 12, it produceth2400
And 40 by 10, produceth400
Lastly, 100 by 5, (which is C time) produceth500
In all3300

Then for A.

If 3300 pay 80, what 2400? answer, 58 600/3300.

Then for B.

If 3300 pay 80, what 400? answer, 9 2300/3300.

And for C.

If 3300 pay 80, what 500? answer, 9 400/3300.

The whole numbers make 79, and the broken numbers make 1. In all 80.

Note

Whereas, hitherto we have considered onely difference of time and money; it may be noted, that there may be difference of other kinds, as persons or place; but whatsoever they are the pow­er of all is found like these by multiplication; and are to be worked like these, with so many Uses of the Golden Rule, as the question requires. I will therefore adde but one question more, which is this.

Quest. 4.

One leaves a Legacy of 900 l. among foure Kinsfolks. A. B. C. D; so as B may have twice as much as A. and C thrice as much as B. and D as much and half as much as C, what is every one to have.

Say, if A be 1, B is 2, C 6, and D 9, adde these Numbers: 1, 2, 6, 9, together, they give 18, then say, If 18 require 900, what 1? answer is 50. So A is to have 50 l. B 100 l. C 3000 l. and D 450 l. which are their just parts; and altogether are e­qual to 900 l. and the work right.

The Rule of Alligation.

THis hath its name from binding, tying, or uniting many particulars in one Masse or Summe, the nature of it will be understood in working some Questions or Examples.

Question 1.

A Corn Master would mix 4 sorts of graine together, viz. wheat at 4 s. the bushel, wheat at 4 s. 0 d. the bushel; Rie, at 3 s. the bushel, and Barly at 2 s. 8 d. the bushel; so as to make 15 quarters in all, to be sold at 3 s. 6 d. the bushel; How much must he take of each?

Place them as in the Margin,

[figure]

so as a greater and lesser may still be together, as 4½, with 2⅔ and 4 with 3, and place the price required by it selfe toward the left hand, as here you see 3½ then in a seperated Column note the difference between the price of a bushel of every one parti­cular given, and a bushel of that required, as the [Page 125]difference betwixt 4½ and 3½ is 1, which must not be placed against 4½, but against that number linked with 4½, that is, against 2⅔, and so must all the differences be ordered, as is easie to be seen in the Margine: then

THE RULE.

Multiply the whole Masse to be made, by any par­ticular difference; and divide the product by the summe of all the differences, the quotient shall be the just quantity of that particular kind, whose price standeth against the difference you worked with.

Example.

First turn the quarters into bushels, by say­ing, 8 times 15 is 120, then for the quantity of the first sort at 4½: multiply 120 by ⅚, the pro­duct is 100, which divided by 2⅚, the quotient is 35 5/7 bushels of that sort at 4 s. 6 d. and working so for every of the other; they will be found to be thus:

At 4 s. 6 d.35 5/7 bushels.
At 4 s.21 3/17 bushels.
At 3 s. 6 d.21 3/17 bushels.
At 2 s. 8 d.42 6/17 bushels.
In all120

Now to prove this right; first multiply the whole Masse 120 bushels, by the desired price 3½ s. omitting the denominations, the summe is 420 s. [Page 126]

Then secondly, multiply 38 5/17 by 4½, it is,158 14/17
And 21 3/17 by 4, it produces84 12/17
And 21 3/17 by 3, it makes63 2/17
And 42 6/17 by 2⅔ is112 16/17
In all420

But so much all should be worth in shillings. And therefore the question in rightly solved.

Quest. 2.

One hath 6 sorts of Fruits at several prices, Dates at 2 s. Almonds at 1 s. 4 d. Currance at 10 d. Raisons at 5 d. Prunes at 4 d. and Figs at 3 d. the pound; and would take of every sort some to make a mixed quantity of 30 l. weight to sell one with another for 9 d. the pound, how much must he take of each?

Having placed the Numbers & their differen­ces,

[figure]

and the sum of those diffe­rences distinctly as hath beene shewed before, and may be seen by the Figure in the Margine; the work is ever­more like that in the former question. So 38 is the first number in the Golden Rule; 30, the se­cond [Page 127](which that may not be forgotten may be set at the right side of the figure) and every par­ticular difference, as 6, 5, 4, &c. is the third, the Rule to be repeated till all the differences have been employed.

So 30, multiplyed by 6, produceth 180, which Divided by 38, the quotient is 4 22/38 of a pound weight: and so much must be taken of Dates, at 24 d.

Secondly, 5 times 30 is 150, which divided by 38, the quotient is 3 36/38 for Almonds. And work­ing after the same manner with 4, 1, 7, 15, their respective quantities will be found to be these;

 pounds 38 parts.
Dates4, 28
Almonds3, 36
Currance3, 6
Raisons0, 30
Prunes5, 20
Figges11, 32
In all26, 152

That is 26 252/32, and the reduction of the fracti­on will make it 30, as it ought to be, and by com­paring the prices of these particulars added, with the price of 30 li. weight, at 9 d. per l. weight, which makes 470 d. this may be proved like the former.

But that the Reader may be perfect in it, I will doe it here also, as followeth.

Say first, 24 times 4 is 96, and 24 times 28 [Page 128]is 672; for the first, set them thus: 96, 672.

And 16 times 3 is 48, and 16 times 36 is 57648, 576
And 10 times 3 is 30, and 10 times 6 is 6030, 60
And 5 times 30 is 15000, 150
And 4 times 5 is 20, and 4 times 20 is 8020, 80
Lastly. 3 times 11 is 33, and 3 times 32 is 9632, 96
In all227, 1634

Now this 1634 being the summe of the Nume­rators of Fractions, whose common Denomi­nator is 38, must be divided by 38, and the quo­tient will be 43, which added to the whole num­ber 227: the summe is 270. And so much is 30, multiplyed by 9, which shewes the work to be right.

The Combination, or linking of Numbers may be varied at pleasure, as whereas above I linked 24, and 3, also 16 with 4, and 10 with 5, it might have been 24 with 5, and 16 with 4, and 10 with 3. Or 14 with 4, and 16 with 3, and 10 with 4, of which diversity of linking would fol­low diversity of solutions, but all true, as the Reader may easily prove by himself.

Likewise, if the Numbers to be linked were 3, 5, 7, or any odde Number, one of them may be linked to two severally, to make the worke even.

Example 3.

If the Numbers were 12, 10, 8, 6, and 4, and the mean or common price required were 9, you might first linke them as you see here, take­ing 12 twice,

[figure]

or else you might take any other twice as you shall like; and so the work will be every way right, though not the same; if the differences be rightly set off and orderly used, as is taught before in the first question.

Question 3.

A Goldsmith would mix 3 sorts of silver, A. B. C. A is 10 d. weight better; B 7 d. weight bet­ter; and C, 4 d. weight better, to make an In­got of 50 li. weight, which should be in finesse 8 d. weight better: How much must be taken of each?

Set them,

[figure]

their diffe­rences, and the summe of their dif­ferences as in the Mar­gine.

[Page 130]

1. Then first 50 multiplyed by 4200
and divided by 9, quotient is9
2. 50 multiplyed by 2, and50
divided by 9, quotient is9
3. 50 multiplyed by 2, and100
divided by 9, quotient is9.
4. And again, the same100
 9
In all450
 9

Which is equal to 50, the quantity required,

Now the first fraction multiplyed by 10, (o­mitting the Denominator) is2000
The second also by 10 gives500
The third by 7 makes700
The last by 4 makes400
In all3600

That is, 1600/9, which is equal to 400, and if the whole Ingot 50, be multiplyed by the betternesse required, namely, by 8, they shall produce 400 also: So this is proved.

In every Alligation, or linking of two num­bers, this is evident, that if the summe of the numbers linked be greater then the mean number required taken so many times as there are num­bers to be linked, the question would be absurd; and the resolution-thereof impossible. And this shall serve for the rule of Alligation.

The Rule of False Position.

THis Rule serves to resolve questions, which are not presently fit for the Golden Rule; and therefore in stead of the true Number which is sought: Suppose any number great or small, and make trial of it, whether it resolve the question without any er­rour; if so, it is the true Number: if not note what errour is at the end of the worke, and whe­ther it be too much or too little: if too much, mark it thus + but if too little, thus −

Then suppose again another Number, (it im­ports not whether it be neerer or further off) and try as before, and mark that errour also;

And, then

THE RULE.

Multiply the first Position by the second Errour, and the second Position by the first errour, and (if the errours be both + or both −) Subtract the lesser product from the greater, and keep the remaine for a Dividend, and the difference of the errours for the [Page 132]Divisor: the quotient of that division is the true Number required.

But if the errours had been one +, the other −, then the summe of the products added toge­ther had been the Dividend: and the summe of the errours, the Divisor; the rest of the work the same as before.

Question 1.

A man is to drive 48 young Turkies 40 miles: and for every Turkie which comes alive to the end of the Journey, he is to receive 3 d. but for every one which dyes by the way, he is to pay 6 d. at the end, he received 72 d. How many dyed by the way?

Let the first supposition be

That by the way there dyed20
For them he was to pay 120 d. and for 28 which lived, he was to receive 84 d. so he paid more then he received36 d.
and should have got clear72 d.
 Adde 108

Wherefore the first errour is−108
Let the second supposition be10
For these he paid 60 d. and for the rest he re­ceived 114 d. the difference is 54, and should be 72: so the second errour is−18
Now 20 multiplyed by 18, produceth360
and 10 by 108 produceth1080

and the difference is 720 for the Dividend, like­wise the difference of the errours is 90 for the Di­visor, and the quotient is 8, which is the true Number of those he lost by the way. As may be proved by trial.

Question 3.

If it was required to make up a pound Sterling of shillings and groats only: and so as the num­ber of groats, may be to the Number of shillings, as 7 to 1, how many shillings must there be?

First, suppose the shillings4
then the groats must be equal to 16 s. that is 48 groats; but the shillings taken 7 times are 28, to which 48 should be equal, but is+20
Secondly, suppose the shillings2
then the groats (making 18 s.) are 25, 
which should be equal to 7 times 2, but is+40
Multiply 4 by 40 product is 160, then 
Multiply 2 by 20, the product is 40, which ta­ken from 160, rests for the Dividend120
And the difference of errours is20
Lastly, 120 divided by 29, quotient is6
The number of shillings therefore is6
And the number of groats is42

But as 7 to 1, so is 6 times 7 which is 42 to 6 times 1, which is 6: so the work is done.

Question 3.

If there be 4 several weights, A. B. C. D. of which D is 24 ounces, and C is double to B, and triple to A, and D with twice A is double to C. and quadruple to B: How much doth every one of these weights weigh?

[Page 134]

First, suppose A to be then D with twice A is 24, and 16, that is 40, of which [...] being the halfe is [...]0, and B 10.8
Now thrice A is 24, to which C should be equal, but is−4
Secondly, let A be supposed4
then D more, twice A is 32, and C 16 and B is 8, but thrice A is 12, to which 16 should be equal, but is+4

Then 8 multiplied by 4, gives 32, and 4 by 4, produceth 16: both these products gives 48 for the Dividend: and the summe of the er­rours (because the first is −, the other +) gives 8, for the Divisor; and the quotient will be 6, to which A is equal, and twice A more D is 36, of which C, (being halfe is 18, and B is 9, and thrice A is equal to C, namely, 18, and all right.

Whereas the first errour is equal here to the second, it followes that the Positions were equal­ly false: and therefore their difference which is 4, being parted into two equal parts, 2 and 2, if 2 be taken from 8, the remaine is the true number 6, or if [...] be added to 4, (which was the second position) the summe will be also 6.

And further, whensoever the erronrs be one +, the other −, though they be not equal; yet then if the difference between the positions be parted into two parts, which are in proportion one to another, as the two errours are one to another respectively: then if the first part be taken from the first position (if that be the greater) or adde to it (if it be the lesse, the same number required is thereby had.

[Page 135]As, let the last question be resumed,

And let the first position for A be15
Then the first errour will be−18
Then let the second position be3
And so the second errour will be+6
And the difference of positions is12

which divided into two parts 9 and 3, which have that proportion one to another as have the er­rours 18 and 6, then if the first part 9, be taken from the first position 15, there remains the true Number 6. Or else if the second part 3, be added to the second position 3: thereby also is made the true Number 6.

The way of parting 12, (or any other) into two parts proportional with the errours is easily done by the Golden Rule, thus:

  • As the summe of the Errours 24,
  • is to the difference of positions 12;
  • So is the greater errour 18
  • to the greater part required, namely, 9.

Many other questions are in other bookes exemplifyed and wrought by this Rule; but see­ing I intend not to write a great book; and al­so because some of those questions may be easily resolved without this Rule, I will adde no more: ondy mention one of those questions.

I there be a Cisterne with 4 Cocks, which holls 8 barrels of water, and the first cock will runit all out in 6 houres, the second in 4, the third in 3, and the last in 2 houres: in what time shal all of them runne it out?

[Page 136]

If the first in 6 houres run8
the second in the same time would [...]un12
the third16
the last24
In all60

Then say, If 60 require 6 [...] what 8?

The answer 48/60, that is 4/5 o [...] an hour; in which time all the 4 Cocks together would run out all the 8 Barrels of water.

The Rule of Ceres and Virginum.

THis is the most uncertaine, and unneces­sary Rule in Arithmetick; being seldome used except in sporting questions to puz­zle young beginners, with easie pro­blems; such as follow.

Question 1.

A Caterer bought 8 birds of two sorts, as Geese and Hens for 20 s. the Geese cost 4 s. a piece, the Hens 2 s. a piece; How many did he buy of each sort?

This may be done by the Rule of False; and also thus: multiply the whole number 8, into the least price 2, it produceth 16, which taken from the whole price 20, rests 4 for a Dividend; which divided by 2, which is the difference of the particular prices, the quotient is 2, for the number of Geese; and 6 must be for the Hens: the proofe is easie.

Question 2.

If 21 Persons, Men, Women, and Children [Page 138]spend 26 shillings; so that every man payes 2 s. every woman 1 s. every child 6 d. How many is there of each sort?

THE RULE.

Multiply the number of Persons by the least expence, and take the product of it from the whole expence, the rest shall be the Dividend; which divided by the difference betwixt the greatest and least particular ex­pences: the quotient is a Number which the Number of men (or they which spend most) comes neer to: but can­not exceed: or of the said Dividend be divided by the summe of the greatest and least expences, the quotient is a Number then which the Number of men or those which spend most) cannot be much lesse.

So here 21 multiplyed by 6 d. that is, by ½, the product is 10½, which taken from 26, rests 15½, for the Dividend: and then taking ½ from 2, rests 1½ for the Divisor, and the quotient is 3⅓, which is something more then 10, the Number of men therefore must be but 9.

Then turn the Dividend, and the Divisor both into whole Numbers, by multiplying them by the common Denominator 2, so they reduced will be 31 and 3, as before is to be seen in the quo­tient.

Multiply the Divisor 3, by 9, (which is the Number of Men) the product is 27, which ta­ken from 31, (which is the reduced Dividend) the remaine is 4, for the Number of Women; and the Children must be 8.

[Page 139]

Example 1.
9 men at 2 s. each18 s.
4 women at 1 s. each4
8 children at 6 d. each4
In all 21In all 26

But the number of men may be also 8, which multiplyed by the reduced Divisor 3, product is 24, which taken from 31, the remaine is 7 for the women; and then the children must be 6.

Example 2.
8 men at 2 s. each16
7 women at 1 s. each7
6 children at 6 d. each3
In all 21In all 26

Or the number of men may be 7, which mul­tiplyed by 3, produceth 21, which taken from 31, remaines 10 for the women; and 4 children.

Example 3.
7 men at 2 s. each14 s.
10 women at 1 s. each10
4 children at 6 d. each2
In all 21In all 26

So here is already seen 3 various solutions of [Page 140]this question, which makes this Rule the lesse to be regarded. But further, the number of men may be 10, and not more, for if you put them 11, that multiplyed by 3, produceth 33, which is greater then 31, from which it should be taken, but I say it may be 10, and then there is only one woman, and 5 children: this confirmes the for­mer part of the Rule.

Now for the later part, if the Dividend 31 be divided by the summe of the two extream ex­pences (reduced by doubling as the Dividend is) 4, the quotient will be 7¾. And the men may be 7, as hath been shewed; but they may be also but 6, and fewer they cannot be: as 6 men, 13 women, and 2 children; for if you put them 5, that multiplyed by 3, produceth 15, which taken from 31, there remaines 16 for the women, and so there should be no children: which is contrary to the supposition.

And further, because the quotient was 7¾, the number of men might be so, if pure Arithmeti­cal division be onely regarded: and then the wo­men also are in number 7¾, and the children 5½, as may easily be tryed; I need not exemplifie it.

Question 3.

If there be an Exhibition of 900 l. per annum to 30 persons: some Clerks, some Messengers, and some Door-keepers, at 60 l. each Clerk, 40 l. each messenger, and 20 l. each door-keeper, how many must there be of each sort?

Multiply (according to the Rule) 30 by 20, the product is 600, which taken from 900, re­mains [Page 141]300 for the Dividend; and 60 want 20, that is 40, for the Divisor: and the quotient is 7½, and more the Clerks cannot be; Also divide by 60 more 20, that is 80, quotient is 3¾, and much fewer the Clerks cannot be.

Not to stand upon the Fractions (in this case of dividing men) the Clerks may be 7, 6, 5, 4, or 3: and the messengers 1, 3, 5, 7, or 9, and the Door-keepers 22, 21, 20, 19, or 18, that the Clerks cannot (in whole Numbers) be more then 7, or lesse then 3, may thus be proved; First, let them be 8, then 8 times 40 is 320, which is more then 300, out of which it should be taken: Se­condly, let them be 2, then 2 times 40 is 80, out of 300 remains 220, which divided by 20, gives the quotient 11, for the messengers, so the Clerks and messengers being 13, the remaine thereof to 30, namely, 17, must be Door-keepers.

but,

2 Clerks at 60 l. each120 l.
11 Messengers at 40 l. each480
17 Door-keepers at 20 l. each340
In all 30In all 940

Which is 40 l. too much, therefore the Clerks cannot be two.

Note.

It may be asked, why the remaine 220 should be divided by 20: whereas the like remaine in the former Example, namely 16, was taken (without any division) absolutely for the num­ber [Page 142]of Women, or middle number? I answer, although the greatest or first Number being found, (as here to be 2) the residue of 2 to 30, might be rightly parted into two fit parts in the same manner as the first question of this Rule was resolved, or else by the Rule of False: yet to give further satisfaction, the cause of this is, the difference betwixt the two lesser expences, was there ½, which (before the division) was reduced to 1, which neither multiplies nor divides any Number, but leaves it the same: whereas, in this last, the middle expence (or exhibition) being 40, and the least 20, the difference of them was 20, by which dividing the Remaine of the last sub­traction: the quotient is ever the Number of the middle persons. Which may serve as an additi­on to the Rule, where the sorts of things are but three.

Question 4.

If there be 10 persons of four several Countries. English, French, Dutch, and Spanish, to pay a Debt of 1000 l. So that every English man payes 50 l. every French man 70 l. every Dutch man 130 l. and every Spaniard 150 l. How many is thereof each?

The Dividend, (according to the former Rule) is 500.

Now to make the Divisor, take his summe that payes least (namely 50,) out of each of the other three 150, 130, and 70, and the Remaines will be 100, 80, and 20.

Adde the first and last for the Divisor, it is 120.

And the quotient will be 4 2/12, and the Spaniards cannot be more.

[Page 143]Secondly, adde the first and second together for the Divisor, it is 180, and the quotient is 2 14/18, and the Spaniards cannot be lesse.

I mean, they cannot be much more than 4, or lesse then 2: and therefore, seeing any one solu­tion will serve, let them be 3, and by that mul­tiply 100, and take the product out of 500, there remaines 200 for a second Dividend, which di­vided by (the second remain) 80, the quotient is 2½: therefore the Dutchmen are 2, which mul­tiplyed by 80 make 160, take that out of 200, there remaines 40 for a third Dividend: which divided by (the third remaine) 20, the quotient is 2 for the French-men also; and consequently the English must be 3, because all of them are 10: But the Spaniards may be also 4 or 2.

Example.
4 Spaniards, at 150 l. each600
1 Dutchman at 130 l.130
1 Frenchman, at 70 l.70
4 English, at 50 l. each200
10 In all1000

2 Spaniards at 150 l. each300
3 Dutch, at 130 l. each390
3 French, at 70 l. each210
2 English, at 50 l. each100
10 In all1000

[Page 144]The reason why the Spaniards and English, as also the Dutch and French are equal in number, is because their payments differ equally from 100, which is the mean summe with which 10 men should pay 1000 l. and making it so, this questi­on, and many other of this nature may be answe­red by the Rule of Alligation: thus,

If 160 give 10,

[figure]

what 50? Answer is 3 20/160, (that is in this case 3) for the Spaniards, and as many for the En­glish, because their respective differen­ces from 100, the one 50 more; the English 50 lesse, are equal.

And also, because the other two differences 30 and 30 are equal: the Number of the French is equal to the Number of the Dutch.

But both those numbers together are 4, because 3 Spaniards, and 3 English, taken out of 10, the remaine must be 4.

Wherefore the Number of the French is 2, and the Dutch also are 2.

[Page 145]Or thus,

Accounting the men in the same order as before.

[figure]

If 160 require 10, what 30?

Answer is 1 160/160, (that is, in this case 2) for the Spani­ards; and conse­quently 2 English, and therefore the French and Dutch each 3.

But where any one of the particular summes, is equal to the mean summe, there this cannot so well be done by Alligation.

Example.

If one should buy 12 loaves of bread for 12 pence, so that some might be two peny; some peny, some half-peny; and some farthing loaves: and it be required to know how many he must buy of each; then because of 12 loaves for 12 pence, the mean price is 1, but one of the particulars being also 1, there should be no peny loaves, because there is no difference betwixt the mean price, and a peny.

[Page 146]But it may be found by the Rule of Ceres and Virginum, to be either.

4 two peny loaves8 pence
2 peny loaves2 pence
2 half-peny loaves1 peny
4 farthing loaves1 peny
In all 12 loaves.In all 12 pence.

Or else,
3 two peny loaves6 d.
4 peny loaves4
3 half-peny loaves1 ½
2 farthing loaves0 ½
In all 12In all 12

Extraction of Roots.

ANd first for the square Root, that is, a square Number being given, to find the root or side of it in a Number: which root or side, being multiplyed into it self, must therefore pro­duce that square Number.

The doing of this is shewed in (almost) all books of Arithmetick; and the reason of it (in some) which is taken from the fourth proposition of the second book of Euclide; which saith,

If a right line be divided by chance; the squares made of the parts, together with the Rectangle made of the parts twice, is equal to the square of the whole.

Example.

Let the line AB,

[figure]

be divided by chance in the point E, it is ma­nifest that the square of AB, that is to say the square AB DC, is e­qual to the square of LO, that [Page 148]is, of A E, and to the square of E B, and to the two rectangle figures A O and D O, that is, the rectangle A O (which is made of the parts AE, and B E) twice according to the Proposition.

Now let AB, be supposed 20, AE 15, and BE 5.

Then the square of AB, (which is made by multiplying the root 20 into it selfe) is equal to 400.

And the square AE, that is, 15 times 15 is e­qual to225
And the square of BE is 5 times 525
And the rectangle AO, 5 times 1575
And the rectangle DO, 5 times 1575
In all400

which is equal to the square of AB, as before.

If therefore a square Number, as 334084 be given to have the root extracted: first, make a point over 4, (the place of Unity) and another o­ver o, the second figure from it; and so another over o, the second figure from o, towards the left hand, observing the like order still, if there were more secondary figures, point them all as in the Margine;

Then look at the figures under the first point towards the left hand which are 33, [...] and take the greatest square root of them, which is 5, (for 6 times 6 is 36) and say 5 times 5 is 25, which take out of [Page 149]of 33, there remaines [...]

Then multiply the root 5 by 20, it gives 100 for a Divisor; and the Di­vidend is 840, namely, the figures which reach to the next point, and so the quotient might be 8, but must be but 7, because the square of the quotient, being now 49, must (toge­ther with 7 times the Divisor) be ta­ken out of 840.

And the remainer will be [...]

Adde this last quotient 7, being put to the former quotient 5, after the manner used in plain division, it will be 57, which multiply still by 20, the product is 1140 for a new Divisor, and the Dividend (because there re­maines but one point) is all the figures 9184.

And the quotient can be but 8, and must be so much; for 8 times 1140 is9120
and 8 times 864
In all9184

Which taken out of the remaine 9184, there now remaines nothing: which shews the square 334084 is justly resolved, and putting the last quotient 8 to the two former, 5 and 7, the whole will be 578, which is the true root required.

[Page 150]And the parts of it, if they be 3, are 500, 70, and 8. Or if but two 570 and 8. And the truth may be either way proved by adding the squares and twice the rectangles of the parts; for the sum of them shall be equal to the whole square 334084, as hath been shewed before.

It is ever certaine that there shall be as many figures, or places of figures in the root, as there are points in the square, ordered as before.

And by reading this little seriously, may any one be able to find the root of any other square whatsoever: the first operation in all being to take the greatest square contained under that point next the left hand, and put the root thereof for the first figure in a quotient.

The Secondary operation, must be repeated so often as there are remaining points; as hath been plainly shewed in the fore-going Example.

I will therefore adde no more, but give you two other squares, and their roots, leaving the Reader to extract them himself.

So, if there were given the square 88̇30̇

The root of it according to the former practice may be found to be 94.

Or let there be given the square 1̇52̇27̇56̇

The root of that will be 1234.

Here followeth a Table of Roots and their Squares from 1, to 1000.

[Page 151]

1 Cent.
R.Square
11
24
39
416
525
636
749
864
981
10100
11121
12144
13169
14196
15225
16256
17289
18324
19361
20400
21441
22484
23529
24576
25625
26676
27729
28784
29841
30900
31961
321024
331089
341156
351225
361296
371369
381444
391521
401600
411681
421764
431849
441936
452025
462116
472209
482304
492401
502500
512601
522704
532809
542916
553025
563136
573249
583 [...]64
593481
603600
613721
623844
633969
644096
654225
664356
674489
684624
694761
704900
715041
725184
735329
745476
755625
765776
775929
786084
796241
806400
816561
826724
836889
847056
857225
867396
877569
887744
897921
908100
918281
928464
938649
948836
959025
969216
979409
989604
999801
10010000
10110201
10210404
10310609
10410816
10511025
10611236
10711449
10811664
10911881
11012100
11112321
11212544
11312769
11412996
11513225
11613456
11713689
11813924
11914161
12014400
12114641
12214884
12315129
12415376
12515625
12615876
12716129
12816384
12916641
13016900
13117161
132174 [...]4
13317689
13417956
13518225
13618496
13718769
13819044
13919321
14019600
14119881
14220164
14320449
14420736
14521025
14621316
14721609
14821904
14922201
15022500
15122801
15223104
15323409
15423716
15524025
15624336
15724649
15824964
15925281
16025600
16125921
16226 [...]44
16326569
16426896
16527225
16627556
16727889
16828224
16928561
17028900
17129241
17229584
17329929
17430276
17530625
17630976
17731329
17831684
17932041
18032400

[Page 153]

2 Cent.
R.Square
18132761
18233124
18333489
18433856
18534225
18634596
18734969
18835344
18935721
19036100
19136481
19236864
19337249
19437636
19538025
19638416
19738809
19839204
19939601
20040000
20140401
20240804
20341209
20441616
20542025
20642436
20742849
20843264
20943681
21044100
21144521
21244944
21345369
21445796
21546225
21646656
21747089
21847524
21947961
22048400
22148841
22249284
22349729
22450176
22550625
22651076
22751529
22851984
22952441
23052900
23153361
23253824
23354289
23454756
23555225
23655696
23756169
23856644
23957121
24057600
24158081
24258564
24359049
24459536
24560025
24660516
24761009
24861504
24962001
25062500
25163001
25263504
25364009
25464516
2556 [...]025
25665536
25766049
25866564
25967081
26067600
26168121
26268644
26369169
26469696
26570225
26670756
26771289
26871824
26972361
27072900
27173441
27273984
27374529
27475076
27575625
27676176
27776729
27877284
27977841
28078400
28178961
28279524
28380089
28480656
28581225
28681796
28782369
28882944
28983521
29084100
29184681
29285264
29385849
29486436
29587025
29687616
29788209
29888804
29989401
30090000
3 Cent.
R.Square
30190601
30291204
30391809
30492416
30593025
30693636
30794249
30894864
30995481
31096100
31196721
31297344
31397969
31498596
31599225
31699856
317100489
318101124
319101761
320102400
321103041
322103684
323104329
324104976
325105625
326106276
327106929
328107584
329108241
330108900
331109561
332110224
333110889
334111556
335112225
336112896
337113569
338114244
339114921
340115600
341116281
342116964
343117649
344118336
345119025
346119716
347120409
348121104
349121801
350122500
351123201
352123904
353124609
354125316
355126025
356126736
357127449
358128164
359128881
360129600
361130321
362131044
363131769
364132496
365133225
366133956
367134689
368135424
369136161
370136900
371137641
372138384
373139129
374139876
375140625
376141376
377142129
378142884
379143641
380144400
381145161
382145924
383146689
384147456
385148225
386148996
387149769
388150544
389151321
390152100
4 Cent.
R.Square
391152881
392153664
393154449
394155236
395156025
396156816
397157609
398158404
399159201
400160000
401160801
402161604
403162409
404163216
405164025
406164836
407165649
408166464
409167281
410168100
411168921
412169744
413170569
414171396
415172225
416173056
417173889
418174724
419175561
420176400
421177241
422178084
423178929
424179776
425180625
426181476
427182329
428183184
429184041
430184900
431185761
432186624
433187489
434188356
435189225
436190096
437190969
438191844
439192721
440193600
441194481
442195364
443196249
444197136
445198025
446198916
447199809
448200704
449201601
450202500
451203401
452204304
453205209
454206116
455207025
456207936
457208849
458209764
459210681
460211600
461212521
462213444
463214369
464215296
465216225
466217156
467218089
468219024
469219961
470220900
471221841
472222784
473223729
474224676
475225625
476226576
477227529
478228484
479229441
480230400
5 Cent.
R.Square
481231361
482232324
483233289
484234256
485235225
486236196
487237169
488238144
489239121
490240100
491241081
492242064
493243049
494244036
495245025
496246016
497247009
498248004
499249001
500250000
501251001
502252004
503253009
504254016
505255025
506256036
507257049
508258064
509259081
510260100
6 Cent.
R.Square
511261121
512262144
513263169
514264196
515265225
516266256
517267289
518268324
519266361
520270400
521271441
522272484
523273529
524274576
525275625
526276676
527277729
528278784
529279841
530280900
531281961
532283024
533284089
534285156
535286225
536287296
537288369
538289444
539290521
540291600

[Page 157]

5 Cent.
R.Square
541292681
542293764
543294849
544295936
545297025
546298116
547299209
548300304
549301401
550302500
551303601
552304704
553305809
554306916
555308025
556309136
557310249
558311364
559312481
560313600
561314721
562315844
563316969
564318096
565319225
566320356
567321489
568322624
569323761
570324900
571326041
572327184
573328329
574329476
575330625
576331776
577332929
578334084
579335241
580336400
581337561
582338724
583339889
584341056
585342225
586343396
587344569
588345744
589346921
590348100
591349281
592350464
593351649
594352836
595354025
596355216
597356409
598357604
599358801
600360000
6 Cent.
R.Square
601361201
602362404
603363609
604364816
605366025
606367236
607368449
608369964
609370881
610372100
611373321
612374544
613375769
614376996
615378225
616379456
617380689
618381924
619383161
620384400
621385641
622386884
623388129
624389376
625390625
626391876
627393129
628394384
629395641
630396900

[Page 158]

7 Cent.
R.Square
631398161
632399424
633400689
634401956
635403225
636404496
637405769
638407044
639408321
640409600
641410881
642412164
643413449
644414736
645416025
646417316
647418609
648419904
649421201
650422500
651423801
652425104
653426409
654427716
655429025
656430336
657431649
658432964
659434281
660435600
661436921
662438244
663439569
664440896
665442225
666443556
667444889
668446224
669447561
670448900
671450241
672451584
673452929
674454276
675455625
676456976
677458329
678459684
679461041
680462400
681463761
682465124
683466489
684467856
685469225
686470596
687471969
688473344
689474721
690476100
8 Cent.
R.Square
691477481
692478864
693480249
694881636
695483025
696484416
697485809
698487204
699488601
700490000
701491401
702492804
703494209
704495616
705497025
706498436
707499849
708501264
709502681
710504100
711505521
712506944
713508369
714509796
715511225
716512656
717514089
718515524
719516961
720518400

[Page 159]

7 Cent.
R.Square
721519841
722521284
723522729
724524176
725525625
726527076
727528529
728529984
729531441
730532900
731534361
732535824
733537289
734538756
735540225
736541696
737543169
738544644
739546121
740547600
741549081
742550564
743552049
744553536
745555025
746556516
747558009
748559504
749561001
750562500
751564001
752565504
753567009
754568516
755570025
756571536
757573049
758574564
759576081
760577600
761579121
762580644
763582169
764583696
765585225
766586756
767588289
768589824
769591361
770592900
771594441
772595984
773597529
774599076
775600625
776602176
777603729
778605284
779606841
780608400
8 Cent.
R.Square
781609961
782611524
783613089
784614656
785619225
786617796
787619369
788620944
789622521
790624100
791625681
792627264
793628849
794630436
795632025
796633616
797635209
798636804
799638401
800640000
801641601
802643204
803644809
804646416
805648025
806649636
807651249
808652864
809654481
810656100
811657721
812659344
813660969
814662596
815664225
816665856
817667489
818669124
819670761
820672400
821674041
822675684
823677329
824678976
825680625
826682276
827683929
828685584
829687241
830688900
831690561
832692224
833693889
834695556
835697225
836698896
837700569
838702244
839703921
840705600
841707281
842708964
843710649
844712336
845714025
846715716
847717409
848719104
849720801
850722500
851724201
852725904
853727609
854729316
855731025
856732736
857734449
858736164
859737881
860739600
861741321
862743044
863744769
864746496
865748225
866749956
867751689
868753424
869755161
870756900
871758641
872760384
873762129
874763876
875765625
876767376
877769129
878770884
879772641
880774400
881776161
882777924
883779689
884781456
885783225
886784996
887786769
888788544
889790321
890792100
891793881
892795664
893797449
894799236
895801025
896802816
897804609
898806404
899808201
900810000

[Page 161]

9 Cent.
R.Square
901811801
902813604
903815409
904817216
905819025
906820836
907822649
908824464
909826281
910828100
911829921
912831744
913833569
914835396
915837225
916839056
917840889
918842724
919844561
920846400
921848241
922850084
923851929
924853776
925855625
926857476
927859329
928861184
929863041
930864900
931866761
932868624
933870489
934872356
935874225
936876096
937877969
938879844
939881721
940883600
941885481
942887364
943889249
944891136
945893025
946894916
947896809
948898704
949900601
950902500
951904401
952906304
953908209
954910116
955912025
956913936
957915849
958917764
959919681
960921600
961923521
962925444
963927369
964929296
965931225
966933156
967935089
968937024
969938961
970940900
971942841
972944784
973946729
974948676
975950625
976952576
977954529
978956484
979958441
980960400
981962361
982964324
983966289
984968256
985970225
986972196
987974169
988976144
989978121
990980100
991982081
992984064
993986049
994988036
995990025
996992016
997994009
998996004
999998001
10001000000

Extraction of the Cube Root.

ACube is a solid, or Body contained with­in six equal squares; and may be fitly re­presented by a Die.

When a Cube is given, as is 435̇519̇512̇; point the Number as you see in the place of U­nity, and every third figure after; then see what is the root of the greatest Cube contained under the first point toward the left hand; that is, in 435, it will be found 7, (for 8 times 8, taken 8 times is 5 12, which is too much) put this 7 for the first figure in the quotient, having taken the Cube thereof, which is 343, out of 435: thus, [...]

[Page 163]And so the first work is done.

For the second, take the square of the quotient, that is 49, which multiply by 300, the product is14700
to which adde 30 times 7, that is∷∷210
which makes the Divisor14910
And the Dividend is92519

So the quotient might be 6, but must be but 5, because the Cube of the new quotient, and 210 times the square of the said quotient, must be allowed in this worke, as followeth.

The remaine is92519̇512̇
multiply the first product 14700 by78875
The second quotient 5, and they produce73500
then multiply 210 by the square of 5, it makes:5250
to which adde the cube of 5...::125
It makes in all78875
Which taken from 92519 there remaines13644512
and the quotient is 75, whose square 5625, multiplyed by 300 the product is1687500
secondly, 30 times 75 is::2250
And the new Divisor is1710000
And the third quotient figure is 8, by which multiply 1687500, it is13500000
Likewise, 2250, multiplyed by the square of 8, (64) is144000
And the Cube of 8 is:::512
Altogether are13644512

Which taken from the second remaine, there remaines now thirdly nothing. And therefore the last quotient 8, put to the former two, it is 75.8 for the whole Root, as may be tryed by multiplying 758 into it selfe; and the product a­gaine by 758, then the last product shall be equal to the whole Cube, which was given at first to be resolved.

R.Cube
11
28
327
464
5125
6216
7343
8512
9729
101000
111331
121728
132197
142744
153375
164096
174913
185832
196859
208000
219261
2210648
2312167
2413824
2515625
2617576
2719683
2821972
2924389
3027000
3129791
3232768
3335937
3439304
3542825
3648656
3750653
3854872
3955419
4064000
4168921
4274088
4379507
4485184
4591125
4697336
47103823
48110592
49117649
50125000
51135651
52140608
53148877
54157464
55167375
56175616
57185193
58195112
59205379
60216000
61216981
62238328
63293047
64262244
65274625
66287496
67300753
68314432
69329199
70333000
71357911
72373348
73389017
74405224
75411875
76438976
77456533
78474522
79493039
80512000
81531441
82550408
83571787
84592604
85614125
86636056
87648303
88681472
89705669
90729000
91753571
92778688
93804357
94830584
95857375
96884736
97915673
98941192
99970299
1001000000
1011030301
1021061208
1031092727
1041124856
1051157625
1061191016
1071225043
1081259712
1091295029
1101331000
1111367631
1121404928
1131442897
1141481544
1151520875
1161560896
1171601613
1181643032
1191685159
1201728000
1211771561
1221815848
1231860867
1241906624
1251953125
1262000376
1272048383
1282097172
1292146689
1302197000
1312248091
1322299968
1332352637
1342406104
1352460375
1362515856
1372570353
1382628072
1392685619
1402744000
1412803221
1422864288
1432924207
1442985984
1453027525
1463112136
1473176523
1483241792
1493307949
1503375000
1513442951
1523511808
1533581577
1543652264
1553723875
1563796416
1573869893
1583944312
1594019679
1604096000
1614173281
1624251528
1634230744
1644410944
1654492125
1664574296
1674657463
1684741632
1694826809
1704913000
1715000211
1725088448
1735177717
1745268024
1755359375
1765451776
1775545233
1785639752
1795735339
1805832000

[Page 167]

2 Cent.
R.Cube
1815929741
1826028568
1836128487
1846229504
1856331625
1866434856
1876539203
1886644672
1896751269
1906859000
1916967871
1927077888
1937189057
1947301384
1957414875
1967529536
1977645373
1987762392
1997880599
2008000000
2018120601
2028242408
2038365427
2048489664
2058615125
2068741816
2078869743
208899 [...]912
2099129329
2109261000
2119393931
2129528128
2139663597
2149800344
2159938375
21610077696
21710218313
21810360232
21910503459
22010648000
22110793861
22210941048
22311089567
224113 [...]9424
22511390625
22611543176
22711697083
22811852452
22912008989
23012167000
23112326391
23212487168
23312649337
23412812904
23512977875
23613144256
23713312053
23813481272
23913651919
24013824000
24113997521
24214172448
24314 [...]48907
24414526784
24514705125
24614886936
24715069223
24815252992
24915438249
25015625000
25115813251
25216003008
25316194277
25416387064
25516581375
25616777216
25716974593
25817173512
25917473979
260175760 [...]0
26117779581
26217984728
26318191447
26418399744
26518609625
26618821096
26719034163
26819248832
26919465109
27019683000
27119902511
27220123648
27320346417
27420571024
27520796875
27621024576
27721253933
27821484952
27921717639
28021952000
28122188041
28222425768
28322665187
28422906304
28523149125
28623393656
28723539903
28823897872
28924137569
29024389000
29124642171
29224897088
29325153757
29425412184
29525672375
29625934336
29726198073
29826353592
29926730899
30027000000
3 Cent.
R.Cube
30127270901
30227543608
30327818127
30428094464
30528372625
30628652616
30728934443
30829218112
30929503629
31029791000
31130080231
31230371328
31330664 [...]97
31430959144
31531255875
31631554496
31731855013
31832157432
31932461759
32032768000
32133076161
32233386248
32333698267
32434012224
32534328125
32634645976
32734965783
32835287552
32935611289
33035937000
33136264691
33236594368
33336926037
33437259704
33537595375
33637933076
33738272753
33838614472
33938958219
34039304000
34139651821
34240001688
34340353607
34440707584
34541063625
34641421736
34741781923
34842144192
34942508549
35042875000
35143243551
35243614208
35343986977
35444361864
35544738875
35645118016
35745499293
35845882712
35946268279
36046656000
36147045881
36247439928
36347832147
36448228544
36548627125
36649027896
36749430863
36849836032
36950243409
37050653000
37151064811
37251478848
37351895117
37452313624
37552734375
37653157376
37753582633
37854010152
37954439939
38054872000
38155306341
38255742968
38356181887
38456623104
38557066625
38657512456
38757960603
38858411072
38958863869
39059319000
4 Cent.
R.Cube
39159776471
39260236288
39360698457
39461162984
39561629875
39662099136
39762570773
39863044792
39963521199
40064000000
40164481201
40264964808
40365450827
40465939264
40566430125
40666923416
40767419143
40867917312
40968417929
41068921000
41169426531
41269934528
41370444997
41470957944
41571473375
41671991296
41772511713
41873034632
41973560059
42074088000
42174618461
42275151448
42375686967
42476225024
42576765625
42677308776
42777854483
42878402752
42978953589
43079507000
43180062991
43280621568
43381182737
43481746504
43582312875
43682881856
43783453453
43884027672
43984604519
44085184000
44185766121
44286350888
44386938307
44487528384
44588121125
44688716536
44789314623
44889915392
44990518849
45091125000
45191733851
45292345408
45392959677
45493576664
45594196375
45694818816
45795443993
45896071912
45996702579
46097336000
46197972181
46298611128
46399252847
46499897344
465100544625
466101194696
467101847563
468102503232
469103161709
470103823000
471104487111
472105154048
473105823817
474106496424
475107171875
476107850176
477108531333
478109215352
479109902239
480110592000
5 Cent.
R.Cube
481111284641
482111980168
483112678587
484113379904
485114084125
486114791256
487115501303
488116214272
489116930169
490117649000
491118370771
492119095488
493119823157
494120553784
495121287375
496122023936
497122763473
498123505992
499124251499
500125000000
501125751501
502126506008
503127263527
504128024064
505128787625
506129554216
507130323843
508131096512
509131872229
510132651000
511133432831
512134217728
513135005697
514135796744
515136590875
516137388096
517138188413
518138991832
519139798359
520140608000
521141420761
522142246648
523143055667
524143877824
525144703125
526145531576
527146363183
528147197952
529148035889
530148877000
531149721291
532150568768
533151419437
534152273304
535153130375
536153990656
537154854153
538155720872
539156590819
540157464000
541158340421
542159220088
543160103007
544160989184
545161878625
546162771336
547163667323
548164566592
549165469149
550166375000
551167284151
552168196608
553169112377
554170031464
555170953875
556171879616
557172808683
558173741112
559174676879
560175616000
561176558481
562177504328
563178453547
564179306144
565188262125
566181221496
567182154263
568183150432
569184220009
570185193000
571186169411
572187149248
573188132517
574189119224
575190109375
576191102976
577192100033
578193100552
579194104539
580195112000
581196122941
582197137368
583198155287
584199176704
585200202625
586201230056
587202262003
588203297472
589204336469
590205379000
591206425071
592207474688
593208527857
594209584584
595210644875
596211708736
597212776173
598113847192
599214921799
600216000000
6 Cent.
R.Cube
601217081801
602218167208
603219256227
604220348864
605221445125
606222545016
607223648543
608224755712
609225866529
610226981000
611228099131
612229220928
613230346397
614231475544
615232608375
616233744896
617234885113
618236029032
619237176659
620238328000
621239483061
622240641848
623241804367
624242970624
625244140625
626245314376
627246491883
628247673152
629248858189
630250047000
631251239591
632252435968
633253636137
634254840104
635256047875
636257259456
637258474853
638259694072
639260917119
640262144000
641263374721
642264609288
643265847707
644267089984
645268336125
646269586136
647270840023
648272097792
649273359449
650274625000
651275894451
652277167808
653278445077
654279726264
655281011375
656282300416
657283593393
658284890312
659286191179
660287496000
661288804781
662290117528
663291434247
664292754944
665294079625
666295408296
667296740963
668298077632
669299418309
670300763000
671302111711
672303464448
673304821217
674306182024
675307546875
676308915776
677310288733
678311665752
679313046839
680314432000
681315821241
682317214568
683318611987
684320013504
685321419125
686322828856
687324242703
688325660672
689327082769
690328509000
7 Cent.
R.Cube
691329939371
692331373888
693332812557
694334255384
695335702375
6963371535 [...]6
697338608873
698340068392
699341532099
700343000000
701344472101
702345948408
703347428927
704348913664
705350402625
706351895816
707353393243
708354894912
709356400829
710357911000
711359425431
712360944128
713362467097
714363994344
715365525875
716367061696
717368601813
718370146232
719371694959
720373248000
721374805361
722376367048
723377933067
724379503424
725381078125
726382657176
727384240583
728385828352
729387420489
730389017000
731390617891
732392223168
733393832837
734395446904
735397065375
736398688256
737400 [...]15553
738401947272
739403583419
740405224000
741406869021
742408518488
743410172407
744411830784
745413493625
746415160936
747416832723
748418508992
749420189749
750421875000
751423564751
752425259008
753426957777
754428661064
755430368875
756432081216
757433798093
758435519512
759437245479
760438976000
761440711081
762442450728
763444194947
764445943744
765447697125
766449455096
767451217663
768452984832
769454756609
770456533000
771458314011
772460099648
773461889917
774463684824
775465484375
776467288576
777469097433
778470910952
779472729139
780474552000
8 Cent.
R.Cube
781476379541
782478211768
783480048687
784481890304
785483736625
786485587656
787487443403
788489303872
789491169069
790493039000
791494913671
792496793088
793498677257
794500566184
795502459875
796504358336
797506261573
798508169592
799510082399
800512000000
801513922401
802515849608
803517781627
804519718464
805521660125
806523606616
807525557943
808527514112
809529475129
810531441000
811533411731
812535387328
813537367797
814539353144
815541343375
816543338496
817545338513
818547343432
819549353259
820551368000
821553387661
822555412248
823557441767
824559476224
825561515625
826563559976
827565609283
828567663552
829569722789
830571787000
831573856191
832575930368
833578009537
834580093704
835582182875
836584277056
837586376253
838588480472
839590589719
840592704000
841594823321
842596947688
843599077107
844601211584
845603351125
846605495736
847607645423
848609800192
849611960049
850614125000
851616295051
852618470208
853620650477
854622835864
855625026375
856627222016
857629422793
858631628712
859633839779
860636056000
861638277381
862640503928
863642735647
864644972544
865647214625
866649461896
867651714363
868653972032
869656234909
870658503000
871660776311
872663054848
873665338617
874667627624
875669921875
876672221376
877674526133
878676836152
879679151435
880681472000
881683797841
882686128968
883688465387
884690807104
885693154125
886695506456
887697864103
888700227072
889702595369
890704969000
891707347971
892709732288
893712121957
894714516984
895716917375
896719323136
897721734273
898724150792
899726572699
900729000000

[Page 175]

9 Cent.
R.Cube
901731432701
902733870808
903736314327
904738763264
905741217625
906743677416
907746142643
908748613312
909751089429
910753571000
911756058031
912758550528
913761048497
914763551944
915766060875
916768575296
917771095213
918773620632
919776151559
920778688000
921781229961
922783777448
923786330467
924788889024
925791453125
926794022776
927796597983
928799178752
929801765089
930804357000
931806954491
932800557569
933812166237
934814780504
935817400375
936820025856
937822656953
938825293672
939827936019
940830584000
941833237621
942835896888
943838561807
944841232384
945843908625
946846590536
947849278123
948851971392
949854670349
950857375000
951860085351
952862801408
953865523177
954868250664
955870983875
956873722816
957876467493
958879217912
959881974079
960384736000
961887503681
962890277128
963893056347
964895841344
965898632125
966901428696
967904231063
968907039232
969909853209
970912673000
971915498611
972918330048
973921167317
974924010424
975926859375
976929714176
977932574833
978935441352
979938313739
980941192000
981944076141
982946966168
983949862087
984952763904
985955671625
986958585256
987961504803
988964430272
989967361669
990970299000
991973242271
992976191488
993979146657
994982107784
995985074875
996988047936
997991026973
998994011992
999997002999
10001000000000

Some uses of the Square and Cube Root.

Uses of the Square Root.

WHat the Square and Cube Roots are, and how to extract them, hath alrea­dy been taght, and for more ease and expedition, there are Tables ready cal­culated, both of the Square and Cube Roots, from 1 to 1000, we come now to shew some uses thereof, which in some measure will appear in the Propositions following.

PROPOSITION. I.

Admit the height of the Wall of a Fort or Castle to be scaled, be 30 foot, and the breadth of the Trench about the Fort be 40 foot, I demand of what length a Scaling Ladder should be, justly to reach from the edge of the Trench, to the top of the Wall.

By the 47th. of the first Book of Euclides Ele­ments, it is demonstrated that the square of the Hypotenusal of all right angled plain Triangles is equal to the squares of the 2 other sides; I there­fore to resolve this Proposition, square the height of the Wall which is 30, facit 900, also I square the breadth of the Trench which is 40, facit 1600, these two added together make 2500, the square root whereof is 50, and so long must Scaling Ladders be made to reach from the edge of the Trench to the top of the Wall.

PROPOSITION II.

There be two Towns, as Chichester and York, which lie North and South one from another, and their di­stance is 220 miles, and Excester lieth directly West from Chichester 120 miles; I desire to know the distance of York from Excester.

Square 120,

Excester

120

Chichester

220

York

the distance of Excester and Chichester, it maketh 14400, likewise square 220, the distance of York and Chi­chester, facit 48400, these two numbers added together make 62800, whose square root extracted (or found in the Tables) will be 250 3/5; neer, and so many miles is Excester distant from York.

Ʋse of the Cube Root.

ONe chief use of the Cube Root, is to find out a proportion between, like Solids, such are Spheres, Cubes, and such like; as in the Proposi­on following.

PROPOSITION.

If a Bullet of Brasse of 4 inches Diameter, weigh 9 pound, what shall a Bullet of Brasse weigh, whose Diameter is 8 inches?

Cube 4 the Diameter of the lesser Bullet, makes 64, like­wise Cube the Diameter of the greater Bullet 8, makes 4608. This done, say by the Rule of proportion; If the Cube num­ber 6 [...], give 9 li. weight, what shall the Cube number 4608 give? Multiply and divide, you shall find 72, and so many pounds will a Bullet of Brasse weigh, whose Diameter is 8 inches.

DECIMAL ARITHMETICK. The Second Part. With the ground and reason thereof, illustrated by divers Examples, in all the most usual Rules of Arithmetick.
LONDON Printed Anno Domini 1659.
[Page] [Page 181] The Second Part, Containing DECIMAL ARITHMETICK.

HAving in the first Part of this Book exemplified the Art of Vulgar Arithmetick both in whole Numbers and Fracti­ons; We come now to treat of Decimal Arithmetick, which teacheth how to per­form (as it were) in whole Numbers all that the former did effect by Fractions. I will not insist upon the antiquity or excellency of this kind of Arithmetick, but come immediately to the pra­ctice thereof, and shall therefore premise these Propositions following.

Proposition 1.

A vulgar Fraction being given, how to reduce the same into a Decimal.

[Page 182] THE RULE.

To the Numerator of the Fraction given, adde what number of Cyphers you please; then divide the Nume­rator by the Denominator, the quotient shall be the Decimal fraction required.

Example 1.

Let it be reuired to reduce 4/17 into a Decimal: first, to the Numerator 4, adde five Cyphers, so will it be 400000, divide this number by the De­nominator 17, and the quotient will be .23529, which is the Decimal required.

¶ And here note, that Decimal fractions are not written in a smaller figure with a line be­tween them, as vulgar fractions are, but of the same figure, only there must be a Comma put between the whole Number and the Fraction, and that is the distinction.

Example 2. If you would expresse 235 4/17 in a decimal way, it must be written as followeth.

By the last example you finde that 4/17 reduced to a decimal was .23529, therefore 235 4/17 must be written thus: 235,23529. In decimal fracti­ons, the Numerator is only expressed, and the Denominator onely intimated; for this is gene­ral, Of how many figures soever the Numerator of a decimal fraction doth consist, of so many Cyphers with a Unite before them, doth the Denominator of the same fraction consist. So this Decimal 12,625, if it [Page 183]were writtenin a vulgar way, would be 12. 625/1000, but in a decimal, only 12,625, the Comma between 12 and 625 distinguisheth the whole number from the fraction, and the fraction 625 consist­ing of three figures, intimates that the Denomi­nator thereof must consist of three Cyphers and an unite before them; So the decimal before ex­pressed, 235,23529, if it were written in the vul­gar way, would be 235 23529/100000.

But it sufficeth to expresse in Decimals the Nu­merators only, and omit the Denominators, the Denominators of all Decimal fractions being ei­ther 10, 100, 1000, 10000, 100000, &c. according to the number of figures contained in the Nume­rators.

According to this Rule, you shall find that

  • ⅘ will be in deicimals by adding 5 Cyphers. .80000
  • 12 3/7 will be in deicimals by adding 5 Cyphers. 12.42857
  • 132 9/17 will be in deicimals by adding 5 Cyphers. 132.52941

And by this meanes, all manner of fractions of Coyns, Weights, and Measures, may be redu­ced from vulgar fractions, to decimal fractions; as by the next Proposition will appear.

Proposition 2. How to expresse English Coin, in Decimal Numbers.

Let it be required to expresse 9 shillines (which is 9/20 of a pound Sterling) in a decimal; To the Numerator 9 adde Cyphers, making it 900, which divide by 20, the quotient is .45, for the decimal of 9 s. So the decimal of 13 s. will be 65 [...], and so for any number of shillings.

[Page 184]¶ Here note, that in the Reduction of Vulgar fra­ctions into Decimals, that many times the first, second or third places of the Decimal fracti­ons are Cyphers, as in the following Table the Decimal of one farthing is 00104167, and the reason is, because if you reduce 1/960 into a De­cimal (for one farthing is the 960 part of a pound Sterling) you shall by adding of Cyphers to the Numerator) find the Quotient to be 104167, but two Cyphers must be placed before it; because dividing 1000000 by 960, the place of unites in the Divisor at the first demand ex­tendeth unto the third Cypher in the Dividend, for in reducing of Vulgar fractions to Deci­mals, this is, A general Rule.

That if the place of unites in the Divisor, at the first demand extend but unto the first of the Cyphers annexed to the Numerator of the fraction, there must be no Cypher put before the Quotient, but if the place of unites extend unto the second Cypher added, then one Cypher must be placed before the Quotient, if un­to the third Cypher, then two Cyphers must be placed before the Quotient, &c.

According to which Rule, if you make trial you shall find that the Decimal of 7 s. will be .35, the Decimal of 5 d. will be .02083333, the Deci­mal of two farthings will be .00208333, as in the Table.

By these rules last delivered are the ensuing Tables of English Money, Weight, and Measure composed, and the like may be done for any for­reign Coin, &c. according as every mans occa­sions shall require,

The TABLE of English Coin in Decimals.
English Coin.
Sh. 1995
189
1785
168
1575
147
1365
126
1155
105
945
84
735
63
525
42
315
21
105
D. 1104583333
1004166667
90375
803333333
702916667
6025
502083333
401666667
30125
200833333
104166667
F. 3003125
200208333
100104167
Troy Weight in Decimals.
O. 1191666667
1083333333
975
866666667
758333333
65
541666667
433333333
325
216666667
108333333
P. 1907916667
18075
1707083333
1606666667
150625
1405833333
1305416667
1205
1104583333
1004166667
90375
803333333
702916667
6025
502083333
401666667
30125
200833333
100416667
Gr. 2300399395
2200381944
2100364583
2000347222
1900329861
18003125
1700295139
1600277778
1500260417
1400243056
1300225694
1200208333
1100190972
1000173611
90015625
800138889
700121528
600104166
500086805
400069444
300052083
200044722
100017361
Averdupois great weight in Decimals.
3 qu.75
2 qu.5
1 qu.25
lib. 2724107142
2623214285
2522321428
2421428571
2320535714
2219642857
211875
2017857143
1916964286
1816071428
1715178571
1614285714
1513392857
14125
1311607143
1210714286
1109821428
1008928571
908035714
807142857
70625
605357143
504464286
403571428
302678571
201785714
100892857
Oun. 1500837053
140078125
1300725446
1200669643
1100613839
1000558035
900502232
800446429
700390625
600334821
500279018
400223214
300167411
200111607
100055804
3 qu.00041853
half00027902
1 qu.00013951
Averdupois little weight in Decimals.
Ounces 15937
14875
138125
1275
116875
10625
95625
85
74375
6375
53125
425
31875
2125
10625
Drams. 1505859375
140546875
1305078115
12046875
1104296875
100390625
903515625
803125
702734375
60234375
501953125
4015625
301171875
20078125
100390625
3 qu.00292969
half00195312
1 qu.00097656
Liquid Measures in Decimals.
P. 7875
675
5625
45
3375
225
1125
3 qu.09375
halfe.0625
1 qu.03125
Dry Measures in Decimals.
Bushels. 7875
675
5625
45
3375
225
1125
Pecks. 309375
20625
103125
3 qu.0234375
halfe.015625
1 qu.0078125
Pintt. 30058594
20039063
10019531
Long Measures, the-In­tegers being yards and ells in 'Decimals.
qu. 375
25
125
Nale 31875
2125
10625
3 qu.046875
halfe.03125
1 qu.015625
Time in Decimals.
Mo. 11916667
10833333
975
8666667
7833333
65
5416667
4333333
325
2166667
1083333
Da. 30082193
29097454
28076714
27073973
26071233
25068495
24065755
23063016
22060274
21057536
20054795
19052055
18049316
17046577
16043837
15041097
14038357
13035617
12032877
11030137
10027397
9024657
8021918
7019178
6016438
5013698
4010959
30082192
20054795
10027397
Dozens in Decimals.
De. 119166667
108333333
975
86666667
75833333
65
54166667
43333333
325
21666667
10833333
Pa. 11076388
100694444
90625
80555555
70486111
60416667
50347222
40277778
30208333
20138889
10069444

The use of the fore-going Tables.

THe Tables preceding are in number nine; The first being of English Coyn, the second of Troy weight, the third of Averdupois great weight, the fourth of Averdupois little weight, the fifth of Liquid Measures, the sixth of Dry Measures, the seventh of Long Measures, the eighth of Time, and the ninth of Dozens: These several Tables are made by the Rules immediately going before them, and their use is to expresse in Decimal numbers either Money, Weight, or Measure, as by the following Propositions will appear.

PROP. I. How to expresse English Coin in Deci­mals.

The first of the nine Tables is for this pur­pose, therefore if you would expresse either shillings, pence, or fartbings, in Decimal numbers, you must repair to the first Table, which is of English Coyn, and there against 13 shillings you shall finde 65, which is the Decimal of 13 shil­lings, also against seven pence you shall finde 02916667, which is the Decimal representing 7 pence: Also against 2 farthings you shall find [Page 192]00208333, which is the Decimal answering to 12 farthings, and the like is to be done for any other number of shillings, pence, or far­things.

But if it be required to find the Decimal of divers Denominations of Coin in one sum, as of shillings, pence, and farthings together, you must adde the Decimals of all the particulars to­gether, and the sum of them shall be the Deci­mal sought.

Example.

If you would know the Decimal of 13 s. 7 d. 2 q. in one number,

65
02916667
00208333
68125000

you must first look in the Table for the Deci­mal of 13 s. which is 65, and set that down, then look for the De­cimal of 7 d. which is 00916667, and set that down also: Lastly, seek the Decimal of 2 q. which is 00208 [...]33, set that down also; then if you adde these three numbers together, as in common Addition, you shall find the sum of them to be 68125000, which is the Decimal belonging to 13 s. 7 d. 2 q. as by tee work in the Margent appeareth.

PROP. II. How to express Troy weight in De­cimals.

The second Table is of Troy weight, the se­verall Denominations whereof are Ounces, Peny-weights, and Grains: So that by the Table you shall find that the Decimal belonging to five ounces is 41666667, the Decimal belonging to 17 peny-weight, is 07083333, and the Decimal belonging to 13 Grains is 00225694, and so of any other number of ounces, peny-weights, and grains severally.

But if it were required to expresse these (or any other) several Denominations in one Decimal Fraction, then you must (as before you did for money) take out of the Table the several Deci­mals belonging to the respective quantities, and adde them together, so shall the sum of that ad­dition be the Decimal sought.

Example.

If it were required to finde a Decimal which should represent 5 ounces, 17 peny-weight, 13 grains, you must first look in the Table for the Decimal belonging to five ounces, which is 41666667, and write it down, then look the Decimal belonging to 17 grains, which is [Page 194]07083 [...]33, and write that down,

41666667
07083333
00225694
48975694

then look for the Decimal of 13 grains, which is 00225694, and write that down, then adding these three num­bers together, you shall find the sum of them to be 48975694, which is the Decimal representing 5 ounces, 17 grains, 13 peny-weight, as by the operation in the margent appeareth.

PROP. III. How to express Averdupois great weight in Decimals.

This Averdupois great weight is the third Ta­ble, the several Denominations whereof are Quarters of Hundreds, Pounds, Ounces, and Quarters of Ounces, thus you shall finde in the Table, that the Decimal of 3 Quarters of a hun­dred is 75, the Decimal of 22 pounds is 19642857, the Decimal of 7 ounces is 00390625, and the Decimal of 3 quarters of an Ounce is 00041853, in this manner by the Table may you finde the correspondent Decimal belonging to any num­ber of quarters, pounds, ounces, and parts of ounces severally.

But if it be required to finde one decimal number which shall represent divers denomina­tions, you must first finde the decimal belong­ing to the several particulars, and add them to­gether, [Page 195]the sum whereof shall be the entire deci­mal required.

Example.

Let it be required to finde a decimal which shall represent 3 quarters,

75
19642857
00390623
00041853
95075335

22 pounds, 7 ounces ¾. First, look in the Table for the decimal of three quarters of a hundred, which is 75, and write it down, then look for the decimal of 22 pound which is 19642857, and write that down, also look the decimal be­longing to seven ounces, which is 00390625, and write that down: Lastly, seek the decimal of three quarters of an Ounce, which is 00041853, and write that down, then adding these four numbers together, you shal find their sum to be 95075335, which is the Decimal representing 3 qu. 22 lb. 7 oun. ¾.

PROP. IV. How to expresse Averdupois little weight in Decimals.

The fourth Table is of Averdupois little weight, the denominations whereof are ounces, drams, and quarters of drams, so that the Decimal of 11. ounces is 6875, the Decimal five drams is 01953125, and the decimal of one quarter of a dram is 00097656.

But if it be required to find one decimal num­ber, [Page 196]which shall represent 11 ounces,

6875
01953125
00097656
70800781

5¼ drams, then you must first look for the decimal belong­ing to 11 ounces, which is 6875, and set it down, then look for the Decimal answering to five drams, which is 01953125, and set that down. Lastly, look the decimal belonging to a quarter of a dram, and set that down; these three numbers being added together, pro­duce 70800781, which is the correspondent Deci­mal belonging to 11 ounces, five drams, and a quarter of a dram.

PROP. V. How to expresse Liquid Measures in Decimals.

Because there is so great variety of Liquid measures, that hardly any two commodities are sold by the same, the difference of the gallon con­tinually making alteration, we have therefore in this fifth Table made the greatest denomination to be one gallon, the next lesse denomination be­ing pints and quarters of pints, so that in the Ta­ble you shall finde the Decimal belonging three pints, to be 375, and the Decimal belonging to two quarters, or half a pint, to be 0625, and so for any other.

But for to expresse pints and parts of pints in one entire decimal number, you must adde the Decimals of the several denominations together, [Page 197]and their summe shall be the entire Decimal.

So if you were to expresse 3 pints and an half,

375
0625
4375

in one entire decimal number, adde the decimal of three pints, which is 375, to the decimal of two quarters, which is 0625, and their sum 4375, shal be the Decimal of three pints and an half.

PROP. VI. How to expresse dry measures in De­cimals.

The sixth Table is of dry measures,

625
0625
0234375
0039063
7148438

the seve­rall denominations whereof are Bushels, Pecks, quarters of Pecks, and Pints, so may you find the decimal of five bushels to be 625, the decimal of two pecks to be 0625, the decimal of three quarters of a peck to be 0234375, the decimal of two pints to be 0039063. Thus are the correspon­dent decimals belonging to the several denomi­nations found: But if you would have one num­ber to expresse five bushels, 2¾ pecks, two pints: you must first find the decimal belonging to five pecks, wich is 625, and write it down, then find the decimal of two pecks, which is 0625, then seek the decimal of three quarters of a peck, which is 0234375, & write that down. Lastly, seek the decimal of two pints, which is 0039063, which numbers be­ing added together, produce 7148438, which is the decimal belonging (or expressing) five bushels, 2¼ pecks, and two pints.

PROP. VII. How to expresse long measures in De­cimals.

The seventh Table is of long measures, the in­tegers being Yards and Ells: and the lesser deno­minations are quarters of Yards or Ells, Nales, and quarters of Nales. So may you find in the Table that the decimal of three quarters of a Yard, or an Ell is 75, the decimal of two Nales, is 125, and the decimal of one quarter of a Nale is 015625.

But if you would have one number to expresse 3 quarters of a Yard,

75
125
015625
890625

or an Ell, two Nales, and one quarter of a Nale, you must first seek the De­cimal of three quarters of a Yard or Ell, which is 75, and write it down, likewise seek the Decimal of two Nales, which is 125, and write that down. Lastly, seek the decimal of one quarter of a Nale, which is 015625, and write that down, these three numbers added together, make 890625, which is the Decimal belonging to 3 quarters of a Yard or Ell, two Nales, and one quarter of a Nale.

PROP. VIII. How to expresse the parts of Time in Decimals.

Time is usually divided into Years, Moneths, [Page 199]and Dayes: So in the eighth table of Time, which consisteth of these two denominations, Moneths and Dayes, you may find that the deci­mal of five moneths is 416667, the decimal of 26 dayes is 071233. These are the particular deci­mals, but the compound decimal number repre­senting five moneths, 26 dayes, is 487900, as you shall find, if you adde 071233, which is the deci­mal of 26 dayes, to 416667, which is the decimal of five moneths.

PROP. IX. How to expresse Dozens in Decimals.

The last table is of Dozens, the integer being a Grosse, and the smaller denominations are Do­zens, and parts of Dozens, so may you find the decimal of seven dozen to be 5833333, and the decimal of five parts of a dozen to be 0347222, and these two numbers added together, make 6220555, which is the number which represent­eth 7 dozen, and 5/12 parts of a dozen.

Hitherto we have shewed the use of the foregoing tables in expressing of Fracti­ons in decimal numbers. It resteth now to shew the use of them in finding what fraction either of money, weight, or mea­sure, any decimal number given doth re­present, and that shall be made evident by the ensuing Propositions.

PROP. X. A decimal number being given, how to find what Fraction it doth represent.

Let 02916667 be a decimal number represent­ing some Fraction part of English Coin, because it is required to find the value of this Fraction in English Coin, you must therefore repair to the table of English Coin, in the second column of which table seek for the number given (viz. 02916667) which you shall find to stand against 7 pence, and so much is the value of the decimal Fraction 02916667, in English Coin.

Also if the decimal Fraction 75 were given, you shall find the value thereof to be 15 shillings, and the value of 003125 to be three farthings.

Likewise in the table of Troy weight, if 41666667 were given, it would signifie five oun­ces, and 05416667 would express 13 peny-weight, and 00173611 will expresse ten grains, &c

After this manner may you find the value of any decimal number given, either in Money, Weight, or Measure, when the number given, may be exactly found in the table: But if the number given cannot be found exactly in the ta­ble unto which it is directed, then you must find in the same table, the neerest number you can lesse than the given number, and take the num­ber that answers unto it in the first column which will be the greatest fraction of the number re­quired: [Page 201]then subtracting the decimal thus found, out of the decimal given, you shall have a remain­der, which remainder seek also in the second co­lumn of the table, if it may be found; if not, seek the neerest lesse, and the number answering thereunto in the first column shall be the next greatest fraction; then subtracting this decimal found out of the former remainder, there will be another remainder, which also seek in the ta­ble, and proceed as in the former: An Example or two will make all plain.

Example 1.

Let 68125000 be a Decimal given, represent­ing some part of English Coin. If you look in the table of English Coin for 68125000, you can­not find it, but the neerest number in the table lesse than it is 65, against which I find 13 s. so that 13 s. is the greatest fraction part of English Coin agreeing to this number.

This done, subtract 65 out of 68125000, and there will remain 03125000, which number also you must seek in the table of English Coin, but being you cannot finde it there, you must take the neerest number lesse than it, which is 02016667, against which I finde 7 pence, which is the next greatest fraction part of English Coin agreeing to this number.

Again, subtract 02916667, out of 03125000, and there will remain 00208333, which number seek in the table, and you shall finde it to stand [Page 202]against two farthings, and so much doth this last remainer signifie in English Coin, and the whole given number 68125000 doth represent in En­glish Coin 13 shillings, seven pence, two far­things, as by the operation following doth ap­pear.

68125000number given.
65......the next lesser number in the table re­presenting 13 s.
03125000first remainer.
02916667the next lesser number in the table re­presenting 7 d.
00208333second remainer, which represents two farthings.

So doth the whole number represent 13 s. 7 d. 2 q.

Example 2.

Let the Decimal 87426934 representing some fraction of a pound sterling be given. If you look in the table of English Coin for 87426934 you cannot finde it; but the neerest number in the table lesse than it is 85, against which I finde 17 shillings, so that 17 shillings is the great­est fraction part of English Coin, agreeing to this number.

Then subtracting 85 out of 87426934, there will remain 02426934, which number also you must seek in the table of English Coin, but seeing you cannot finde it there, you must take the nee­rest [Page 203]number lesse than it, which is 02083333, against which I finde five pence, which is the next greatest fraction part of English Coin.

Lastly, subtract 02083333, out of 02426934, and there will remain 00343601, which num­ber you must also seek in the table of English Coin; but not finding it exactly there, you must take the neerest number lesse, which is 003125, against which you shall finde three far­things, which is the next greatest fraction part of English Coin, and the Decimal 87426934, doth in value signifie 17 shillings, 5 pence, 3 far­things, and something more, for 003125 is the decimal of 3 farthings; and the number you are to look for in the table is 00343601, greater than the decimal of 3 farthings; wherefore, if you subtract 003125 out of 00343601, there will remain 31101, which is the 31101/100000 part of a far­thing, which is inconsiderable. See the follow­ing operation.

87426934Decimal given.
85......Decimal of—17 s.
02426934First remainer.
02083333Decimal of—5 d.
00343600Second remainer.
003125..Decimal of—3 q.
00031101Decimal part of a Farthing.

[Page 204]¶ And here note, that whatsoever hath been here said concerning the uses of the table of En­glish Coin, the same order is to be observed in the use of the other tables of Weight, Measure, Time, &c. as by the following Examples (if you make trial) will appear.

Examples.

1 If this decimal 48975694, were given to know the value thereof in Troy weight, you shall finde it to containe 15 ounces, 17 peny­weights, and 13 grains.

2 Also if 95075335 were a Decimal given, and it were required to find the value thereof in Averdupois great weight, you shall find it to con­tain 3 quarters of a hundred, 22 pound, 7 ounces, and 3 quarters of an ounce.

3 Likewise, if 70800781 were a decimal Fra­ction give, you shall finde the value thereof in Averdupois little weight to be 11 ounces, 5 drams and one quarter of a dram.

4 If 4375 were a Decimal, whose value were required in Liquid Measure, you shall finde it to contain 3 pints and an half.

5 Let 7148438 be a Decimal given, whose va­lue is required in dry measures, you shall find it to contain five bushels, 2 pecks, 3 quarters of a peck, and 2 pints.

Thus have I shewed you the use of these deci­mal tables in expressing of the fraction parts of Money, Weight, Measure, &c. But because these [Page 205]tables may not be alwayesat hand when there is need of them. I will here shew you how the va­lue of any Decimal given may be known by Multiplication onely; and this is

THE RULE.

Multiply the Decimal given by the number of known parts of the next inferiour Denomination, which are equall to the Integer, the Product is the value of the Decimal proposed in that inferiour Denomination; and if there happen to be any Decimal in the Product, you may in like manner finde the value thereof in the next inferiour Denomination, and so proceed till you come to the least known parts of the Integer.

Example.

Let .67395834 be a Decimal given, re­presenting the fraction of a pound sterling. First, multiply .67395834 by 20, (the number of shillings in a pound sterling) and the Product will be 1347916680, from which cutting off the last eight figures with a dash of the pen (because there were eight figures in the given Fraction) there will stand before the line towards the left hand 13, which are shillings, and the remainer 47916680 standing behind the line, will be the fraction part of one shilling sterling, which num­ber .47916680, you must multiply by 12 (the number of pence in one shilling) and the Pro­duct will be 575000160, from which number cut [Page 206]off the last eight figures as before, and there will be five left to the left hand, which are five pence, and the figures on the right hand of the line, viz. 75000160 are the fraction part of one penny ster­ling, which therefore multiply by 4 (the num­ber of farthings in one peny) and the Product of that multiplication will be 300000640, from which cut off the last eight figures to the right hand, and there will be left three towards the left hand, which representeth 3 farthings, and the re­maining figures towards the right hand are but the fraction part of a farthing, which we there­fore reject. And thus you finde by Multiplication onely, that this fraction .67395834 doth repre­sent in the known parts of English Coin 13 shil­lings, 5 pence, 3 farthings, as by the following operation appeareth.

[...]

In like manner, if this fraction .94809028 were given, representing some fraction part of [Page 207]Troy weight, you shall finde the value thereof to be 11 Ounces, 7 peny weight, 13 grains, as by the operation following appeareth.

[...]

In this manner may any Decimal given be re­duced into the known parts of the Integer by Multiplication onely.

¶ Here note, that whereas in the preceding ta­bles the decimal fractions consist of seven or eight Figures, we shall in the prosecution of our work make use onely of four or five of the first of them, which will be sufficient in ordinary practice, and come neer enough to the truth in any ordinary question whatso­ever.

So if in stead of 02916667, which is the fraction part of 7 pence, you take onely 02916, it will be sufficient.

  • [Page 208]Also for 05833333 take 05833 in Troy w. in dry measure. in Time.
  • Also for 0058594 take 0058 in Troy w. in dry measure. in Time.
  • Also for 5833333 take 5833 in Troy w. in dry measure. in Time.

Thus much concerning the construction and use of the decimal Tables, we shall now come to the praetice of Decimal Arithmetick, which shall be taught in the Rules following.

Of Notation of Decimals.

NOtation of Decimals is contrary to that of whole numbers; for whereas in whole numbers their values are increa­sed ten-fold by continuall addition of Cyphers towards the left hand: so on the contra­ry, the values of the places of Decimals do de­crease in the same proportion.

And whereas in whole number, Cyphers in the first place towards the left hand are unneces­sary, yet in Decimals they are absolutely neces­sary to discover the true denominator. Also Cyphers at the end (or towards the right hand) of decimal numbers are of no value, for one single Figure in decimals signifies as much as the same Figure would do, if there were Cyphers placed behind it, so 7 is equivalent unto 70, 700, or 7000, &c. For the denominators of decimal Fractions are alwayes Cyphers with a unite to­wards [Page 209]the left hand, as hath been already intima­ted. So 70/100 being reduced to its least terms will be 7/10, and 7000/10000 will be reduced to 7/10 also, and so of any other, as by the Table following doth evi­dently appear.

987654321 12345678
100000000 .00000001
 10000000 .0000001 
  1000000 .000001  
   100000 .00001   
    10000 .0001    
a thousand1000 .001or 1/1000
a hundred 100 .01 or 1/100
Ten  10 .1  or 1/10

Addition of Decimals.

IN addition of Decimals, the same order is to be observed as in addition of numbers of one denomination before taught in the first part, in which there is no difficulty: But in decimal numbers the chief care to be taken is in placing your whole numbers and fractions in their due order, which you shall easily and certainly do, if you observe this generall rule, viz. to place your whole numbers and fractions one under another, so that the points of separation which (in decimal num­bers) distinguish the whole numbers from the fractions, [Page 210]stand directly one under the other, then are you to proceed in the addition of them in all respects, as you did in whole numbers:

Example 1.

Let it be required to add together into one sum these several sums following, in a decimal way, viz. 36 li.—2 s.—8 d. 29 li.—0 —2 d. 21 li.—16 s. 9 d. and 6 li. 2 s. 5 d.

First, set down 36 li. and a point or Comma af­ter it, then for the fraction part of 2 s. 8 d. look in your Table of English Coin, where you shall finde the decimal fraction of a sh. 8 d. to be, 1333 therefore fot 36 li. 2 s. 8 d. set down 36. 1333.

Secondly, for your 29 li. 0 s. 2 d. set down 29.00831.

Thirdly, for your 31 li. 16 s. 9 d. set down 31.8375.

Lastly, for your 6 li. 2 s. 5 d. set down 6. 1208 as you see done in the operation following.

li.s.d.  
3628set down36,1333
For 2902set down19,0083
31169set down31,8375
625set down6,1208
10320set down103,0999

Your decimal numbers being thus placed in due order one under another, proceed to the ad­ding of them together, as if they were whole numbers, and you shall finde the sum or totall of them to be 103.0999.

[Page 211]Now the 103 which stands towards the left hand are 103 pounds, and the 0999 which stands towards the right hand of the Comma, is the fraction part of one pound sterling, the value whereof you may finde (by the Proposition be­fore going) to be two shillings fer [...], which should be two shillings exact, but it wanteth some-what, viz. the / 1000 part of a farthing, which is insensible; for if by the fore-mentioned rule you seek the value of the decimal fraction, .0999, you shall finde it to be 1 shilling, 11 pence, 3 farthings and the 2/1000 part of a farthing, which you may call in all 2 shillings, for deci­mal numbers will seldome happen to give the exact value of fractions, but will be either great­er or lesser than they ought to be; but in such a sum as this is, the thousand part of a farthing is not to be regarded.

Example 2.

Let it be required to adde together in a deci­mal way these sums following, viz. 29 li. 18 s. 7 d. 3 q. 63 li. 11 s. 2 d. 1 q. 129 li. 4 s. 0 d. 2 q. and 3 li 7 s. 10 d. 1 q.

First, for 29 li. 18 s. 7 d. 3 q. set downe 29.93229.

Secondly, for 63 li. 11 s. 2 d. 1 q. set downe 63.55937.

Thirdly, for 129 li. 4 s. 2 q. set downe 129.20208.

Lastly, for 3 li. 7 s. 10 d. 1 q. set downe 3.39271, as you see here down in the Margent.

[Page 212]Your decimal numbers thus placed in order,

29.93229
63.55937
129.20208
3.39371
226.08645

adde them together, as if they were whole numbers, and you shall finde the sum of them to contain 226.08645.

Now the 226 which stands towards the left hand of the Comma, are 226 pounds, and the o­ther figures towards the right hand, viz. 08645 are the fraction of a pound sterling, which if you reduce by the fore-mentioned Propositi­on, you shall finde the value thereof to be 1 shil­ling, 8 pence, 3 farthings, so the whole summe is 226 li. 1 s. 8 d. 3 q.

And here note, that what hath been said, as concerning Money, the same is also to be under­stood of Weight, Measure, Time, &c. as by the following Examples will appear.

Other Examples for practice.

Example 1. In Money.
135.8833
95.5583
3.2875
234.7291
234 li. 14 s. 7 d.

Example 2. In Troy weight.
7.97413
6.65330
3.62187
18.24930
18 li. 2 ou. 19 p.w. 20 gr.

Example 3. In Averdupois little w.

12.7227
76.3594
32.625
91.4883
32.8398
246.0352
246 li. 00 oun. 9 dr.

Example 4. In Averdup. great weight.
37.9442
9.3053
33.6786
10.0000
12.8142
103.7423
103 C. 2 q. 27 l. 3 oun.

Subtraction of Decimals.

THe Subtraction of decimals differeth no­thing from the subtracting of one whole number from another, and the decimal numbers to be subtracted one from ano­ther, must be placed in the same order, as in Ad­dition of decimal numbers, the practice of Sub­taction shall be seen in the following Examples.

Example 1.

Let it be required to subtract 31 li. 16 s. 9 d. out of 36 li. 2 s. 8 d.

First, for your 36 li. 2 s. 8 d. set down the de­cimal thereof, which is 36.1333.

Secondly, for your 31 li. 16 s. 9 d. set down the Decimal thereof 31.8375.

[Page 214]This done draw a line under them,

36,1333
31,8 [...]75
4,2958

& subtracting the lesser from the grea­ter, you shall finde the remainer to be 4. 2958. the 4 on the left side of the Comma are four pounds, and the 2958 which standeth towards the right hand, is the fraction part of a pound, the value whereof being sought, will be found to be 5 s. and 11 pence▪ So that if you subtract 31 li. 16 s. 9 d. out of 36 li. 2 s. 8 d. there will remain 4 li. 5 s. 1 [...] d.

But if divers sums be to be subtracted out of one greater sum, then you must first adde all the several smaller sums together, and subtract the sum of them from the greater given sum, so shall the residue be the sum desired.

Examples for practice.

Example 1.
In Money.
Lent2684.8375
Paid at several times36.1333
 29.0083
 31.8375
 6.1208
paid in all103.0999
rests to pay2581.7376
2581 li. 14 s. 9 d.

Example 2.
  In Averdupois great weight.
Bought 103.7423
Sold 37.0442
Unsold 65.7981
65 C.3 q.5 l. 7. oun.

Example. 3.
 in Troy weight.
Delivered to a Goldsmith of old Plate7. 97413
Received of new Plate5.59670
Rests in the Goldsmiths hands2.7743
2 li. 4 oun. 10 p w. 14 gr.

Multiplication of Decimals.

Multiplication of Decimals differeth no­thing at all from the multiplication of whole numbers, for making the great­er number the multiplicand, and the lesser number the multiplyer, the number issu­ing from that multiplication shall be called the Product.

Now in the multiplication of decimal num­bers one by another, if there be any Fraction ei­ther in the multiplicand or multiplyer, or Fra­ctions in both: So many figures as the Fractions contein, so many figures must be cut off from the Product towards the right hand, which shall be the Fraction of the Product, and the figures towards the left hand of the Comma in the Pro­duct, shall be the Integers of the Product.

Example 1.

Let it be required to multiply 34 pound, five shillings, three pence, by 16 pound, six shillings, six pence.

First, seek the Decimal of 34 li. 5 s. 3 d. which you shall finde to be 34. 2625, make this num­ber [Page 216]your Multiplicand, then seek the Decimal of 16 li. 6 s. 6 d. which you shall find to be 16. 325, [...] make this decimal number your Mul­tiplyer, then draw a line, and multi­ply these two num­bers together, as if they were whole numbers, and you shall finde the Pro­duct of them to be 559. 3353125. Now because there are four fi­gures in the Multiplicand which are Fractions, namely, these four towards the right hand, viz. 2625, and there are also three figures in the multiplyer, which are Fractions, namely, these three towards the right hand, viz. 325, that is in all seven figures representing Fractions, I there­fore cut off from the product the seven figures towards the right hand, by making of a Comma there to distinguish the whole number from the fraction: So is 559 the Integer or whole num­ber, and 3353125, the fraction of this multipli­cation.

Example 2.

If there be Fractions in the multiplicand, and none in the multiplyer, yet that work is still the same, for you must cut off onely so many figures from the product, as there are Fractions either in multiplicand, multiplyer, or both: So if it were required to multiply 5767 yards, and 3 quarters [Page 217]of a yard, by 235 yards, you must first set down 5767. 75 for your 5767 yards, [...] and three Quarters, which number must be your multi­plicand, also set down 235 yards for your multiplyer, then multiplying them toge­ther, as if they were whole numbers, you shall finde the product to be 1255421, 25, and because there are onely two Fraction Figures, both which are in the multiplicand, namely, the two last thereof .75, and none in the multiplyer, I therefore cut off onely two figures of the product, namely, the two last, which are .25, so is the product of this multiplication 1355421. 25; which is 1355421 square yards, and one quarter of a yard, And so if a garden or other piece of land lying square, should contein in length 5767 yards, and three quarters, and in breadth 235 yards, the whole piece would contein 1355421 square yards, and one quarter of a yard.

Example 3.

If decimal Fractions be to be multiplyed by decimal Fractions, you must then (as before) multiply them as whole numbers, and from the Product cut off so many Figures towards the right hand, as there are Figures in the multipli­cand and the multiplyer: So if it were required to multiply .953 by 782, you shall finde their [Page 218]product to be .745246, [...] which being but six figure sin all, I cut them off and that fraction .745246 is the product of the multiplication of the two gi­ven fractions.

Example 4.

If any two Decimal fractions being multipli­ed together, the product thereof doth not consist of so many places as are required (by the former rules) to be cut off, you must then supply that defect by prefixing a Cypher, or Cyphers before the product towards the left hand: So if these Decimal fractions .063 and .0752 were to be multiplyed, [...] their pro­duct would be 47376. Now (by the former rules) you should cut off seven figures of the product towards the right hand, but this product 47376 consisteth but of five figures; wherefore to make it seven figures, I prefix two Cyphers before the product on the left hand, making it .0047376, and that is the true product produced by this multiplication.

Example 5.

If you would multiply any Decimal (either fraction onely, or whole number and fraction together) by 10, 100, 1000, &c. You must adde so many Cyphers to the multiplicand, as there [Page 219]are Cyphers in the multiplyer, and cut off so many Figures as there are fractions in the mul­tiplicand, and that number shall be the product required: So if 7,856025 were a Decimal given to be multiplyed by 100, you must (be­cause there are two Cyphers in 100) add two Cyphers to the number given, making it 785, [...]02500, then because there were six fraction Figures in the multiplicand, if you cut off the six figures of this number towards the right hand, it will be 785,602500, which is the true product required.

Examples for practice.

Example 1. [...]

Example 2. [...]

Example 3. [...]

Example 4. [...]

Division of Decimals.

AS Division of whole numbers is the hard­est of the four Species of Vulgar Arith­metick, so the Division of Decimals is the most difficult of the four kinds of Decimal Arithmetick, but I hope to make it plain to the understanding of the meanest capa­city.

The several varieties that may happen in Di­vision, are principally (if not onely these) four. Namely, first, to divide whole numbers and fra­ctions, by whole numbers and fractions. Se­condly, to divide whole numbers by mixt, or mixt numbers by whole. Thirdly, to divide a greater fraction by a lesse, and lastly, to divide a lesser fraction by a greater.

In Division of Decimals this Rule is general, If the dividend be greater than the Divisor, the quo­tient will be either a whole number or a mixt, but if the Dividend be lesse than the Divisor, the quotient will be a Decimal. And (for convenience in working, if there be need) any number of Cyphers may be annexed to the Dividend that thereby the quo­tient may extend to as many places as the tenour of the question shall require.

The manner of the working of Division in Decimals, is the same with that before delivered [Page 221]in whole numbers, as will appear by the Exam­ples following, in every of the four premised varieties.

The Rule for the first varietie.

The Dividend and the Divisor, being both mixt numbers, or one of them being a whole number and the other a mixt; or the Dividend being a Decimal, and the Divisor a whole number or a mixt, the first figure in the quotient will be of the same place or de­gree, with that figure or Cypher of the Dividend, which at the first demand standeth, or (at least) is supposed to stand directly over the place of Ʋnits in the Divisor.

Example 1. Where the terms given, are both of mixt numbers.

Let it be required to divide 559.3353125 by 16.325. Here the terms given, are both of mixt numbers, which being, placed accord­ing to the Rules delivered before, for the Division of whole numbers, the figure in the Dividend, which at the first demand, stand­eth over 6, the place of Units in the Divisor is 5, and because this standeth in the place of tenths, therefore the first figure in the quotient is in the place of tenths also, and the whole number consisteth of two of the foremost pla­ces, and the rest is a Decimal, thus the quoti­ent sought in our present example is 34.2625, of which 34 the two first figures is the Integer [Page 222]or whole number, 2625 the Decimal fraction.

[...]

Example 2. One of the terms given, being a whole number the other mixt.

The mixt number 1355421.25 being divided by the whole number 235, the quotient will be 5767.75, and the first figure in the place of Thousands, as by the operation it doth ap­peare.

[Page 223] [...]

Example 3. The Dividend being a Decimal, and Divisor the whole number.

The Decimal fraction .35673 being divided by the whole number 26, the quotient will be 001372, and the first significant figure in the place of thousands or fourth place from Unitie, as by the operation it doth appear.

[Page 224] [...]

The Rule for the second varietie.

When the Dividend is a whole or mixt number and the Divisor a Decimal, add as many Cyphers to the Dividend as there are places in the divisor for the in­tegral part of the quotient will consist of as many pla­ces as the Divisor, and the places arising from the in­tegral parts of the Dividend added together.

Example 1.

Let 348.75 be the mixt number given, to be di­vided by the Decimal .25, to the number given, I add to Cyphers, the number of places in the Divisor, and then it will be 348.7500, which being divided by .25; the integral part of the quotient will be 1395. because the whole part of the Dividend 348, being divided by .25 giveth two places, and the number of places in the [Page 225]Divisor being two, giveth two more; and so the Integral part consisteth of four figures, as by the operation.

[...]

Example 2.

Let the mixt number 72.5 be divided by 075, [...] the number of places in the in­tegral part of this Quotient will be 966, be­cause there are 3 places in the Divisor; & but 3, because the [Page 226]integral part of the dividend is lesse than the sig­nificant figures in the Divisor, as by the opera­tion it doth appear.

The Rule for the third variety.

When the Terms given, are both Decimals, the Di­vidend being the greater the integral part of the quo­tient will consist of as many places as the Divisor doth.

Example.

Let the Decimal .73958 be divided by the De­cimal .32 the inte­gral part of the quo­tient will be 23, [...] be­cause the Divisor doth consist of two places, as by the o­peration in the mar­gine it doth appear.

The Rule for the fourth varietie.

When the terms given are both Decimals, consisting [Page 227]of equal places, the Dividend being the lesser term, place the dividend as a Numerator, and the Divisor as 'Denominator; so is such vulgar fraction the quo­tient fought: But if the terms given consist not of e­qual places, supply the place or places wanting in ei­ther of the terms, by annexing a Cypher or Cyphers on the right hand, and then proceed as before. Thus if .27 be given to be divided by .93 the quotient will be 27/93. Also if .35 be given to be divided by .78563, the quotient by annexing 3 Cyphers to .35 the lesser Decimal given will be 35000/78563, which vulgar fractions may be reduced into Decimals if need be, by the first Proposition in this Second part of Decimal Arithmetick.

Examples for Practice.

[...]

[...]

The Rule of three in fracti­ons Vulgar and Decimal.

WHat the Rule of three is, and the manner of working, is already shew­ed in the first part, that which we here intend is onely to adde some Examples in fractions vulgar as well as Decimal; that by comparing the work in both, the excel­lent use of Decimal Arithmetick might the better appear.

And how to convert the known parts of mo­ny, Weight, or Measures English, into De­cimals hath been already shewed, both Arith­metically and by Tables; yet to prevent the se­veral additions and subtractions in those Tables, we have here annexed another Decimal Table, for the more speedy Reduction of English money under two shillings, all summes of money above, not having pence or farthings annexed, being as easily reduced by memory as by Tables; and this we have the rather done because the same Table, will also reduce the Coins of France, and the parts of Troy weight, if an ounce be made the Integer, which in point of practice is much more useful then the pound.

[Page 229]

The Table of Reduction.
Peny001042 
 002083Gr. 1
 003125 
10041662
 005208 
 0062503
 007291 
20083334
 009375 
 0104165
 011458 
30125006
 013541 
 0145827
 015625 
40166668
 017708 
 0187509
 019791 
502083310
 021874 
 02291611
 023958 
602500012
 026014 
 02708313
 028125 
702916614
 030208 
 03125015
 032291 
803333316
 034375 
 03541617
 036458 
903750018
 038541 
 03958319
 040625 
1004166620
 042708 
 04375021
 044791 
1104583322
 046875 
 04791623
 048958 
1205000024

[Page 230]

The Table of Reduction.
 051042 
 052083G 1
s.d.053125 
t 10541662
 055208 
 0562503
 057292 
20583334
 059375 
 0604165
3061458 
 0625006
 063542 
 0645827
 065625 
40666668
 067708 
 0687509
 069792 
507083310
 071874 
 07291611
 073958 
607500012
 076014 
 07708313
 078125 
707916614
 080208 
 08125015
 082921 
808333316
 084375 
 08541617
 086458 
908750018
 088541 
 08958319
 090625 
1009166620
 092708 
 09375021
 094791 
1109583322
 096875 
 09791623
 098958 
1200000024

[Page 231]These things premised; we will now shew the use of the Table in some practical questions be­longing to the Rule of three direct.

1 Question.

Ifof a yard of cloth, cost 9/12 of a pound: what shall 17 yards cost at the same rate?

If ⅞ cost 9/12 what shall 17 cost? Ans. 14 li. 4/7.

First, multiply 9/12 by 17/1 the product is 153/12, then divide 153/12 by ⅞, the quotient is 1224/84: again, if you divide 1224 by 84, the quotient is 14 [...]/ [...]4 or in the least terms 14 pound 4/7 of a pound.

And the value of this fraction 4/7 of a pound, will be found by the third Rule of Reduction page 107, to be 11 shillings, 5 pence, and 4/7 of a peny, which is somewhat above two farthings: for it is 2 farthings, and 2/7 of a farthing.

The same Question in Decimals.

Ifof a yard of cloth, cost 9/12 of a pound, what shall 17 yards cost at the same rate?

To answer this question ⅞ of a yard, and 9/12 of a pound must first be reduced into Deci­mals, either by Division, or by the Tables of Reduction: by both which wayes of Reduction the Decimal of ⅞ will be 875, and the Decimals of 9/12 will be .75 and then the terms of the que­stion will stand thus;

If .875 parts of a yard cost .75 parts of a pound, what shall 17 yards cost at the same rate?

If 0.875 — 0.75 — 17. Here if you mul­tiply [Page 232]the second terme 0.75 by 17 the third term given, the product will be 12.75. and this pro­duct divided by .875 gives in the quotient 14.57142, that is 14 pound .57142 parts of a pound, or 145 Decades, that is 14 pound 10 shil­lings, and .7142 parts of a Decade (or two shil­lings) which by the preceding Tables is 1 s. 5 d. 2 farthings, and .0059 parts of a farthing.

2 Question.

If a piece of Gold plate weighing 19 ounces 3 peny weight and 5 grains, be worth 62 pounds, 10 shillings, 6 pence, what is one ounce of the same gold worth.

This question in vulgar fractions must be ex­pressed thus.

If 1 l. 3437/5260 Troy weight, cost 62 l. 126/240, what shall 1/12 of a pound Troy cost? at the same rate.

To answer this question the fractions 1 l 3437/5760, and 6 [...] l. 126/240, must be first reduced into impro­per fractions, and the fraction 1/12 into the least known parts of a pound Troy, and then the que­stion will stand thus.

If 9197/5760 give 15006/240 what shall 480/5760 give?

Now because it is necessary the terms given be reduced into their least Denomination, be­fore the question be resolved, therefore the an­swer may be found, by using the terms given thus reduced, as whole numbers, not having any regard to the Denominators of these fractions; Saying thus,

If 9197 grains, cost 15006 pence, what shall 480 grains.

[Page 233]And here if you multiply 15006 by 480 the product will be 7202880 which being divided by 9197 the quotient will be 783 pence 1629/9197 parts of a peny, and dividing .783 by 12 it will be 65 shillings 3 pence 1629/9197, or 3 l. 5 s. 3 d. 1629/9197. And although this question is thus more easily an­swered then it would have been, if the terms had been wrought as vulgar fractions, yet the same terms being reduced to Decimals, the answer of the question will yet be found with more ease, as shall appear by the operation following.

The same question in Decimals.

If a piece of gold plate weighing 19 ounces 3 peny weight and 5 grains, be worth 62 l. 10 s. 6 d. what is an ounce of the same gold.

The Decimal of 19 ounces 3 peny weight and 5 grains, making an ounce the Integer is by this Table 19.16041, for that 19 ounces are 19 Inte­gers, 2 peny weight is one tenth of an ounce, and the Decimal of one peny weight 5 grains is by this Table 06041; and the Decimal of 62 l. 10 s. 6 d. by the same Table is 62.525, and because an Unite or Integer is the third term given, there needs no multiplication, if therefore you divide 62.525 the second term, by 19.16041 the first term propounded, the quotient will be 3.2632 that is 3 pounds, 5 shillings, 2 pence, and somewhat more, as by the operation in the margine it doth appear.

[Page 234] [...]

If 5 Els and a quarter of linnen cloth, cost 2 l. 16 s. 8 d. 3 q. what shall 278 els and a half cost at the same rate?

If you would work this Question by whole numbers your easiest way is first to reduce all the terms into their least Denominations, that is to say, the els into quarters, and the pounds, shillings, pence, and farthings all into farthings, so shall your 5 els and a quarter be 21 quarters, and your 278 els and an half will be 1114 quar­ters, and your 2 l. 16 s. 8 d. 3 q. will be 2723 far­things, and then will your question stand thus in whole numbers.

If 21 quarters — cost — 2723 farthings — what will — 1114 quarters cost

Then multiplying the second number by the third, that is, 2723, by 1114, the product will [Page 235]be 3033422, which divided by 21, the quotient will be 144448 farthings, which being again reduced into pounds, shillings, and pence, giveth 150 l. 9 s. and 4 pence, as by the operation follow­ing doth appear.

The Operation.

[...]

[Page 236] [...]

But if you would work the same question by Decimal numbers you may save the labour of reducing the terms to their least Denominati­ons, for 5 Els and a quarter is in Decimal num­bers 5.25, and 278 Els and an halfe is 278.50, and 2 pound 16 s. 8 d. 3 q. is in Decimals 2.8364, [Page 237]and then your question in Decimals will stand thus.

If 5.25 Els cost 2.836 pounds what 278.50 Els

If you multiply (according to the Rule) the second term by the third, that is 2.8364 by 278.50 the product of that multiplication will be 789.937400, which divided by the first term 5.25, the quotient will be 150.4642, which Decimal (by the Example in page 177) repre­senteth 150 l. 9 s. 4 pence, and so much in money will 278 Els and a half cost.

The Operation.

[...]

[Page 238] [...]

I have been the larger in this Rule and espe­cially in this Example, which is incumbred with fractions sufficient, because I would have the Reader the better discern the difference be­tween the Vulgar and the Decimal way, and also to see how expeditious the one is over the other. Now this Example being thus largely ex­plained, I shall with the more brevity passe over the Rules following giving one Example, or two at the most in each Rule; And thus much shall suffice for the Golden Rule, or Rule of three di­rect in Fractions.

An Example in the Rule of Three Reverse.

A Lends B 233 l. 6 s. 8 d. for a year without in­terest, upon condition that B should do the like courtesie for A when required. A hath occasion for money 7 moneths, how much money ought B to lend A, to requite his courtesie, and save himself harm­lesse?

I will not in this place tell you what the Rule of Three reverse is, nor the manner of working thereof, that being already sufficiently declared in the first part, but give you the Example, and the working thereof, which take as followeth: So will the Question be thus stated.

[...]

[Page 240]Here you see that 12 moneths and 7 moneths are whole numbers, and so we let them alone without any reduction, but the Decimal of 233 l. 6 s. 8 d. will be found by the fore-menti­oned Tables and Rules to be 233.33, which is the middle term in the question, and of the same quality with that, must the fourth term sought be, therefore if (according to the Rule delivered in the first part) you multiply 233.33 by 12, the product will be 2799.96, which divided by 7 gi­veth in the quotient 39999, which is the Deci­mal of 400 l. and so much mony ought B to lend A for 7 moneths.

Examples in the Rule of Proportion, consisting of five numbers.

Question 1.

If 100 l. in 12 moneths, yields 6 l. interest, what interest shall 264 l. 16 s. 5 d. yield in 15 moneths at the same rate?

Set down your numbers in Decimals, as in the Example following appeareth, so shall you find the Decimal of 264 li. 16 s. 5 d. to be 264.8208, all the rest being whole numbers having no fra­ctions joyned with them we neglect, and work [Page 241]with them as they are, so will the several num­bers of your question (if rightly disposed) stand as followeth.

[...]

Your numbers being thus orderly disposed, you must according to the Rule before delivered [Page 242]in the first Part, Page 92) multiply the first and second terms together, which in this Ex­ample are 100 and 12, whose product is 1200, which is your Divisor; Then multiply the three last terms one into another, as 264,8208 (which is the Decimal of 264 li. 16 s. 5 d.) by 6, and the product thereof will be 1588.9248, which number again multiplyed by 15, (which is the last term,) the product will be 23833.8720 which is your Dividend, and this number being divided by your former product, giveth in the quotient 19.8615, which is the Decimal of 19 l. 17 s. 2 d. 3 q. ferè, and so much doth the simple interest of 264 l. 16 s. and 5 d. amount unto in 15 moneths, after the rate of six per centum for a year.

Quest. 2.

If the carriage of 23 hundred and 3 quarters of any thing 127 miles, cost 4 li. 13 sh. 6 d. what shall the carriage of 47 hundred and an half of such like com­modity cost, being carried 381 miles.

Place your numbers in order as in the follow­ing Example doth appear, then multiply the first and second terms together for your Divi­sor, and the three last one into another for your Dividend, and so will the quotient of this division answer the question demanded, and the work will stand as followeth.

[Page 243] [...]

Here you see that the first and second terms multiplied together produced 3016.25, which must be your Divisor, and the three last terms being multiplyed one into another, produce 84605.81250, which number divided by 3016.25, giveth in the quotient 28.050, which Decimal representeth 28 l. one shilling, and so much will the carriage of 47 hundred and a half cost, being carried 381 miles?

[Page 244]Thus have I shewed the use of decimal Arith­metick in such questions as concern the Golden Rule, or Rule of Three; both Direct, Reverse, and Compounded, by an Example or two in each rule, & those compounded of fractions sufficient, I should now proceed to questions in Fellowship, with and without Time, as also Barter, Alligation; the Extraction of the Square and Cube Roots, &c. but forasmuch as these last mentioned Rules depend only upon the Rule of Three, as by Exam­ples in the first part doth plainly appear, I shall therefore save that labour, and spare my Reader the pains of practising questions which wholly depend upon that which I (by this time,) suppose him perfect in; Yet if the Reader be desirous to make trial of any such question for his own satis­faction, he may either make trial of those que­stions in the former part of this Book in those se­veral Rules, reducing the numbers there given into Decimals, or if he please, he may frame que­stions according to his own fancy. And thus I shall conclude this second Part.

The end of the second Part.

THE THIRD PART Conteining INSTRUMENTAL ARITHMETICK.

THe Arithmetick, of which we now come to treat, and which I call Instrumental Arithmetick, is not any new kind of Arith­metick, but is indeed the same with Decimal Arithmetick be­fore taught; onely, whereas in Decimal Arith­metick, there were certain Tables made of Mo­ney, Weight, and Measure, by help of which the Decimal of any fraction of Money, Weight or Measure might be set down (as it were) in whole numbers, here in this Instrumental part, we have contrived certain Scales of Money, Weight, and Measure, equally divided into the several De­nominations into which the several Weights and Measures, can be equally divided, unto every [Page 246]of which Scales there is joyned a Scale of 100, 1000, or 10000, equal parts, according to the length of the Scale, so that by inspection onely you may readily and exactly without addition (as in using the fore mentioned Tables you must necessarily do) set down the Decimal fraction of any part of Money, Weight or Measure, with great celerity and exactnesse, if the Scale be any thing well divided, though the Scale be but of a reasonable length.

Now the Scale, which I have chiefly made choice of in this Work, as being of most use with English men; (though other Scales may be made for the Coyns, Weights or Measures of any other Countrey as well, and upon the same ground) are chiefly these, viz.

  • 1 Of Money.
  • 2 Of Troy Weight.
  • 3 Of Averdupois great Weight.
  • 4 Of Averdupois little Weight.
  • 5 Of Liquid measures.
  • 6 Of dry measures.
  • 7 Of Long measures.
  • 8 Of Dozens.

Unto every of these Scales, is joyned another Scale of 100 or 1000 equal parts, these Scales are made to face one another, so that if you look upon any one Division, in the one, you shall also dis­cern plainly what Division or part of a Division answereth thereunto in the other.

[Page] [Page]

English Coyn, Two shillings being the Integer

Troy weight, Two peny weight being the Integer

Averdupois great weight, 28 lib: or one quarter of an hundred being the Integer

Averdupois little weight, 16 ounces, or one pound being the Integer

Place this betweene pag 246 & 247

Dry measure, S Bushels being the Integer

Liquid measure, 36 Gallons, or one Barrell being the Integer

Long measure, one Ell or one Yard being the Integer

[Page 247]These Scales being thus disposed, as they may easily be upon any Ruler of Silver, Brass, or Wood; but best of all upon a square Ruler, made in form of a Parallelepipedon, will by inspection onely give you any Decimal fraction required without Addition, or (on the contrary) reduce the fraction into the know parts of the Integer, by inspection also, without subtraction.

Let thus much suffice for a general description of what I mean by Scales, the particular descrip­tion of them, will more plainly appear, when we treat of Numeration upon the Scales, unto which we shall now proceed. But first take a view of the Scales as they are here disposed, and as they may be set upon such a Ruler as I have here mentioned.

Numeration upon the Scales.

THe Scales here to be described are in number eight, as hath been already shewed, and as by the figure of them appear. Now Numeration upon a Scale is to find upon what part of the Scale, any number upon the same Scale will fall.

We will begin with the first, and so proceed till we have given an Example in every one.

1 The first Scale is of English money, and is divided into 24 equal parts, which represent 24 pence or 2 shillings, these parts are numbred [Page 248]with Arithmetical figures, from the beginning thereof, by 1, 2, 3, 4, 5, &c. to 24, each division representing one peny, and the whole 24 divisi­ons representeth 24 pence or 2 shillings, So that where the figure 1 standeth, that part of the Scale representeth one peny, where the figure 2 stand­eth, it representeth two pence, where the figure 18 standeth, it representeth 18 pence, or one shil­ling six pence, and so of any other figure of the same Seale.

Then because there are four farthings con­tained in a peny, each of these pence (or divi­sions) is sub-divided into 4 other equal parts by short lines, every one of these representing one farthing, so is the whole Scale divided in all in 96 equal parts, which are the number of far­things conteined in two shillings. This if you look into the Scale of Money for 8 pence 3 far­things, you shall find it at the letter a, which letter is here put onely for example sake, Also if you would find in the Scale the place of 18 d. half peny, you shall find it at c, and thus may you find the place of any number of pence and far­things under two shillings upon the Scale.

Unto this Scale of money (as to all the rest of the Scales,) there is joyned another Scale which I shall alwayes hereafter call the Scale of 1000, the use of which Scale is this. When you have found any Number of pence or farthings upon the Scale of Money, you shall find upon the Scale of 1000, what parts of a thousand is the Decimal of those pence and farthings: Thus [Page 249]when in the Scale of money you find at the letter a 8 pence 3 farthings, if you cast your eye dire­ctly crosse to the Scale of 1000, you shall find 364 to stand directly against 8 d. 3 q. which 364 is the Decimal of 8 d. 3 q. Also, if you find upon the Scale of money 18 d. half penny, which is at the letter c, you shall find against it in the Scale of 1000, this number 771, which is the Deci­mal of 18 d. half penny. And in this manner may the Decimal of any number of pence or far­things under two shillings be most easily and ex­actly obteined.

Now on the contrary, suppose a Decimal fra­ction were given, representing some part of English Coin if you look in the Scale of 1000 for your Number given, right against it in the Scale of Money you shall find what Number of pence and farthings is representeth thereby. As for Example. Suppose .364 were a Decimal given, and it were required to find what part of Coin it doth represent. Look in the Scale of 1000 for the Number 364, and right against it you shall find 8 pence 3 farthings. Also if .771 were a Decimal given, if you look in the Scale of 1000 for .771, you shall find against that number 771, 18 pence 2 farthings. And thus of any other.

By what hath been already said, it may be ea­sily discerned of what exceeding expedition these Scales thus disposed are of, for I dare af­firme, that I will set down 2 (if not 3) numbers by the Scale, as soon as one by the Tables, and [Page 250]if the Scale be but of any reasonable length, al­together with as much exactnesse, but if I should vary an unite in the last place, in my estima­tion in the Scale of 1000, it is not any thing ma­terial.

I have been very tedious in shewing the use of these Scales to finde the fraction parts of money, but the reason is because I intend to be the briefer in the rest, for weight and measure, the manner of working (when the division of the Scale is known,) being the same in all respects without the least alteration.

2 The second Scale is of Troy weight, two Peny weight being the Integer, which Scale is divided into 48 equal parts or divisions, each of which divisions contains one grain, and are num­bered by Arithmetical figures at every three grains by 3, 6, 9, 12, &c. to 24, and at the place where 24 should stand, there standeth PW, which signifieth one peny weight, or 24 grains, this PW standeth in the middle of the line. Then is the same Scale continued farther by Arithmeti­cal figures 3, 6, 9, 12, &c. as before to 24, and there is written PW again, representing two peny weight, or 48 grains.

The Scale being thus divided, it is easie to finde the place where any number of Grains un­der 48 shall be upon the Scale; As for example, if it be required to finde where 8 peny weight shall fall, look upon the Scale of Troy weight, from the beginning thereof, and count the fi­gures 3 and 6, then count also two of the smaller [Page 251]Divisions, and that makes 8 grains, which you shall finde to stand at the letter d, which is the place of 8 grains; Also if upon the Scale you would finde the place of one peny weight 10 grains, you shall finde it at the letter e, and so of any other number of grains under 48, or two pe­ny weight.

But if you had a Decimal given, and would know what number of grains it representeth, if you seek your Decimal given in the Scale of 1000, right against it in the Scale of Troy weight, you shall finde the number of grains represented thereby.

Example. Let .167 be a Decimal fraction given, If you look in the Scale of 1000, for 167, right against it in the scale of Troy weight, you shall finde 8 peny weight.

Also if .708 were a Decimal given, if you seek 708 in the scale of 1000, right against it you shall finde 1 peny weight 10 grains.

3 The third scale is of Avoirdupois great weight, 28 pounds, or one quarter of an hundred, being the Integer, this scale is numbered by 1, 2, 3, 4, &c. to [...]8, which 28 representeth 28 l. or a quarter of a hundred, and each of those is subdi­vided into four small parts, each representing one quarter of a pound.

Now if you would know what is the Decimal of any number of pounds or quarters under 28, if you seek the number of pounds in the scale of Avoirdupois great weight, right against it in the scale of 1000, you shall finde the decimal thereof.

[Page 252]Thus if it were required to find the Decimal of 8 pound and an half, if you look upon the Scale for 8 pound and an half you shall find it at the letter g, and right against it in the Scale of 1000 you shall find 304, which is the Decimal of 8 pounds and an half.

On the contrary, suppose 304 were a Decimal given, and it were required to find what part of Avoirdupois great weight were represented thereby, if you look in the Scale of 1000 for 304, right against it in the Scale of Avoirdupois great weight, you shall find 8 pound and an half.

4 The fourth Scale is of Avoirdupois little Weight, 16 ounces or one pound being the In­teger, This Scale is first divided unto 16 equal parts, and numbred by 1, 2, 3, 4, &c. to 16, each Division representing one ounce. Then again, each of these ounces is sub-divided into 8 other smaller parts or divisions, each of which divi­sions representeth two Drams; but if your Scale be large enough, you may have each ounce di­vided into 16 equal parts, or divisions, each di­vision representing one Dram.

Now to find the Decimal belonging to any number of Ounces and Drams, repaire to the Scale of Avoirdupois little weight, and on it find the quantity of ounces and drams required, and right against it in the Scale of 1000, you shall have the Decimal thereof.

Thus if it were required to find the Decimal of 6 ounces, and 6 drams, if you look this in the Scale of Avoirdupois little weight, you shall [Page 253]find it at the letter h, and right against it in the Scale of 1000, you shall find 398, which is the Decimal of 6 ounces and 6 drams.

But if 398 were a Decimal given, and it were required to find the value thereof in Avoirdupois little weight, if you look for 398 in the Scale of 1000, right against it in the other scale you shall find 6 ounces 6 drams.

5 The fifth Scale is of Dry Measures, one Quarter or 8 Bushels being the Integer, This Scale is first divided into 8 equal parts, and num­bred by 1, 2, 3, &c. to 8, each of which divisions re­presenteth a Bushel, & each of those parts is again sub-divided, first into 4 equal parts or divisions each representing one peck, and then those again sub-divided into 4 other smaller parts, represen­ting quarters, halves, and three quarters of a peck.

Now if you would know the Decimal belong­ing to any number of Bushels (under 8 Bushels or one quarter) Pecks and parts of a Peck, if you seek the number of Bushels, Pecks, and parts of a Peck in the Scale of Dry Measures, right against it in the Scale of 1000, you shall have the Decimal required.

As for Example, if it were required to find the Decimal belonging to 5 Bushels 2 Pecks, and half a Peck, if you look into the Scale of dry Measures you shall find 5 bushels, 2 pecks, and an half, to stand at the letter k, and right against it in the Scale of 1000, you shall find 702, which is the De­cimal answering to 5 bushels, 2 pecks and an half.

But if 702 were a decimal given, and it were [Page 254]required to find what number of bushels, pecks and parts of a peck it representeth, if you look in the scale of 1000 for 702, you shall find against it in the scale of Dry measures, 5 bushels 2 pecks and an half.

6 The sixth Scale is of Liquid Measures, the Integer being 36 Gallons or one Barrel, this Scale is divided first into 36 equal parts or divi­sions, and numbered by 5, 10, 15, &c. to 36, then every of these divisions is again sub-divi­ded into 4 other smal divisions, each representing a quart, but (if the Scale be large enough) you may sub-divide each Gallon into 8 parts, so will every part represent one pinte.

Now to find the Decimal belonging to any number of Gallons (under 36 Gallons or one barrel) quarts or pintes, repaire to the Scale of Liquid measures, and seek there upon the Scale, the number of gallons, quarts, or pintes, and a­gainst it in the Scale of 1000, you shall find the Decimal thereunto belonging.

So if it were required to find a Decimal repre­senting 10 gallons and two quarts, or 4 pintes, which is all one, if you seek in the scale of Li­quid measures for 10 gallons 2 quarts, you shall find it at the letter m, against which in the scale of 1000, you shall find 292, which is the Decimal of 10 gallons 4 pintes.

Likewise if 292 were a Decimal given, and it were required to find what number of gallons, quarts or pintes were represented thereby, if you look in the scale of 1000 for 292, right [Page 255]against it in the scale of Liquid measures, you shall find 10 gallons, 2 quarts, or 4 pintes.

7 The seventh scale is of Long Measures, the Integer being Yards or Ells, this scale is divided into 4 equal parts, and numbred by 1, 2, 3, 4, repre­senting 1 quarter, 2 quarters, 3 quarters or 4 quar­ters of a Yard or Ell, these are again sub-divided first into 4 other equal parts, representing nails, and those may be again sub-divided at pleasure if need be.

Now if you would know what decimal belon­geth to any number of quarters or nailes of a yard if you seek the number of quarters and nailes in the scale of Long measure, the scale of 1000 will give you the Decimal thereof.

Thus if it be required to find the Decimal be­longing to 1 quarter and 3 nails, if you seek this in the scale of long measure, you shall find it to stand at the letter o, against which in the scale of 1000 you shall find 437, which is the decimal an­swering to 1 quarter & 3 nails of a Yard or Ell.

Also if 437 were a decimal given, & it were re­quired to find what quantity of yards or ells were represented thereby, if you look in the scale of 1000, for 437, you shall in the scale of long mea­sure find against it one quarter and 3 nails.

8 The eighth and last scale is of representative inches, the whole scale being divided into 12 e­qual parts and numbred by 1, 2, 3, &c. to 12, and those parts are again subdivided into halves, quarters, and half quarters, as Carpenters-Rules are usually divided.

[Page 256]Unto this Scale, (as unto all the other) there is joyned a Scale of 1000, this Scale will readily dis­cover what is the Decimal belonging to any number of Inches, halves or quarters, and the use is the same with the Scales before mentioned

Thus have I given you a briefe description of these Scales and the uses of them, and do now suppose my Reader to be perfectly acquainted with the way of numbring or counting upon them, wherefore I intend onely to give you a question or two in the most usual Rules of Arith­metick, and so conclude; for Decimal Arith­metick being already sufficiently explained, I shall not need to repeat the Rules (or the manner of working them) again, but give you one Ex­ample, by which the exactnesse and expedition of these Scales may the more evidently ap­pear, for when we work by Scales, it is supposed that we do not use Vulgar, but Decimal Arith­metick, and Addition, Subtraction, Multiplica­tion, Division, the Rule of Three, and indeed, all the other Rules of Arithmetick, are to be perfor­med, as is before taught, the Scale serving only to avoid Reduction.

Addition.

WHat Addition is, and the manner of working of it hath been already taught, both in the first and second Parts, we will now come to an Exam­ple, which let be in Addition of English Coin, and let the sums to be added be 36 l. 8 s. 8 d. 29 l. 0 s. 2 d. 31 l. 16 s. 9 d. and 6 l. 2 s. 5 d.

First,

36  
29  
31  
6  

set down 36 l. 29 l. 31 l. and 6 l. one un­der another, in such order as you see here in the mar­gent, drawing a line by the side of them as you see here done, and also a line under them.

This done, seeing that your first number to be set down to 36 l. is 8 s. 8 d. you must for the 8 s. (because two shillings, which we called a Decade, or the tenth part of a pound, is made the Integer, in the Scale of Money) set down 4, which is done by memorie, and after it make a comma. Then your next number to be set by 29 l. being 0 s. 2 d. for the 0 s.

364, 
290, 
318, 
6  

set down a Cypher, thirdly, for your number to be set by 31 l. being 16 s. 9 d. for the 16 s. set down 8 decades, with a comma [Page 258]after it, and lastly, the number to be set by 6 l. being 2 s. 5 d. for the 2 s. I set down 1 decade with a comma after it, and then will your work stand, as here you see.

Then take your scale in hand, and seeing your first number of pence are 8 d. look in your scale of money for 8 d. and against it in the scale of 1000, you shall finde 333, which set to 36 l. 8 s. behinde the comma, then your next number of pence being 2 d. look in your scale for 2 d. and against it in the scale of 1000, you shall finde 082, which set to 29 l. 0 s. behinde the com­ma.

364,333
290,083
318,376
61,208

Then your third number of pence being 9 d. look in your scale for 9 d. and against it in the scale of 1000, you shall finde 376, which set to 31 l. 16 s. and last­ly, your last number of pence being 3 d. look in your scale for 5 d. and against it you shall finde 208, which set to 6 l. 2 s. and then will your whole work stand, as here you see.

Your sums being thus set down, which is done with more facilitie then you can imagine, till you make trial and be something perfect therein, you must then adde all the numbers to­gether, as in Addition of Decimals, and you shall finde the sum of them to be 103|4,000, Now to know what this is in money, is as easie as it was to set the several sums down, for the figures 103, which stand behinde the down [Page 259]right line, are 103 l. and the figure 4 which stands between the down right line and the com­ma,

l.  
364,333
290,083
318,376
61,208
1034,000

are 4 decades or 8 s. and being the rest to the right hand are all Cy­phers, they signifie nei­ther pence nor farthings, so is the total of this addition 103 l. 8 s. 0 d.

That the manner of working may appear more plain. I will give you another short Ex­ample as difficult as I can invent, which I per­formed by a Scale of Wood but of 8 inches long. Let the sums to be added together, be these following.

li.t.d.q
3321741
159681
217533
709941

First, set down your several sums of pounds one under another as before, and draw a line by the side of them,

332 
159 
217 

and an­other under them. So will they stand as here you see.

1 Your sums of pounds being thus orderly placed and lines drawn, repair to your Scale and seeing your first number of shillings, pence and farthings is 17 s. 4 d. 1 q. for your 17 s. set [Page 260]down 8 Decads, which is 16 s. with a comma after it, then will there rest to be set down 1 s. 4 d. 1 q. or 16 d. 1 q. which if you seek in your scale of money, you shall find to stand against it in the scale of 1000 this number 677, which is the Decimal of 1 s. 4 d. 1 q.

2 Your second number of shillings pence and farthings is 6 s. 8 d. 1 q. for your 6 s. set down 3 decads, which is 6 s. and then there will re­main 8 d. 1 q. which if you seek in your scale of money, you shall find to stand against it in the scale of 1000 this number 344, which is the deci­mal of 8 d. 1 q.

3 Your third number of shillings, pence, and farthings, is 5 s. 3 d. 3 q. for your 5 s. set down 2 decads, which is 4 s. with a comma after it, then will there rest to be set down 1 s. 3 d. 3 q. or 15 d. 3 q. which if you seek in your scale of money, you shall finde to stand against it in the scale of 1000,

l. 
3328,677
1593,344
2172,656
7094,677

this num­ber 656, which is the De­cimal of 15 d. 3 q. or 1 s. 3 d. 3 q. and the three sums to be added together will stand as here you see.

These sums being added together, according to the rule for Addition of Decimals, you shall finde the sum of them to be 7094,677, now to know what this is in money, take notice that the 709 which stands to the left hand of the downright line are 709 pounds, and the figure 4, [Page 261]which stands between the downright line and the comma, are 4 decads or 8 s. but (because the first figure next after the comma is above 5, viz. 6) you must add 1 s. to the 4 decads, making them 9 s.; then will there remain 177, wherefore if you look in the Scale of 1000 for 177, you shall find against it in the Scale of Money 4 d. 1 q. So is the whole sum of this Addition 709 l. 9 s. 4 d. 1 q. as by the preceding work doth appear.

¶ Here note, that when you had set down your 709 l. 4 decads or 8 s. there remained beyond the comma 677, which if you had sought in your Scale of 1000, you should have found a­gainst it in the Scale of money 15 d. 1 q. or 1 s. 4 d. 1 q. (which is all one) as before, for it ap­peareth plainly by the Scale that 500 in the line of 1000 is equal to one shilling.

I might proceed farther in giving you Exam­ples in Weight and Measure answerable to the Scales, but that would only make the Reader spend his time to little purpose, for being before acquainted with Decimal Arithmetick, and (as by this time I suppose he is) with Numeration upon the Scales, he cannot be deficient in the ap­plying of the other Scales of Weight and Measure to the same purpose for which they were contrived, I having so largely exemplified the use of the Scale of money.

Subtraction.

SUbtraction (as hath been before said) is the taking one or more smaller sums out of one greater, I shall onely give you an Ex­ample or two, as I have taken the num­bers from a Scale.

Example.

Delivered to a Gold-smith of old plate 297 ounces 13 peny weight, 19 grains.

Received of the same Gold-smith first 165 ounces, 11 peny weight and 7 grains, and after that received of the same Gold-smith 32 onnces 19 penny weight, and 23 grains, what plate remains in the Gold-smiths bands?

Take your numbers out the scale of Troy weight, and set them down, as here you see.

 ounces 
Delivered1655,646
Received more329,979
Received in all1985,625
Reses in the Goldsmiths hands991,271

or 99 ounces — 2 peny w. — 13 gr.

[Page 263]Then adde the several weights of plate recei­ved, together, and they make 1985, 625, or 198 ounces, 11 peny weight, 6 grains, which if you subtract from 2976,896, or 257 ounces, 13 peny weight 19 grains, which was the quantity of plate delivered, there will remain 991271, or 99 ounces 2 peny weight, 13 grains, and so much plate is still in the Gold-smiths hand. And let thus much suffice for subtraction.

Now we should proceed to Multiplication and Division, but being when the numbers are taken from the scale and set down, the manner of wor­king doth not at all differ from Multiplication and Division of Decimals before taught. But that we are now a treating of Instrumental A­rithmetick (because I will not, alter the method before used, in the first and second parts of this Book;) I will therefore here shew you how Multiplication and Division, may be Instrumen­tally performed by Nepers Bones.

Multiplication by Nepers Bones.

MUltiplication and Division are account­ed the hardest parts of Vulgar Arith­metick; but by the invention of the Right Honorable John Lord Nepeir Baron of Merciston, they are both of them made very easie, and more certain then in the ordi­nary way, the memory not being charged at all, but as in Addition and Subtraction, what this Artifice is, is shewed by himself in his Trea­tise of Rabdologia, as also in that of Mr. Seth Pa­tridge and others, and therefore it will not be necessary to make any large discourse of it now, it will be sufficient here, onely to give you a de­scription of them, and how to use them.

These Rods or Bones are nothing else, but the ordinary Multiplication Table, commonly called Pythagor as his Table, cut into pieces, with Diagonal lines between the figures of every pro­duct in single multiplication; as by the figure following it doth appear.

[Page 265]

A Figure of the Bones.

The Bones are usually made of small pieces of Box, containing in length about 1 inch and 8/10 of an inch, and in breadth about 2/10 of an inch, and in thicknesse neer 1/10 of an inch. One set of these Bones consisteth of five pieces of Box, and are graduated with Diagonal lines and fi­gures, as you see here in the figure of them. On one of the five pieces there is 0 and 9, the 0 being on one side of the Wood, and the 9 on the other side, another hath 1 and 8. the third 2 and 7, the fourth 3 and 6, the fifth 4 and 5. [Page 266]Now these five pieces of Box contein one Ta­ble of Multiplication cut in pieces, but so few are not sufficient for use, for you cannot well have lesse then six sets of them, that is 30 pieces, and so many are usually made.

For the placing of these Bones, when you use them, there belongeth a piece of Wood about four inches in length, and two inches in breadth, having 2 ledges of the same thicknesse with the Bones, on the thin legde which is at the end of this board, there is figured the nine Digits, 1, 2, 3, 4, 5, 6, 7, 8, 9. This board is called a Tabulat, because we usually say when the bones are laid thereon for any the Bones are Tabulated. See the figure thereof.

[figure]

[Page 267]Now these Bones thus ordered and Tabulated doth not onely give you the product of every single Digit, multiplyed by it self, or any one of the Digits, but the several products of all the 9 Digits together by every of the 9 Digits, as if it were demanded how much 123456789, will a­mount unto, being multiplyed by 6, in this Table it will be found to be 740740734, for the product of 9 multiplyed by 6 is 54, the first figure whereof towards the right hand, viz. 4, is placed under the Diagonal line, and 5 being in the place of tens is placed above it, again 8 multiplyed by 6 is 48, the first figure whereof towards the right hand, viz. 8 is placed under the Diagonal line as in the other product, and these 2 figures 5 in the upper part of the former product, and 8 in the lower part of this must be added together, and being so they make 13, therefore next unto 4 in the first pro­duct, I set down 3, and carry 1 to 4 the figure pla­ced above the Diagonal in this second product, which together with the lower figure in the next product, viz. 2, do make 7, for 4 and 1 that I car­ried is 5 and 2 the figure under the Diagonal in the next product are 7, and so the 3 first figures of this product towards the right hand are 734, in like manner must the figures remaining between the Diagonals be added together, and their sum if it amount not to ten must be set down in his proper place, but if they make or exceed ten, set down the overplus, and carry the ten in mind to the next figures between the Diagonals as in the last product; So likewise in that which followes 4 [Page 268]and 6 the figures between the Diagonals being 10, I set down a Cypher, and carry one to the next figures, which are 3 and 0, now 4 and 1 that I carried is 4, and 4 and 0 is 4 still, and there­fore next to the Cypher I set down 4 in the paper; the figures in the next Diagonal are 3 and 3. that is 7, wherefore I set down 7 in the paper, the next are 2 and 8 that is 10, therefore I set down 0 in the paper and carry 1, the next are 1 and 2 which with 1 that I carried make 4, therefore I set down 4 in the paper, the last are 1 and 6 which make 7 and so at last the whole product of 123456789 multiplyed by 6 is 740740734.

Now as the multiplication Table thus ordered doth give the product of all the 9 Digits placed orderly by every of the 9 digits, so being cut in pieces or slips, and having many of them made in Past-board of Wood, any other sum may be placed on the Tabulat, and the several pro­ducts thereof by every of the 9 digits will be found in the like manner; thus if it were requi­red to multiply this number 5767 by every of the 9 digits, placing the slips with these figureson the top orderly as, they are in the following Figure, product of 5767 by 3 will be found to be 17301, the product thereof by 8 is 46136, and so for any other.

[Page 269]

[figure]

Also if it were required to multiply 5767 by 749, the product of 5767 by 9 is 51903, and the product thereof by 4 is 23068, and the product by 7 is 40369, [...] and these products being set down and added to­gether, as in ordina­ry Multiplication, the product of 5767 mul­tiplyed by 749 will be 4319483, as by the o­peration in the mar­gine it doth appear.

[Page 270]

[figure]

Of Division by the Rods.

LEt it be required to divide 4319483 by 5767 placing this Divisor on the Tabulat, as was directed for Multiplication, the pro­ducts of your Divisor by every of the 9 Digits is given you, and may be deducted from the Dividend, as hath been already shewed Pag. 55, and according to the Rules there delivered, I ask how often is 5767 in 43194, and the an­swer by my Table is 7 times, and therefore [Page 271]I subtract 40369 the product of 5767 by 7 from 43194, and there rests 2825, which with the next figure in the Dividend, viz. 8, is 28258, in which I find 5767 to be conteined 4 times, and therefore I subtract 23068 the product of 5767 by 4 from 28258, and there rests 5190, which with the next figure in the Dividend, viz. 3, is 51903, in which I find 5767 to be conteined nine times, [...] and therefore I subtract 51903 the product of 5767 by 9, from 51903 the remaining figures of the Dividend, and there resteth nothing, and therefore the quotient is 749, as by the operation it doth appear.

And thus may you doe with any other sum (having a competent number of Bones or Rods ready for the work) without any charge to the memory, which in other wayes cannot be avoided, and by reason of which many mi­stakes may happen, which are in this prevent­ed. Also by the Bones there is this commodity, [Page 272]that when it is doubtfull what figure to set in the Quotient, the Bones will certainly direct you, so that you cannot possibly mistake.

These Rods, or Bones (as they are common­ly called) are of excellent use in extracting the Square and Cube Roots, but that being already taught in the first Part, I shall not here insert it again, for he that can extract the Square and Cube Roots any way, cannot be ignorant to ef­fect the same by the Bones, if he can multiply and divide by them.

Having done with the four Species of Arith­metick in this instrumental way, I should now proceed to the Golden Rule of three, and conse­quently to the other Rules of Arithmetick, but having in the former parts of this Book perform­ed that already, both in whole numbers, Fracti­ons, and Decimals, I shall onely give you an Example in the most necessary Rules, and so conclude this third part.

Example in the Rule of Three Direct.

IF 37 Ells and a half of linnen cloth, cost 24 l. 7 s. 9 d. what shall 283 Ells and an half cost?

First, set down your 37 Ells, then if you look in your Scale for your half Ell, you shall find it to stand against 500 in the Scale of 1000, which 500 may be called 5 onely, for the two Cyphers may be omitted. Then set down your 24 l. and for your 7 set down 3 decads, which is 6 s. and look in your Scale for the Decimal of 1 s. 9 d. which you shall find to be 875; Lastly, set down your 183 Ells, and for your half Ell set down 5 as before, so will your numbers stand thus, [...]

[Page 274] [...]

Your numbers thus taken out of your Scale, and placed as here you see, if you multiply the second and third together, you shall finde the product of that multiplication to be 4475.10625, which divided by the first number 37.5, giveth in the quotient 119.3361, which is 119 pounds, [Page 275]3 decads, or 6 shillings, and 361, which reduced by the Scale giveth 8 pence 2 farthings, and something more.

Example in the Rule of Three Reverse.

IF when the price of a Quarter of Wheat is 1 li. 5 s. 6 d. the penny white loaf shall weigh 12 ounces 16 penny weight, I demand what the penny white loaf shall weigh, when the price of the Quarter of Wheat is 7 s. 6 d?

If you place your numbers according to the tenor of the Question, they will stand as fol­loweth.

1 li. — 5 s. — 6 d. — 12 où. — 16 pw. — 7 s. — 6 d.

But being taken out of the Scales of Money an Troy Weight, they will stand thus, [Page 276] [...]

Here if you multiply 1.275 which is the Deci­mal of 1 l. 5 s. 6 d. by 12.8. which is the Deci­mal of 12 ounces 16 peny weight, you shall find the product of that multiplication to be 16.3200, which being divided by .375, which is the Deci­mal of 7 s. 6 d. the quotient will be 43.5, which is the Decimal of 43 ounces, 10 peny weight 3 grains, and so much ought the peny white loafe to weigh, when the quarter of Wheat is sold for 7 s. 6 d.

Example in the Double Rule of Three.

IF 24 yards of stuffe of three quarters broad cost 4 l. 14 s. what shall 328 yards of the same stuffe cost be­ing 5 quarters broad.

If you place your numbers according to the directions of this Rule, they will stand thus, If 24 yards of 3 quarters cost 4 li. 14 s. what shall 328 yards cost of 5 q.

But if you take your fraction numbers out of their proper Scales, they will stand thus, [...] [Page 278] [...]

First, multiply the two first numbers, as 24 and 3 together, they make 72 for Divisor, then mul­tiply 4.7, which is the Decimal of 4 l. 14 s. by 328, and the Product is 15416, which again mul­tiplied by 5 the last number, giveth 77080, unto this Product, (that there may be a competent number of figures in the quotient,) I adde two Cyphers, making it 7708000, which I divide by 72, and the quotient is 107.055, which is 107 l. 1 s. 1 d. and so much is 328 yards of stuffe worth, being 5 quarters broad.

Example in Barter.

TWo Merchants having two several commodities, are willing to barter, or exchange the one with the other. The one hath Indigo, which he will sell at 4 s. the pound, for ready money, but in Barter he will have 4 s. 9 d. the pound; the other Merchant hath Kersies, which for ready money he will sell for 3 s. 6 d. the yard. Now the question is, at what price he must rate his Kersies in Barter, to equalize the 9 d. advance upon the pound of Indigo.

The tenor of the Question is this.

If 4 s. in Barter require 9 d. what shall 3 s. 6 d. re­quire?

Your numbers placed will stand thus, 4 s. — 9 d. 3 s. 6 d.

But being taken out of your Scale they will stand thus, [Page 280] [...]

Say then by the Rule of three Direct, If 2 decads or 4 s. in Barter require. 357, which is the Decimal of 9 d. what shall 1.57 require? which is the Decimal of 3 s. 6 d.

First, multiply .375 by 1.57, the product is .65625, (but being it is a fraction I cut off the two last figures because we require onely three figures in the quotient, which divided by 2 giveth in the quotient .328, which is the Decimal of 7 d. 3 q. this 7 d. 3 q. added to his 3 s. 6 d. maketh 4 shillings 1 peny 3 farthings, and so much ought he to rate his Kersies at by the yard in Bar­ter to save himself harmlesse.

Example in Fellowship.

THree Persons A, B, and C bought 4000 Sheep, which cost 48 [...] li. 6 sh. 8 d. of which money A paid 203 li. B paid 165 li. 15 sh. 8 d. and C paid 114 l. 11 s.

First, say by the Rule of three Direct.

1 If 483 l. 6 s. 8 d. buy 4000 sheep, how many sheep shall 203 l. (which is A share) buy? Answer 1680.

2 Say, if 483 l. 6 s. 8 d. buy 4000 sheep, how many sheep shall 165 l. 15 s. 8 d. (which is B share) buy? Answer 1372.

3 Say again, if 483 l. 6 s. 8 d. buy 4000 sheep how many sheep shall 114 li. 11 s. (which is C share) buy? Answer 948.

Your numbers being taken out of your Scale, proceed as followeth.

[Page 282] First for A.

[...]

Secondly for B.

[...]

[Page 283] Thirdly for C.

[...]

The manner of Work.

1 For A multiply 203 l. (which is A share) by 4000, (which is the number of sheep bought) and the product is 812000, which number should be divided by 483.3333, but being it is greater then 812000, I therefore add four Cyphers thereto, that I may have four figures in the quotient, and it makes 8120000000, which divided by 483.3333 giveth in the quotient 1680, and so many sheep belong to A.

2 For B, multiply 165.7833 (which is the De­cimal of B share) by 4000, (the number of sheep bought) and it produceth 6631332000, which divided by 483.3333, giveth in the quotient 1372, and so many sheep belong to B.

3 For C, multiply 114.55, (which is the De­eimal of C share,) by 4000, (the number of sheep bought,) it produceth 45820000, which number should be divided by 483.3333, but be­ing it is not large enough to give figures enough in the quotient, I therefore adde two Cyphers making it 4582000000, which divided by 483.3333, giveth in the quotient 948, and so many sheep ought Cto have.

Now for proof,

A1680
B1372
C948
 4000

if you add the number of sheep that A, B and C should severally have, you shall find them in all to make 4000 which demonstrates the work to be true.

Examples in Losse and Grain.

IF one yard of Stuffe cost 6 s. 8 d. and I sell the same again for 8 s. 6 d. what shall I gain in laying out 100 l. upon such commodity?

Take the difference between the price that your commodity cost, and the price for which you sell it, that is, in this Example, the diffe­rence between 6 s. 8 d. and 8 s. 6 d. which is 1 s. 10 d, then say by the Rule of Three Direct.

If 6 s. 8 d. gain 1 s. 10 d. what will 100 li. gain?

If you place your numbers according to the Rule of Three Direct, as they are here given, they will stand thus, 6 s. 8 d. — 1 s. 10 d. 100 li.

But being taken out of your Scale and placed, they will stand as followeth.

[Page 286] [...]

Your numbers being placed, multiply 425 which is the Decimal of 1 s. 10 d. by 100 li. and the Product is 42500, to which I adde two Cyphers (that I may have a competent number of figures in the quotient) and it makes 42500| [...]0, which divided by .3333 the decimal of 6 s. 8 d. giveth in quotient 127.5, which is 127 li. 5 livers or 10 s. so there is 27 li. 10 s. gained in lay­ing out of 100 l.

I will here prove this question by the con­verse.

If one yard of Stuffe which is sold for 8 s. 6 d. [Page 287]there was gained 27 li. 10 s. in laying out of 100 li. I demand what the said stuffe cost a yard at the first hand?

Adde 100 li. and 27 li. 10 s. together, and they make 127 li. 10 s. Then say by the Rule of Three Direct,

If 1 27 li. 10 s. give 100 l. what shall 8 s. 6 d. give?

Take your numbers out of your Scale and place them as here you see.

[...]

Here if you multiply 425, which is the De­cimal of 8 s. 6 d. by 100, you shall have 42500, to which if you adde a Cypher, you make it 425000, this number being divided by 127.5, which is the Decimal of 127 l. 10 s, giveth in the quotient 3333, and if you had added more Cy­phers to the Dividend, you should have had [Page 288]more threes in the quotient, and no other fi­gures, but these four threes are enough, and are a Decimal fraction representing 6 s. 8 d. and so much did the yard of Stuffe cost at the first hand.

Examples in Losse and Gain upon Time wrought by the Double Rule of Three.

IF one Ell of Lockeram, cost me 2 s. 8 d. ready money, and I sell the same again for 2 s. 10 d. the Ell to be paid at the expiration of three moneths, I demand what I shall gain in 12 moneths, laying out 100 l. upon that commodity.

This and such like questions, although they may be wrought by the Rule of Three Direct, at two operations, yet they are best performed by the Double Rule of Three compounded of five numbers, wherefore the question may be thus stated.

If 2 s. 8 d. in three moneths, gain 2 d. what shall 100 l. gain in 11 moneths?

[Page 289]If you take your numbers out of your scale▪ and place them according as was directed in the first part of this Book, you shall finde them to stand thus, [...]

Your numbers being placed according to the tenor of the question, if you multiply 1.333, which is the Decimal of 2 s. 8 d. by 3 moneths, the product will be 3999, which must be your Divisor, then multiply .83 which is the Deci­mal of 2 d. by 100 l. and it makes 8300, that [Page 290]again multiplied by 12 moneths, giveth for the product 99600 for your Dividend, wherefore if you divide 99600 by 3999, it will give you in the quotient 25 almost, which is 25 li. for the Decimal fraction remaining is so small, that it wanteth not neer a farthing of 25 li. and there­fore we call it 25 li, and so it is exactly, as you may try, if you reduce all the numbers to their least denominations, and work as is before taught in Vulgar Arithmetick.

I will prove this question by the converse.

If one Ell of Lockerum cost me 2 s. 8 d. ready mo­ney, for what price shall I sell the same again to be paid at the end of three moneths. So that I may gain 25 li. in 100 li. for 12 moneths?

Say by the Rule of Three Direct.

If 100 li. in 12 moneths gain 25 li. what shall 2 s. 8 d. gain in 3 moneths?

If you take your numbers out of your Scale, and place them according to the Double Rule of Three, they will stand as followeth.

[Page 291] [...]

Your numbers being thus placed, if you multiply 100 li. by 12 moneths, you shall finde the product to be 1200, which is your Divisor. Then multiply 25 li. by 1.333, which is the De­cimal of 2 s. 8 d. and the product thereof will be 33325, which multiply again by 3, and the product will be 99975 for your Dividend, this 99975 divided by 1200, giveth in the quotient. 83, which is the Decimal of 2 d, which 2 d. added to 2 s. 8 d. the price which the Ell of Lockeram cost, giveth 2 s. 10 d. and at that price must you sell the same at 3 moneths [Page 292]time so that you may gain 25 l. in the 100 li. in 12 moneth [...].

I might further proceed to shew you Exam­ples in divers other Rules; As in Alligation, Po­sition, &c. but those Rules being already hand­led in the First Part of this Book, it will be easie to apply them to the scales, And (as I inti­mated at the beginning of this Third Part,) that although I have onely made choice of the eight scales there mentioned and described, yet it was easie to contrive scales for the Coins, Weights, and Measures of other Countries, and not onely so, but when the value of the Money, Weight, or Measure of one Countrey, and the Money, Weight, or Measure of another Countrey is known, it is easie to contrive two scales, which facing one another, shall immediately tell you how many pounds in one place shall be equal to so many Crowns, or other Coin in another place, but this I do onely intimate, that such as are de­sirous, may fit themselves with scales answerable to their most necessary occasions. And thus shall I conclude this third part, referring some things necessary to such a work, to the Appendix fol­lowing.

An Appendix.

Section 1. Of Exchange of the Coins, Weights, and Measures of one Countrey, with the Goins, Weights and Mea­sures of another Countrey.

TO perform this work, there is nothing required more then the Golden Rule, if first the Rate or Proportion between the Coins, Weights, and Measures of any two Countries be known which is best obteined by experience, rather then taken upon trust, all that in this place I shall do, is onely to instruct the ingenious in the manner of Work, and make use of such Rates or Pro­portions as I find set down by Mr. Lewis Roberts Merchant, in his Map of Commerce.

Question 1.

How many Riders (each Rider containing [Page 294]1 l. 1 s. 2 d. 2 q. Sterling) shall I receive for 251 l. 6 s. 4 d. 2 q. Sterling?

Facit 237 Riders.

If 1 l. 1 s.d. give 1 Rider how many Riders shall 251 li. 6 s.d. give?

Here if you reduce your numbers to their least Denominations, or set them down in De­cimals, and multiply and divide according to the Golden Rule, you shall find in your quoti­ent 237, and so many Riders ought to be recei­ved for 251 l. 6 s. 4 d. 2 q. Sterling.

Question 2.

How many French Crowns (each French Crown being valued at 6 s. Sterling) shall I receive for 492 l. 18 s. Sterling?

Facit 1643 French Crowns.

If 6 s. give 1 F.C. what shall 492 li. 18 s. give.

Multiply and divide according to the Golden Rule, and you shall have in your quotient 1643, and so many French Crowns are to be received for 492 l. 18 s. Sterling.

Question 3.

A Merchant delivered at Paris 1643 Crowns of 6 s. Sterling the piece, how many pounds Sterling ought to be received at London?

Answer 492 l. 18 s. Sterling.

If 1 Crown give 6 s. what shall 1643 Crowns give?

Multiply and divide, and you shall have in your quotient 492 l. 18 s. and so much Sterling Money ought to be delivered at London for 1643 French Crowns of 6 s. the Crown Sterling.

Question 4.

If 3 yards at London, be 4 Ells at Antwerpe, how many yards at London make 84 Ells at Ant­werpe?

Ells Antwerpyards LondonElls Ant.
4384

Facit. 63.

And so many yards at London, are equal to 84 Ells at Antwerp.

Question 5.

How many yards of London make 27 Ells of [Page 296]Antwerpe, when 100 Ells of Antwerpe make 60 Ells of Lions, and 20 Ells of Lions make 25 yards of London?

The first work.

[...]

75 facit 75

That is 75 yards of London is equal to 100 Ells of Antwerp.

The second work.

[...]

Question 6.

If 100 li. Sterling be 104 l. 6 s. 4 d. Flemish what is one pound Sterling worth?

[Page 297] [...]

Facit 322 ponce 36/100 of a peny.

Question 7.

How many Ells of Frankford make 42¼ Ells of Vienna in Austria, when 35 Ells of Vienna make 24 at Lions; 3 Ells of Lions, 5 Ells of Antwerp, and 100 Ells of Antwerp, 125 Ells at Frankford.

First work.
Ells Ant.Ells Frank.Ells Ant.
1001.255
 5 
 625 

Facit 6.25 or 6 Ells and a quarter of Frankford equal to 3 Ells of Lions.

[Page 298] Second Work.

[...]

Facit 50 Ells of Frankford, equal to 35 Ells of Vienna.

[...]

[Page 299] Facit 60.35 or 6 5/14 Ells of Frankford, equal to 35 Ells of Vienna.

Thus have I given you a few Examples of Exchanges, I will now insert some few Tables derived from Mr. Lewis Roberts his Map of Commerce afore-said, of the truth of which, I am not a competent Judge, but shall leave that to the scrutiny of such as have occasion to trade into Forreign Countries.

A Table shewing what one pound of Avoirdupois weight at London, maketh in divers eminent Cities, and other remarkable places.

One pound of Avoir­dupois weight at London, makes at

  • ANtwerpe .9615 lb.
  • Amsterdam .9 lb.
  • Abbeville .91 lb.
  • Ancona 1.282 lb.
  • Avignon 1.12 lb.
  • Burdeaux .91 lb.
  • Burgoyne .91 lb.
  • Bollonia 1.25 lb.
  • Bridges .98 lb.
  • Calabria 1.3698 lb.
  • Callais 1.07 lb.
  • Constanti­nople .8474 lb.
  • Constanti­nople Loder; lb.
  • Deepe .91 lb.
  • Dansick 1.16 lb.
  • Ferrara 1.3333 lb.
  • Florence 1.282 lb.
  • Flanders in general 1.06 lb.
  • Geneva .9345 lb.
  • [Page 301]Genoa 1.4084 suttle lb.
  • Genoa 1.4285 grosse lb.
  • Hamburg .92 lb.
  • Holland .95 lb.
  • Lixborne .881 lb.
  • Lions 1.07 common weight lb.
  • Lions .98 silke weight lb.
  • Lions .9 customers weight lb.
  • Legorn 1.3333 lb.
  • Millan 1.4285 lb.
  • Mirandola 1.3333 lb.
  • Norimberg .88 lb.
  • Naples 1.4084 lb.
  • Paris .89 lb.
  • Prague .83 lb.
  • Placentia 1.3888 lb.
  • Rochell 1.12 lb.
  • Rome 1.27 lb.
  • Rovan .875 by Vicont lb.
  • Rovan .9017 common weight lb.
  • Sivill 1.08 lb.
  • Tholousa 1.12 lb.
  • Turin 1.2195 lb.
  • Venetia 1.5625 suttle lb.
  • Venetia .9433 grosse lb.
  • Vienna .813 lb.
The use of the preceding Table.

How much weight at Bollonia, will 655 l. Avoir­dupois make?

Look in the Table for Bollonia, and right a­gainst it you shall find 1.25, which sheweth that one pound Avoirdupois at London is equal to 1.25 l. at Bollonia, Therefore say by the Rule of Three:

If 1 l. Avoirdupois give 1.25 l. at Bollonia, what shall 655 li. Avoirdupois give? Answer 818.75. As by the operation following doth appear.

[...]

A Table shewing what one pound Weight in divers forreign Cities, and remarkable places, maketh at London of Avoirdupois Weight.

  • One pound weight in ANt werpe makes at London of Avoirdupois weight. 1.04 lb.
  • One pound weight in Amsterdam makes at London of Avoirdupois weight. 1.1111 lb.
  • One pound weight in Abbeville makes at London of Avoirdupois weight. 1.0989 lb.
  • One pound weight in Ancona makes at London of Avoirdupois weight. .78 lb.
  • One pound weight in Avignon makes at London of Avoirdupois weight. 8928 lb.
  • One pound weight in Burdeaux makes at London of Avoirdupois weight. 1.0989 lb.
  • One pound weight in Burgoyne makes at London of Avoirdupois weight. 1.0980 lb.
  • One pound weight in Bollonia makes at London of Avoirdupois weight. .8 lb.
  • One pound weight in Bridges makes at London of Avoirdupois weight. 1.0204 lb.
  • One pound weight in Calabria makes at London of Avoirdupois weight. .73 lb.
  • One pound weight in Calais makes at London of Avoirdupois weight. .9345 lb.
  • One pound weight in Deepe makes at London of Avoirdupois weight. 1.0989 lb.
  • One pound weight in Dansick makes at London of Avoirdupois weight. .862 lb.
  • One pound weight in Ferrara makes at London of Avoirdupois weight. .75 lb.
  • One pound weight in Florence makes at London of Avoirdupois weight. .78 lb.
  • One pound weight in Flanders in general makes at London of Avoirdupois weight. .9433 lb.
  • One pound weight in Geneva makes at London of Avoirdupois weight. 1.07 lb.
  • One pound weight in Genoa suttle makes at London of Avoirdupois weight. .71 lb.
  • One pound weight in Genoa grosse makes at London of Avoirdupois weight. .7 lb.
  • [Page 304] One pound weight in Hamburg makes at London of Avoirdupois weight. 1.0865 lb.
  • One pound weight in Holland makes at London of Avoirdupois weight. 1.0526 lb.
  • One pound weight in Lixborne makes at London of Avoirdupois weight. 1.135 lb.
  • One pound weight in Lions common weight makes at London of Avoirdupois weight. .9345 lb.
  • One pound weight in Lions silk weight makes at London of Avoirdupois weight. 1.0204 lb.
  • One pound weight in Lions custom weight makes at London of Avoirdupois weight. 1.1111 lb.
  • One pound weight in Legorne makes at London of Avoirdupois weight. .75 lb.
  • One pound weight in Millan makes at London of Avoirdupois weight. .7 lb.
  • One pound weight in Mirandola makes at London of Avoirdupois weight. .75 lb.
  • One pound weight in Norimberg makes at London of Avoirdupois weight. 1.136 lb.
  • One pound weight in Naples makes at London of Avoirdupois weight. .71 lb.
  • One pound weight in Paris makes at London of Avoirdupois weight. 1.1235 lb.
  • One pound weight in Prague makes at London of Avoirdupois weight. 1.2048 lb.
  • One pound weight in Placentia makes at London of Avoirdupois weight. .72 lb.
  • One pound weight in Rotchel makes at London of Avoirdupois weight. .8928 lb.
  • One pound weight in Rome makes at London of Avoirdupois weight. .7874 lb.
  • One pound weight in Rovan by Vicont, makes at London of Avoirdupois weight. 1.1428 lb.
  • One pound weight in Rovan common weight makes at London of Avoirdupois weight. 1.1089 lb.
  • One pound weight in Sivil makes at London of Avoirdupois weight. .9259 lb.
  • One pound weight in Tholousa makes at London of Avoirdupois weight. .8928 lb.
  • One pound weight in Turin makes at London of Avoirdupois weight. .82 lb.
  • One pound weight in Venetia suttle, makes at London of Avoirdupois weight. .64 lb.
  • One pound weight in Venetia grosse: makes at London of Avoirdupois weight. 1.06 lb.
  • One pound weight in Vienna makes at London of Avoirdupois weight. 1.23 lb.
The use of the foregoing Table.

In 7652 li. weight at Mirandola, how many pound Weight at Avoirdupois.

Look in the Table for Mirandola, and right against it you shall find .75, which sheweth that one pound Avoirdupois is equal to the 75 or ¾ of a pound at Mirandola, wherefore say by the Rule of Three.

If 1 l. at Mirandola, gives .75 or ¾ of a pound Avoirdupois, what shall 7652 l. of Mirandola give? Answer 5739, as by the operation doth appear.

[...]

A Table reducing English Ells to the Measures of divers forreign Cities and remarkable places.

  • One Ell at London makes at AMsterdam 1.6949 Ells
  • One Ell at London makes at Antwerpe 1.6666 Ells
  • One Ell at London makes at Bridges 1.64 Ells
  • One Ell at London makes at Arras 1.65 Ells
  • One Ell at London makes at Norimberg 1.74 Ells
  • One Ell at London makes at Colen 2.08 Ells
  • One Ell at London makes at Lisle 1.66 Ells
  • One Ell at London makes at Mastrich 1.57 Ells
  • One Ell at London makes at Frankford 2.0866 Ells
  • One Ell at London makes at Dansick 1.3833 Ells
  • One Ell at London makes at Vienna 1.45 Aulns
  • One Ell at London makes at Paris .95 Aulns
  • One Ell at London makes at Rovan 1.03 Aulns
  • One Ell at London makes at Lions 1.0166 Aulns
  • One Ell at London makes at Callais 1.57 Aulns
  • One Ell at London makes at Venice linnen 1.8 Braces
  • One Ell at London makes at Venice silke 1.96 Braces
  • One Ell at London makes at Lucques 2. Braces
  • One Ell at London makes at Florence 2.04 Braces
  • One Ell at London makes at Milan 2.3 Braces
  • One Ell at London makes at Legorn 2. Braces
  • One Ell at London makes at Madera 1.0328 Braces
  • One Ell at London makes at Isles 1.0328 Braces
  • [Page 307] One Ell at London makes at Sivill 1.35 Vares
  • One Ell at London makes at Lisbone 1. Vares
  • One Ell at London makes at Castillia 1.3875 Vares
  • One Ell at London makes at Andoluzia 1.4625 Vares
  • One Ell at London makes at Granado 1.3625 Vares
  • One Ell at London makes at Genoa 4.8083 Palmes
  • One Ell at London makes at Saragosa .55 Canes
  • One Ell at London makes at Rome .56 Canes
  • One Ell at London makes at Barselona .7125 Canes
  • One Ell at London makes at Valentia 1.2125 Canes
The use of this Table.

In 632 Ells at London, how many Braces at Florence?

Look in the Table for Florence, and right a­gainst it you shall find 2.04, which sheweth that one Ell at London, maketh at Florence 2.204 Bra­ces, wherefore say by the Rule of Three.

If one Ell at London, give 2.04 Braces at Flo­rence, how many Braces shall 632 Ells give? An­swer 1289.28, as by the operation following doth appear.

[Page 308] [...]

A Table reducing the Measures of divers forreign Cities and remarkable places, to English Ells.

  • One Ell at AMsterdam makes at London .59 Ells
  • One Ell at Antwerp makes at London .6 Ells
  • One Ell at Bridges makes at London .6097 Ells
  • One Ell at Arras makes at London .606 Ells
  • One Ell at Norimberg makes at London .5474 Ells
  • One Ell at Colen makes at London .4807 Ells
  • One Ell at Lisle makes at London .6024 Ells
  • One Ell at Mastright makes at London .6369 Ells
  • One Ell at Frankford makes at London .4792 Ells
  • One Ell at Dansick makes at London .7228 Ells
  • One Ell at Vienna makes at London .9896 Ells
  • One Auln at Paris makes at London 1.0526 Ells
  • One Auln at Rovan makes at London .9708 Ells
  • One Auln at Lions makes at London .9836 Ells
  • One Auln at Callais makes at London .6369 Ells
  • One Brace at Venice linnen makes at London .5555 Ells
  • One Brace at Venice silke makes at London .5102 Ells
  • One Brace at Lucques makes at London .5 Ells
  • One Brace at Florence makes at London .4901 Ells
  • One Brace at Millan makes at London .4347 Ells
  • One Brace at Leghorn makes at London .5 Ells
  • One Brace at Madera Isles makes at London .9681 Ells
  • [Page 310] One Vare at Sivil makes at London .7409
  • One Vare at Lisbone makes at London 1.
  • One Vare at Castilia makes at London .7207
  • One Vare at Andolusia makes at London .7339
  • One Vare at Granado makes at London .7339
  • One Palm at Genoa makes at London .2079
  • One Cane at Saragosa makes at London 1.8181
  • One Cane at Rome makes at London 1.7857
  • One Cane at Barselona makes at London 1.4035
  • One Cane at Valentia makes at London .8247
The use of this Table.

In 5727 Braces at Ligorn, how many Ells English.

Look in the Table for Ligorn, and right a­gainst it you shall find .5, which sheweth that one Brace at Ligorn maketh at London. or half an Ell, wherefore say by the Rule of Three.

If one Brace at Ligorn give .5 Ells at London, what shall 5727 Braces give? Answer 2863.5, as by the work appeareth.

[...]

Section 2. Concerning Interest and Annuities.

The first Table shewing what one pound being forborn any number of years under 31, will amount unto, accounting interest upon interest, after the rate of 6 per cent.

Y.6 per cent.
11,06
21,1236
31,19101
41,26247
51,33822
61,41851
71,50363
81,59384
91,68947
101,79084
111,89829
122,01219
132,13292
142,26090
152,39655
162,54035
172,69277
182,85433
193,02559
203,20713
213,29056
223,60353
234,81975
244,04893
254,29187
264,54938
274,82234
285,11168
295,41838
305,74349

[Page 312]The first Column of this Table having Y at the top thereof, beginning at 1, and so proceeding to 30, signifie years, and the number in the next Column answering thereunto do shew what one pound is worth, being forborn any number of years under 31, which Table is made according to this proportion.

As 100 to 106, so is 1 to 1.06 and again, As 100 — 106 — 106 — 1.1236 and thirdly. As 100 — 106 — 1.1236 — 1.19101 Et sic ad infinitum.

The use of this Table.

What 136 l. 15 s. 6 d. amount unto, being forborn 20 years, after the rate of 6 per centum, interest upon enterest.

Look in the Table for 20 years, and right a­gainst in the broader Column, you shall find 2.20713, which shewes that one pound being forborn 20 years will be augmented to 3.20713. Then if you reduce your 136 li. 15 s. 6 d. into a Decimal, either by the Tables in the Second Part, or by the Scales in the Third Part of this [Page 313]Book, you shall find it to be 136.775. Where­fore say by the Rule of Three Direct.

If one pound being forborn 20 years will be aug­mented to 3.20713, to how much will 136.775 li. be augmented to in the same time. Answer, to 438 li. 13 s. 1 d. 1 q. as by the operation following doth appear.

[...] Or 438 li. 13 s. 1 d. 1 q.

The Second Table, sheweth what one pound Annuity will amount unto, being forborn any number of years under 31, at 6 per cent. interest upon interest, the Annuity being to be paid yearly.

Y.6 per cent.
11,00000
22,06000
33,18360
44,37461
55,63709
66,97531
78,39383
89,89746
911,49131
1013,18079
1114,97164
1216,86994
1318,88213
1421,01506
1523,27596
1625,67252
1728,21287
1830,90565
1933,75999
2036,78559
2139,99272
2243,39228
2346,99582
2450,81557
2554,86451
2659,15638
2763,70576
2868,52810
2973,63979
3079,05818
The use of this Table.

What will an Annuity of 20 li. payable yearly, be augmented unto in 12 years, being all that time forborn [Page 315]accounting interest upon interest at 6 per cent. per annum.

Look in the first column of the Table for 12 years, and right against it in the next column you shall find 16,86994, which shewes that 1 li. Annuity payable yearly, being forborn 12 years will amount unto 16.86994, wherefore say by the Rule of Three Direct.

If 1 pound Annuity forborn 12 years give 16.86994 what shall an Annuity of 20 pound a year give? being forborn the same term of 12 years? Answer 337 li. 7 s. 11 d. 3 q. fere, as in the operation doth appear.

[...]or 337 l. 7 s. 11 d. 3 q. ferè.

The third Table sheweth what one pound being forborn any number of years under 31 is worth in ready money, rebating yearly, after the rate of 6 per cent. interest upon in­terest.

Y6 per cent.
1,943396
2,809996
3,839619
4,792093
5,747258
6,704960
7,665057
8,627412
9,591898
10,558394
11,526787
12,496989
13,468839
14,442300
15,417263
16,393646
17,370364
18,351343
19,330512
20,311804
21,294155
22,277505
23,261797
24,246978
25,232998
26,219810
27,207367
28,195630
29,184536
30,174110
The making of the Table.

As — 106 — 100 — .943396 — 889996 [Page 317]and again, As 106 — 100 — .889996 — 839619

Et sic ad infinitum.

If 356 l. be payable at the end of 7 years, what is it worth in ready money, discounting or rebating after the rate of 6 per cent. interest upon interest.

Look in the Table for 7 years, and against it you shall find .665057, being the ready money which 1 l. is worth payable at 7 years end, wherefore say by the Rule of Three.

If 1 l. in 7 years rebate or decrease to .665057, to what will 356 l. rebate or decrease in the same time? Answer to 170 l. 5 s. 1 d. as by the operation doth appear.

[...]

The fourth Table sheweth the present worth of one pound Annuity; to continue any number of years under 31, and payable yearly after the rate of 6 per cent. interest upon interst.

Y.6 per cent.
10,94339
21,83339
32,67301
43,46510
54,21236
64,91732
75,58238
86,20979
96,80169
10736008
217,88687
228,38384
238,85268
249,29498
259,71224
2610,10589
2710,47725
2810,82760
2911,15811
3011,469 [...]2
2111,76305
2212,04158
2312,30337
2412,55035
2512,78335
2613,00316
2713,01053
2813,40616
2913,59071
3013,76482
The use of this Table.

What is the present Rent or Annuity of 25 pound per annum worth payable yearly, for 21 years, [Page 319]accounting Interest upon Interest at 6 per centum.

Look in the Table for 21 years, and right a­gainst it you shall find 11.76407, which is the present worth of one pound Annuity, for 21 years, wherefore say by the Rule of Three.

If an Annuity of 1 l. per annum for 21 years be worth 11.76407 ready money, what is an Annuity of 25 li. per annum worth in ready money for the same time? Answer 294 l. 2 s. 0 d. 1 q. as by the ope­ration following doth appear.

[...] or 294 l. 2 s. 0 d. 1 q.

The fifth Table sheweth what Annuity payable yearly, one pound will purchase for any num­ber of years under 31, after the rate of 6 per cent. compound interest.

Y6 per cent.
11.06000
2.54363
3.37411
4.28859
5.23739
6.20336
7.17913
8.16103
9.14702
10.13586
11.12679
12.11926
13.10297
14.10758
15.10296
16.09895
17.09544
18.09235
19.08962
20.08718
21.08500
22.08304
23.08127
24.07967
25.07822
26.07690
27.07569
28.07459
29.07357
30.07264
The use of this Table.

What Annuity to begin presently and to continue 28 years, pay­able at yearly payments, will 640 l. purchase, accounting com­pound Interest after the rate of 6 per cent.

Look in your Table for 28 years, and right a­gainst it in the next column you shall find .07459, which shewes that one pound ready money will purchase an annuity worth .07459, and to conti­nue 28 years, wherefore say by the Rule of three.

If one pound ready money, will purchase an Annuity worth .07459 to continue 28 years, what Annuity shall I purchase for the same time, paying 640 l. ready money? Answer 47 l. 14 s. 9 d. as by the operation doth appear.

[...]

FINIS.

The fourth Part: BEING AN ABRIDGEMENT OF THE PRECEPTS OF ALGEBRA
With divers Questions added, which were not in the Original. LONDON, Printed Anno, MDCLIX.
[Page] [Page 323] ALGEBRA BREVIS.

I Have been alwayes of opinion, that the practice of Algebra should not be entangled with a great number of precepts: This Science is of it self dark enough, without adding unto it new obscurity, by the confusion of many diffe­rent operations. You have here an Abridgement which hath pleased many of good judgements, and I hope, such as will with attention read it, shall from thence receive both satisfaction and profit.

I shall in the first place set forth a Table of three ranks: In the first of which there is a pro­gression natural, of which the terms are disposed in that order, that immediately under them you have the Cossick Characters, of which they are Exponents, and in the lowest Cell, a progressron Geometrical, which beginning with an Unite, may be doubled, trebled, quadrupled, &c. we have for the greater ease onely doubled them. Observe then that R is a Coffick Character sig­nifying a Root, and that the Exponent thereof is marked above in the uppermost rank, and C is [Page 324]the Cossick Character of a Cube, whose Expo­nent is 3, and so of the rest. I call those terms, which are in the upper Cell, Exponents; because they expound the Cossick Characters, and the Numbers of the Geometrical Progression which are below. Mark well this manner of speaking, consider diligently this Table,

Expo­nents Chara­cters CossickProgr. Geo­metr.
&c. &c&c
8QQ QSquare quare quare256
7S2Solide second128
6QC.Square Cube64
5SSur so­lide32
4QQBlqua­diate16
3CCube8
2QSquare4
1RRoor2
0NNum­bers abso­lute1

and for the pre­sent content your self with this.

You may continue this Table, if you please, infinitely in this man­ner. Take two num­bers, which if you mul­tiply, they will produce some Exponent, you shal presently see what a cha­racter is to be put under that Exponent. For ex­ample, If you would know the character of the Exponent 6, take the numbers 2 and 3 (be­cause these multiplied together make 6) after that adde their chara­cters which are Q and C, and you shall have Q C for the chara­cter of the Exponent 6. In like manner the cha­racter [Page 325]of the Exponent 8 is QQQ, because un­der the exponents 2 and 4, which multiplied to­gether produce 8, are contained the characters Q and QQ. Also the character of the exponent 12 will be Q Q C, because 12 is a number pro­duced by the multiplication of 2 and 6, or of 3 and 4.

But if the exponent be a first number (that is to say, a number not produced by the multiplicati­on of any two other) mark in what order it is af­ter the exponent 5, and call it sursolide second, third, or fourth, &c. according to its rank. The character of 5 is solide, of 7 solide second, of 11 solide 3; and so consequently under such expo­ponents as are first numbers, under which onely are found such solides.

CHAP. I. The Alegorithm of Cossick Numbers, simple, compounded, or diminished.

BY the word Alegorithm, I mean all the opera­tions comprehended under these four kinds, Addition, Subtraction, Multiplication, and Divi­sion. By the word Cossick simple, I understand such as have not this + (which signifies plus, nor this − signifying minus, expressed before them. On the other side, by Numbers composed, are meant such as have the signe +, and by dimi­nished, such as have the signe −. Note such [Page 326]numbers as have no signe expressed, are supposed to have this of +.

Sect. 1. Addition of simple Cossick numbers.

All simple cossick numbers are of the same de­nomination (that is to say, have the same chara­cter) or of different: If they be of the same deno­mination, the Addition is as in common Arith­metick. Example, 5Q added with 3Q, makes 8Q

If they be of different denominations, they must be added by the interposition of the figure +, as 6R added to 4Q makes 6R+4Q, in like manner 3 added to 4R makes 3+4R.

Sect. 2. Subtraction of simple Cossick numbers.

Either they are of the same, or different deno­minations: if of the same, you must subtract as in ordinary Arithmetick; for example, 3Q sub­tracted from 8Q, there rests 5Q.

If of different, you must subtract by the inter­position of the signe − as 6R subtracted from 4Q there rests 4Q−6R, so 3C subtracted from 66, there remains 66−3C.

Sect. 3. Multiplication of simple Cossick numbers.

You must here have regard both to the abso­lute number, and to the Cossick characters: If therefore a Cossick number be to be multiplied [Page 327]by an absolute, you must multiply the absolute numbers, and unto the product give the same character, as 5R multiplied by 12 produce 60R.

But if you multiply Cossick numbers by Cos­sick, you must multiply the absolute numbers to­gether, and to the product give the character of that exponent, which is made by the addition of the exponents belonging unto the foresaid Cos­sick characters. For example, 2R multiplied by 3Q make 6C, because the exponent of Q is 2 added to one the exponent of R, make 3 the ex­ponent of C, which in this respect ought to be gi­ven to the product. In like manner 5R multi­plied 4C it makes 20QQ for the reason above given.

Sect. 4. Division of simple Cossick numbers.

The speculation of this is marvellous. But the practice of it is by putting the Divisor under the number to be divided, drawing between them a little small line in manner of common fra­ctions. For example, 13Q divided by 7R, the quotient is 13q / 7R and 6QQ divided by 5C the quotient is 699/5C.

Sect. 5. Addition of numbers composed and diminished.

Some order is to be observed in this, in which the numbers are to be disposed in such manner, that those that are of the same denomination must be put right under one another. After you [Page 328]have done this, if they have the same signe, they are added as in common Arithmetick, and to the product give the same signe. As for exam­ple, 7Q−4C added to 3Q−2C, the Sum is 10Q−6C.

But if the numbers be of different signes, the lesser must be subtracted from the greater, and to the residue you must give the signe of the greater number as 6Q+7R added with 7Q−12R, give for the Sum total 13Q−5R.

6Q+7R
7Q−12R
13Q−5R

6Q−7R
7Q+12R
13Q+5R

Sect. 6. Subtraction of numbers composed and dimi­nished.

There is nothing more intricate to beginners than the precepts commonly given for subtracti­on. You have here an order plain, sure, and very casie to practise.

Change the sign of the particulars of that num­ber you desire to subtract, and after this change adde them with the number from which the sub­traction is to be made, and you shall have the re­sidue: As if from 6Q−10R you would sub­tract 18Q−15R by the 5 Sect. the residue will be 5R−12Q. So likewise if you would sub­tract −8R−9Q from 16R+6Q, the resi­due will be 24R+15Q

[Page 329]

6Q−10R Adde
−18Q−15R Adde
−12Q+5R Resid.

16R+6Q Adde
8R+9Q Adde
24R+15Q Residue.

Sect. 7. Multiplication of numbers composed and di­minished.

Mark what I said of Cossick simple in the 3 Sect. and remember that the same signes have alwayes the signe + in the product and diffe­rent − & there is no difficulty in multiplication, so as you multiply every particular of the mul­plicand by every particular of the multiplicator, as in common Arithmetick, as if you multiply 3Q−2R by 8R, multiply 2R by 8R, it makes 16Q, and because the multiplicator and the mul­tiplicand have different signes, the product must have the signe of −; and therefore that shall be 16Q, farther 3Q by 8R make 24C, to which you must give the signe +, because the multi­plicator and the multiplicand have the same sign, so that the product of this multiplication will be 24C−16Q. So 2R+4Q multiplied by 3Q−5 the product is 6C+12QQ−10R−20Q

3Q−2R multiplicand
8R multiplicator
24C−16Q Product.

2R+4Q mult.
3Q−5 multip.
−10R−20Q

6C+12QQ/6C+12QQ−10R−20Q

Sect. 8. Division of numbers composed and diminished.

There is no great difficulty in this, onely put a line between the Dividend and the Divisor, and you have the quotient, as 4C−3Q+2R di­vided by 5R−4C make for this quotient 4C−3Q+2R / 5R−4C

Sect. 9. Algorithm of Fractions.

I shall not give any particular precepts, because if a man understand the Fractions of common Arithmetick, and practise according to what is before said, there will be no need of them.

CHAP. II. The rule of Algebra, with the expli­cation thereof.

IT was meerly necessary by the precedent rules, to trace out insensibly the way to Alge­bra, which cannot be practised without Addi­tion, Subtraction, Multiplication and Divisi­on. Having therefore made plain these difficul­ties, we will proceed in the proposition of the rule of Algebra, and the explication of every part of it briefly and plainly.

Sect. 1. The rule of Algebra.

You must first for the number unknown put 1R, and after examine this root according to the tenour of the question, until you come to an Equation. Secondly, this Equation is to be redu­ced, if need require. Thirdly, you must divide every part of the Equation by the number of the greatest cossick character. After which, either the quotient, or some root of the quotient, will give the root unknown: This is the rule of Al­gebra, let us now explain it.

Sect. 2. How the Equation must be found.

The rule saith, this is done by examining the question propounded, according to the tenour of the same. That is to say, you must well observe all the conditions of the question propounded, to the end you may fully accomplish it. For after you have gone thorow it, you shall find an equa­tion between two numbers. As if I search a num­ber, which added with its square shall make 20, I suppose this number unknown to be 1R, the square thereof is 1Q. (because every number multiplied by it self, makes its square) then 1Q+1R is equal to 20. See thus an equation found between 1Q+1R and 20.

Sect. 3. How your Equation must be reduced.

Your Equation being found, it is reduced by adding the same number to both the terms of the equation, or by subtracting from them the same number. So is it performed by multiplying, or dividing both the terms by the same number. For by this means your equation shall remain the same after these things done. As for exam­ple, 1R+1Q=20, adding throughout 2C, you shall have also an equation between 1R+1Q+2C and 20+2C. So subtracting 1R from the termes, you have 1Q=20−1R. Likewise multiplying, or dividing both your terms by 3, you shall have by the multiplication 3R+3 Qequal to 60, and by the division the equation with be between on ⅓R+⅓Q and 20/3.

Now to make your reduction judiciously and profitably, you must take care alwayes that your greatest character remain alone on one side of your equation. As of all the reductions before made, there is none useful but the second, because in that onely you finde on the one side alone 1Q=20 − 1R which is the onely end of your reduction.

I said in the rule of Algebra, that your equa­tion must be reduced, if it be necessary; because it sometimes happens, that there is no need of it: as when your equation fals out between two sim­ple collateral numbers, I call those numbers col­lateral, whose exponents do not surpasse one an­other by more than an unite.

Sect. 4. When you must extract the root.

When your cossick numbers are simple and collateral, you must not extract any root; but if you divide by the number of the greatest cossick character, the quotient shall shew you the value of the root, which is all you seek for in Algebra. For example, If you finde an equation between 2R and 28, dividing simply 28 by 2 the quotient, shall be the value of 1R.

But when the terms of your equation are not collateral, you must extract some root; either square, cube, squared square, &c. according to the cossick character which remains after your hypobibasme.

Now hypobibasme is nothing else but an abate­ment, or depression of the character, and is done by subtraction of the letter exponent from the greater. As if you finde an equation between 10QC and 90 QQ take notice of the exponent of QC in the Table inserted at the beginning, the which exponent is 6; afterward look to the exponent of QQ which is 4, subtract 4 out of 6 there rests 2, of which the cossick character is Q. From hence I conclude, that 10Q are equal to 90, after dividing 90 by 10, and finding 9 in the quotient, I conclude, that the square root of 9 must be extracted by reason of the character Q.

Sect. 5. How to extract the square root of numbers, compound and diminished.

No man hath yet perfectly found out the way to extract the root of numbers compound and di­minished, unlesse the exponents of three terms of the equation keep between them in some situa­tion or other an Arithmetical proportion, that is to say, the same distance. As if the equation be between 1Q and 20−1R, you may now ex­tract the root of 20−1R, because the exponents of the three numbers which make the equation, are 2, 0, 1, which thus place, 0 1, 2, keep the same distance.

The greatest cossick character left after the Hy­pobibasme shews the root to be extracted; as in this example, before the square root is to be extracted, because the greatest character is Q. The method to be followed in the extraction, see here propounded in general terms.

Take first the half of the number of roots. Se­condly, to the square of this half adde, or from it subduct your absolute number according to the figne of + or −. Thirdly, extract the square root of this sum, or of the residue. Fourthly, to this root adde, or from it subtract half the num­ber of roots, and this last sum or residue shews you the value of the root unknown. For exam­ple, I would find a number, the double whereof added to its square, should be equal to 24, I shall find an equation between 2R+1Q and 24 by [Page 335]the second Section. Moreover; I shall reduce this equation after this manner 1Q equal to 24−2R by the third Section. Then if I di­vide 24−2R by 1, the number of the greatest cossick character there still remains 24−2R, because an unite doth neither multiply nor di­vide. Then in as much as the three terms of the equation do keep an Arithmetical proportion, I extract the square root of 24−2R in this man­ner. I take first half the number of roots, which is 1. Secondly, the square of 1 is 1, to which I adde the absolute number, which is 24, because of the signe + before it, that makes 25. Thirdly, I extract the square root of 25, which is 5. Fourth­ly, from this root I extract the moity of the number of roots which is 1, because of the signe − the residue will be 4; whence I conclude that to be the value of one root, and that the number sought is 4, whose double is 8, added to the square (of 4,) 16, makes 24.

Here note that numbers diminished, where the absolute number hath the signe —, have two roots. The greater is extracted, as before we have taught; the lesser is found out by subtracting the square root of the residue from the half sum of the roots. As if I seek a number whose octuple diminished by 12, shall be equal to its square, you will find an equation between 1Q and 8R−12. The greatest root is 6, the lesser is 2, here both the roots answer the question. But this happens not often.

But if you be to extract the biquadrate root: [Page 336]First, extract the square root, as is taught, and a­gain extract the square root of this, and this shall be your biquadrate root. As if the equation be between 1QQ and 2Q+8, you may finde the square root to be 4 by the method taught, first, ta­king half the number of the squares, &c. and af­terward you must extract the square root of 4, which is 2, this shall be the value of the root. In like manner, if your equation be between 1QC and 2C+48, first, I extract the square root of 2C+28, which is 8, of which extract the cube root 2, because the root to be extracted is the square cube, as the character QC, which is one of the terms of the equation denoteth.

Sect. 6. How to know if the question be impossible, vain, or ill propounded.

You may know the question to be impossible, if you come to an equation impossible: As if fol­lowing the conditions of the Probleme, you meet with an equation between 6R and 24R, or be­tween 3Q+5 and 4+ [...]Q.

Secondly, the question is vain, when the equa­tion is between two equal numbers of the same denomination, as between 6Q and 6Q.

Thirdly, the question is ill propounded, when without any difficulty many numbers will an­swer the probleme propounded.

CHAP. III. Algorithme of second roots with their use.

ALgebraists sometimes use more than one root to finde out divers numbers propoun­ded, and then to the end they may proceed with lesse confusion they usually help themselves with second roots, which they ex­presse thus, 1A, 1B, &c.

Sect. 1. Addition of second roots.

If your second roots be of the same denomina­tion, adde the numbers, and to the sum give the same denomination, as 5A added to 4A make 9A, if they be of different denomination, adde them with the signe of + as 5 A added to 6B make 5A+6B.

Sect. 2. Subtraction of second roots.

If they be of the same denomination, subtract one number from the other, and to the residue give the same denomination, as 5A taken from 9A, there remains 4A, if different, they are sub­tracted with the signe − as 6B taken from 8A, there rests 8A−6B.

Sect. 3. Multiplication of second roots.

If they be of the same denomination, do as you do with the first roots, as 4A multiplied by 7A make 28AQ, if of different, both denominations are reteined in the product, as 3R multiplied by 5A make 15RA.

Sect. 4. Division of second roots.

Division is onely performed by the interposi­tion of a little line, as is before taught, notwith­standing it is to be observed, that if for example 3AR be to be divided by the Divisor 1R the quotient shall be 3A, because in such case there needs nothing else but to take from the Divi­dend the character of the Divisor.

Sect. 5. The extraction and use of second roots.

After you have found and reduced your equa­tion, according to the manner of working in se­cond roots, you must extract the root after the manner taught in the precedent Chapter. As if 1AQ be equal to 25, I say that 5 is the value of the second root, and if 1AQ be equal to 4A+12, you must take the moity of the number of roots, &c. As is said in the 5. Sect. of the prece­dent Chapter, and you shall find 6 to be the value of 1A.

Now since the end of the second roots is to be [Page 339]reduced to first, you must not forget after you have found the value to begin again your work, and to put in first roots that which you have found to be the value of the second, as I shall shew you in some example in a Chapter following.

CHAP. IV. The Algorithme and extraction of the roots of surd & irrational numbers.

SUrd roots are those that have a radical signe before them, and which in propriety of speech ought to be called absolute num­bers, notwithstanding they cannot be ex­pressed by any common number, neither whole, nor broken, we will hereafter expresse the radical signe by this character ℛ.

There are many sorts of surd roots, some are simple, as ℛQ5, that is to say, the root square of 5, others are compound, as ℛQ5+ℛC6, that is to say, the root square of 5, plus the root cube of 6, some are universal, whose radical cha­racter, extends to all the particulars following it, and for that end are enclosed in a Parenthesis in this manner ℛQ (14+ℛQ4) the root uni­versal of 14 joyned with the root square of 4, all which number is 4 for 14+, the root square of 4 which is 2, maketh 16 whose root is 4.

Sect. 1. Reduction of surd roots simple to the same de­nomination.

First, you must put the radical signes under the numbers to which they belong. Secondly, you must multiply the numbers by the signes acrosse, for to get new ones. Thirdly, you must adde the signes together, which is done by multiplying their exponents, & give the character of the pro­duct common to the two new products, as if you would reduce to the same denomination ℛQ5 and ℛC4, you must first place them as followeth. [...] Secondly, you must multiply the numbers 4 and 5 by their signes acrosse, that is to say, you must take the square of 4, and the cube of 5, which are 16, and 125. Thirdly, the exponents of the signes ℛQ and ℛC, which are 2 and 3, ought to be multiplied together, the product is 6. I look then in the table what cossick character is under the exponent 6, and finding QC, I take that for my common denominator, and in stead of my two first surd roots, which were of different denomi­nation, that is to say ℛQ5, and ℛC4. I have two new ones of the same denomination, that is to say, ℛQC of 125, and ℛQC of 16.

Sect. 2. Multiplication and Division of surd simple roots.

If the roots be of the same denomination you must onely multiply and divide the numbers by themselves, and to the product and quotient givse the same radical signe as ℛQ7 by ℛQ2 give for the product ℛQ14. In like manner ℛQ36 divided by ℛQ12, gives for the quotient ℛQ3.

But if the roots be of different denomination, you must reduce them to the same denomination by the precedent Paragraph, and after multiply and divide as shall be shewed, for example, ℛQ3 multiplied by 2, the product is ℛQ12, and ℛQ12 divided by 2, the quotient is ℛQ3.

Sect. 3. How to know whether two surd roots be com­mensurable or not.

You must divide the greatest root by the les­ser, if the quotient be rational, the two roots are commensurable: if otherwise, they are not. As because ℛQ24 divided by ℛQ6, gives for the quotient ℛQ4, which is 2 a rational number, I conclude these two root: ℛQ24 and ℛQ6 to be commensurable. In like manner, since root square 24 divided by ℛQ8, the quotient will be 3, a surd number and irrational, you may con­clude those two roots, ℛQ24, and ℛQ8, to be incommensurable.

Sect. 4. Addition of simple irrational roots.

If the roots be incommensurable, you must adde them onely by the signe+as ℛQ24 added unto ℛQ8 makes ℛQ24+ℛQ8. But if they [Page 342]be commensurable, you must adde a unite to their quotient rational, and you shall have a summe, which being multiplied by the lesser of the two roots to be added, will give a product which shall be the summe sought. As ℛQ24 added with ℛQ6 makes ℛQ54, because ℛQ24 divided by ℛQ6, gives 2 for the quotient rational, to which I adde a unite, and it is 3 by which (alwayes re­ducing them to the same denomination) I multi­ply ℛQ6, which is the lesser of my two roots, and I finde for my summe ℛQ54.

Sect. 5. Subtraction of simple irrational roots.

If they be incommensurable, you must sub­tract them by prefixing the signe−as ℛQ8 sub­tracted out of ℛQ24, the residue shall be ℛQ24−ℛQ8.

But if they be commensurable, you must take away a unite out of the quotient rational, and you shall have the residue, the which being multiplied by the lesser of the roots given, shall give a pro­duct which shall be the residue fought, as if I be to subtract ℛQ [...] from ℛQ4, dividing the greater by the lesser, the quotient rational is 2, from which if you take 1, there remains 1, by which (reducing them first to one denomination) if you multiply the lesser root, that is to say, ℛQ6, the residue will be ℛQ6.

Sect. 6. Addition and Subtraction of surd numbers, composed and diminished.

I have here no new precepts, onely advertise you, that you may remember what I have said before of Cossick numbers, touching the signes of + and − in the fifth and sixth Paragraph of the first Chapter, and what is delivered in the fourth and fifth of this Chapter, touching the Addition and Subduction of simple surd numbers, and these will be no difficulty, as if you be to adde 5+ℛq. 24 with 3+ℛ q. 6, you will find 8+ℛ q. 54. In like manner, if you subduct 3−ℛ q. 6, from 5+ℛ q. 24, there rests ℛ q. 54−42.

Sect. 7. Multiplication of numbers surd, composed and diminished.

This multiplication hath no great difficulty, nor needs new precepts. Remember onely that the same signes have in the Product+and diffe­rent−with this, that your Multiplication is not good, if the particulars to be multiplied be not first reduced to the same denomination. For ex­ample. 5+ℛ q. 24 by [...]−ℛ q. 6, 'tis to be done after this manner:+ℛ q. [...]4 by−ℛ q. 6, ma­keth−ℛ q. 144, or−12 after+ [...] by−ℛ q. 6, maketh−ℛ q. 150. Further, +ℛ q. 24 by 3 is +ℛ q. 216. Lastly, +5 by 4 [...] is+15, then the whole Product will be 15+ℛ q 216−ℛ q. 150−ℛ q. 144, or 3+ℛ q. 216−ℛ q. 150 be­cause [Page 344] ℛ q. 144 is a rational number, to wit 12, which being subducted out of 15, because of the signe−leaveth 3.

Example: [...]

Sect. 8. Division of surd numbers, compounded or di­minished.

If the Divisor be simple, the division is made by the interposition of a line between the Divisor and the compound number to be divided, as if ℛ q. 2+ℛ q. 5 be divided by 8, the quotient will be ℛ q. 2+ℛ q. 5,/8 and so of others.

But because it sometimes may fall out (though very seldome) if the Divisor also will be a Bi­nome, [Page 345]or compound number, that is to say, a surd number compounded of two particulars with the signe +, or a Trinome that is compounded of three particulars, &c. See here the manner to di­vide in such a case.

If the Divisor be a Binome, you must multi­ply by his Apotome, as well the number to be di­vided, as the Divisor (and if the Divisor be an A­potome, you must divide by the Binome, as well the Dividend, as the Divisor) by means of this Multiplication you shall have a new Dividend, and a new Divisor. Now this new Divisor will be alwayes rational, and therefore needs onely to to be set under the Dividend with a line between. As for example: ℛ q. 6−2 by ℛ q. 5+ℛ q. 3, I take the Apotome of my Divisor (that is) ℛ q. 5 −ℛ q. 3. by which I multiply both my Divi­dend and my Divisor, by one of the multiplica­tions is produced ℛ q. 30−ℛ q. 20−ℛ q. 18 +ℛ q. 18 +ℛ q. 12, for my new Dividend, and by the other is produced 2 for my new Divisor. So that the quotient of my Division will be ℛ q. 30−ℛ q. 20−ℛ q. 18+ℛ q. 12/4, because 2 must be first squared, and then 4 put underneath the Dividend.

If your divisor be a Trinome, you must observe the same method, multiplying the dividend and the Divisor by the Apotome of the Divisor, that is to say, by the same Divisor onely, changing the signe of the last particular. After this is done, you shall have a new Dividend and a new Divi­sor, [Page 346]which shall be a Binome. Then you must again seek a new Dividend and Divisor, which now will be simple and rational.

Last of all, if you will not take this pains, the Division is good, if under the Dividend you sub­scribe the Divisor with a line between.

Sect. 9. Multiplieation of roots universal.

You must reduce the root to be multiplied, and the Multiplicator to their squares or cubes, ac­cording to the radical signe prefixed, and after­ward perform you multiplication, as is taught in Sect. 7 of this Chapter. Afterward you must affix the signe radical, and inclose all in a Parenthesis.

This is better understood by an example, as if you multiply ℛ q. (7+ℛ q. 3) by 2, the squares of the one and other numbers are 7+ℛ q. 3 and 4 then the first being multiplied by the last, maketh 28+ℛ q. 48, and therefore if you close this number within a Parenthesis, and put before it the same radical signe, the product will be ℛ q. (28+ℛ q. 48.)

In like manner, if you would multiply this number 7/2−½R+ℛ q. (49/4−3/4 q.−7/2 R) by it self, to get the square of it, you must call to mind the fourth proposition of the second book of Eu­clide, which sheweth that a line being divided in­to two parts, the square to the whole is equal to the square of the parts, and to double their Rect­angles, you must therefore conceive this num­ber, as divided into two parts, of which the first [Page 347]is 7/2−½ R, and the last ℛ q. (29/4−3/4 q.−7/2 R) take then the squares of the parts, which are 49/4+3/4 q.−7/2 R, and 49/4−3/4 q.−7/2 R, the double of the Rectangle of the parts is ℛ q. (9604/16+220/16 q.−2744/8R+7C+12/16 q. q.) then the square of the number proposed is the summe of these three numbers, that is to say 49/2−½ q.−7 R+ℛ q. (9604/16+220/16 q.−2744/8 R+7 C−11/16 q. q.) The square of the Apotome is the same number, put­ting only the signe−before the universal root, & the sum of the two squares is 49−1 q.−14 R.

Sect. 10. Division of roots universal.

You must reduce the roots to be divided, and the Divisor to their Squares, Cubes, &c. And after divide them as is taught in the 8 Sect. and when this is done, enclose all in a Parenthesis, with the same radical signe which was before. As if you divide ℛ q. (13+ℛ q. 17) by the root square of 5, their squares are 13+ℛ q. 17 and 5, then the first being divided by the last, the pro­duct will be 2 3/5+ℛ q. 17/25, then the quotient of the division proposed will be ℛ q. (2 3/5+ℛ q. 17/25.)

Sect. 11. Addition and Subduction of roots uni­versal.

Many trouble themselves to give precepts in­tricate enough, the short and most certain is to [Page 348]adde them with the signe+, and subduct them with the signe−. As for example, ℛ q. (3+ℛ q. 2) added with ℛ q. (ℛ q. 5+6) will be ℛ q. (3+ℛ q. 2)+ℛ q. (ℛ q. 5+6) and the same first root subducted from the last, the residue is ℛ q. (ℛ q. 5+6)−ℛ q. (3+ℛ q. 2.)

Sect. 12. Extraction of the roots of Binomes and Apotomes.

1 Take the difference of the squares of the one and other part of the Binome. 2 Adde and sub­duct the square root of this difference from the greatest part of the Binome. 3 Conjoyn the square root of the moity of the summe, with the square root of the moity of the residue, by the signe+, if it be a Binome, and by the signe−, if it be an Apotome; and thus the extraction is fi­nished. As if you would extract the square root of this Binome 3/2+ℛ q. 5/4, you shall first take the square of the first part, which is 9/4, and the square of the second which is 5/4, the difference of these two is 4/4, that is to say 1. Secondly, you must ex­tract the square root of this difference, which is 1, you shall adde and take it away from the first part of the Binome, by the addition you shall have 5/2 for your sum, and by subduction ½ for your residue. Thirdly, joyn the root square of the sum, with the root square of the residue by the sign+& you shall have ℛ q. 5/2+ℛ q. ½ for the square root of your Binome propounded, and consequently ℛ q. 5/2−ℛ q. ½ shall be the root square of the Apotome 3/2−ℛ q. 5/4.

CHAP. V. The Ʋse of Algebra.

'TIs much to have taken the pains to learn all that we have hitherto shewen or taught: but I dare boldly say, that those that shall rest here, do as yet know no­thing to the purpose, although they may know all the precepts; it behoveth us then to make a step further, to apply and bring those precepts in­to use and exercise. 'Tis that which I desire to demonstrate in this Chapter, by some questions, the solution of which will give great light to the atteining of perfection in this Art. Wherefore I intreat thee (Reader) not to omit this Chapter, in which I pretend to yeeld thee some pleasure and delight, as also an illustration of what hath been before treated of.

Sect. 1. Questions resolved by one simple Equation.

Question I.

A Lexander one day told Ephestion, that he was elder than him by two years; thereupon Cli­tus tels them, that he was as old as both of them (their ages added together) and four years over and above. The Philosopher Callisthenes being present at this discourse (saith he) I well remem­ber that my father, who was 96 years old, had the [Page 350]age of you three. It is demanded here, how old Alexander was when he held this discourse, as al­so how old Clitus and Ephestion were. I put for the age of Ephestion 1 R of years, whence it fol­lows that Alexander had 1 R+2; therefore Clitus had, R+6, and those three together, ac­cording to the condition of the question, ought to be equal to 96: therefore there is an equality between 4 R+8 (which is the sum of the three ages) and 96, take away 8 from both parts of the equation so there will remain on one side 4 R equal to 88, which divide by the number of the greatest Cossick Character, that is to say by 4, the quotient gives 22 for the value of one root, which was supposed for the age of Ephestion. Therfore Ephestion was at that time aged 22 years, Alexander 24, and Clitus 50, which all together make 96 years.

Question II.

A Hare is 100 Geometrical paces distant from a Dog that swiftly pursues him, and the Dog run­neth two times and an half faster than the Hare: It is demanded how many Geometrical paces the Hare will have run when the Dog overtaketh her; I put for these Geometrical paces 1 R; therefore the Dog which runs 100 paces more than the Hare, will have run 100+1 R: and for that the Dog runs twice and an half swifter than the Hare, I take two numbers in like porportion to one another, that is to say 5 and 2, and con­clude that there is the same proportion between 100+1 R to 1 R, as between 5 and 2; therefore [Page 351]the product of the first number 100+1 R mul­tiplied by the last number 2 (which is 200+2 R) is equal to the product of the two means 1 R and 5 (which will be 5 R) therefore if you take from both parts 2 R, there will remain 200 equal to 3 R, and therefore I divide 200 by 3, which is the number of the greatest Character, and find in the quotient 66⅔ which will be the value of the root. I say therefore that the Hare will have run 66 Geometrical paces, and ⅔ when the Dog shall have overtaken her, and the Dog will have run 166⅔ paces, which make twice and an half more than 66⅔.

Question III.

The Architect Vitruvius in his ninth Book Chap. 3. tels us, that Archimedes found the quau­tity of silver that a Goldsmith had mixed in a golden Crown which he had made for the King Hiero (who was obliged by vow to present it to the gods) weighing 100 pounds. It is demanded by what means Archimedes could arrive to the knowledge of that secret. The common opinion is, that he took two masses, the one of gold, the other of silver, which weighed as much as the Crown; afterward he filled a vessel up to the brim with water, which vessel was placed in some grat bason, that the water that should be forced out of the first vessel, might be preserved and not lost. Thirdly, he gently put in the two masses and the Crown, one after the other, into the prepared vessel, taking exact notice of the quantity of water that issued out of the vessel at [Page 352]each time, and concluding from thence, that the Goldsmith had mingled 6 pounds and ⅔ parts of silver. We will suppose then that the masse of gold weighing 100 pounds, did cast out of the ves­sel 60 pounds of water, & that the masse of silver also weighing 100 pounds, cast forth of the vessel 90 pounds of water, and that the Crown cast forth 65 pounds. I put afterward for the silver mixed in the Crown 1R, and constitute twice the Rule of three after this manner.

If 100 pounds of gold give me 60 pounds of water, how much will 100−1.R? and I finde 6000−60R / 100 for my fourth number.

Secondly, if 100 pounds of silver give me 90 pounds of water, how much shall 1R? and I find 90R / 100. Now these pounds of water 6000−60R / 100 and 90R / 100 added together do make 6000+30R / 100 pounds of water cast out, which ought to be equal to 65 pounds of water, cast forth by the Crown, and therefore if we reduce them, we shall finde 6000+30R equal to 6500 (this reduction is made by multiplying the denominator 100 by 65, for see­ing that this fraction 6000+30R / 100 is equal to 65 it shall be also equal to 65/1, Mark well this kinde of reducti­on once for all. and therfore there will be the same proportion of the numerator 6000+30R to the de­nominator 100, as of the second nu­merator 65 to 1. Therefore the pro­duct under the extreams 6000+30R is equal to the product of the means 6500) take away therefore from both parts 6000, and there [Page 352]will remain an equation between 500 and 30 roots, and therefore divide 500 by 30, the num­ber of the greatest character, you shall have the value of the root 16⅔ for the pounds of silver mingled by the Goldsmith in the Crown:

Sect. II. Questions resolved by an Equation com­pounded.

Question I.

TO divide 8 into two such numbers as their squares being added together, may make 34. I put for the first 1R, therefore the second shall be 8−1R, their squares are 1Q, and 64+1Q−16R, which added together, do give for their sum 64+2Q−16R, the question imports that the sum of the squares is 34. Therefore there is an Equation between 64+2Q−16R, and 34, which being reduced by addition and subtraction, there will remain also an equation between 2Q and 16R−30, and the whole divi­ded by 2, which is the number of the greatest Cossick Character, there will yet remain an E­quation between 1Q and 8R−15, from which I extract the root, as hath been shewn in the fifth Section of the second Chapter. The half of the roots is 4, his square 16, from which take the ab­solute number 15, rest 1, whose square root 1 ad­ded to the half of the number of roots, gives for its sum 5, which is the value of the root; there­fore the two numbers sought shall be 5 and 3.

Question II.

To find two numbers whose product may be 12, and the difference of their squares 32, I put for the one of them 1R; therfore seeing that the product is 12, the other number shall be 12/1R (for if the pro­duct of two numbers be divided by one of those two numbers, the quotient shall be the other num­ber) their squares are 1Q and 144/1q whose diffe­rence is 144/19−1Q equal to 32, as appears by the question; therefore there will be an equality be­tween 144/1q and 32+1Q, & therefore if we make the reduction, as in the third Question of the first Section, we shall also finde an Equation between 144 and 32Q+1QQ, also between 144−3Q and 1QQ; it behoveth then to extract the square root of the number 144−32Q. The half of 32 is 16, whose square is 256, to which adde 144, makes 400, whose square root is 20, from which take the half of 32, to wit 16, there re­mains 4. See here the square root but seeing that the squared square root ought to be taken, I take again the root of 4, and I finde 2 for the va­lue of the root. Therefore seeing that the second number hath been put 12/2R the same second num­ber shall be 12/2, that is to say 6.

Question III.

Two Merchants joyn in company, and toge­ther bring in the sum of 165 Crowns; but the first mans money hath been exposed twelve moneths intire, and the second mans money onely eight moneths: it happens that they gain but 28 [Page 355]Crowns, which added to 165 make 193, which they distribute to one another in such sort, as the first takes 67 Crowns, as well for his prin­cipal money as for his profit, and the second takes 126 Crowns, the question is what each of those Merchants brought into stock. I put for the mo­ney of the first man 1R, therefore seeing that the sum of both was 165 Crowns, the second mans money is 165−1R. Now if you take away 1R, which is the sum the first man brought in, from the sum he received, which was compoun­ded of the principal and profit, you will find that the first mans profit will be 67−1R, and by the same argument you shall finde, that the second mans profit will be 1R−39. Now you must finde what one root gaineth in eight moneths, which will be done by the Rule of Three, thus, If in 12 moneths there be gained 67−1R, how much will there be gained in 8 moneths, and the fourth number shall be 134/3−2R / 3 for the first mans profit in 8 moneths; after that I seek what the se­cond man hath gained by another operation of the rule of three, saying; If 1R gain 135/3−3R / 2, what will 165−1R gain, and I finde for my fourth number 7370+2/3Q−464R / 3,/1R which is equal to the second mans profit, which we have already found to be 1R−39, and therefore by reduction there will be an equation between 1Q−39R and 7370+2/3Q−464/3R, all which 39R being added, and ⅔Q taken away, there will be an [Page 356]equation between 7370 and 1/3Q+347/3R and consequently between 7370−347/3R and ⅓Q, and therefore multiplying all by ⅓, which is the number of the greatest character, there will be yet an equation between 1Q and 22110−347R out of which the square root must be extracted. The half of the number of roots is 347/2, whose square is 170409/4, which added to 22110, make 208849/4, whose square root is 457/2, from which if you take half the number of roots, there will re­main 110/2, that is to say, 55 for the value of one root, and was the money the first man put in bank; and for that we have found in the pursuit of these operations, that the first mans profit was 67−1R, that same profit will be 67−55, that is to say 12, by the same reason the second mans stock shall be 110, and his profit 16.

Sect. III. Questions resolved by surd numbers.

Question I.

TO divide any given number (as for exam­ple 4) according to mean and extream rea­son, that is to say, to divide 4 into two numbers, in such manner as that the whole 4 may bear the same proportion to its greatest part, as the greatest part bears to the least. I put for the greatest part 1R, therefore the least shall be 4−1R. There­fore there is the same proportion of 4 to 1R, as of 1R to 4−1R, and therefore the square of the [Page 357]middle part 1Q is equal to the product of the extreams 16−4R, from which I extract the root according to the rule before prescribed. The half of the number of roots is 3, whose square 4, added to 16, makes 20; out of which the square root ought to be extracted according to the pre­cept: but seeing it is no square number, I must content my self by putting the radical signe be­fore it thus, ℛQ 20, from which I take the half of the number of roots, and I have for Residue ℛQ 20−2, which is the value of the root, by which I shall find with facility, that the other part wil be 6−ℛ Q 30. For the proof of this ope­ration, it behoveth that these two parts added to­gether make 4, and that the lesser 6−ℛQ 20, being multiplied by 4, make the product equal to the square of the greater part ℛ Q 20−2.

Question II.

To divide 8 into two parts, between which, 2 may be a mean proportional. I put for the first part 1R, therefore the lesser shall be 8−1R, and seeing that 1R and 2, and 8−1R ought to be proportional, it behoveth that the square of 2, which is 4, be equal to the product of the ex­treams, which is 8 R−1Q, and therefore after the reduction it will be found, that 1Q is equal to 8R−4, whose square root is ℛQ 12+4. For the one part of 8, and for the other part 4−ℛQ12, both the one and the other root do re­solve the question, as you will find, if you take the pains to examine it.

From this practice may be framed an universal [Page 358]Canon, which may serve for the resolution of an infinite number of Algebraical Problemes, which may be conceived after this manner. The sum given, which conteineth the two extreams, ought to be distributed into two equal parts, that the square of the half may be taken, from which the square of the mean proportional given, must be taken, and the square root of the Residue ad­ded, and taken from the half of the given sum, will shew the two parts sought. As for example, I take the half of 8, which is 4, whose square is 16, from which I take 4, the square of 2, which is the given mean, and there remains 12, whose square root added to the same half, makes 4+ℛQ12▪ and taken from the same half, makes 4−ℛQ 12.

Question III.

To divide any given number (as for example 4) into three numbers continually proportional, in such sort as that the squares of the extreams joy­ned together, may be triple the square of the mean.

I put 1R for the number of the middle part, then seeing all three ought to make the same sum of 4, we shall have for the sum of the ex­treams 4−1R. Now seeing that of three num­bers continually proportional, the square of the sum of the extreams is equal to the square of the extreams, and to the double of the square of the middle part. I take the square of this sum 4−1R, which is 16+1Q−8R, from which I take 2Q, which is the double of the square of the [Page 359]middle part, and there will remain 16−1Q−8R for the sum of the squares of the extreams: Therefore seeing that the condition of the que­stion requireth a triple proportion, there will be an equation between 16−1Q−8R and 3Q, adde therefore 1Q on both sides of the equation, and you shall have 4Q equal to 6−8R, and di­viding the whole by 4, which is the number of the greatest character, the equation will be 1Q equal to 4−2R, whose square root is ℛQ5−1 for the middle number sought, and the sum of the extreams shall be 5−ℛQ5, which being divided into two parts, by the Canon of the pre­cedent question, in such sort as ℛQ5−1, be the middle proportional, you will finde that the ex­treams are 2 and 2−ℛQ5, therefore the three numbers are 2 and ℛQ5−1, and 3−ℛQ5, which all together make 4, and are in continual proportion, and the squares of the extreams are triple the square of the mean.

Sect. IV. Geometrical questions resolved by Algebra.

Question I.

THere is a piece of ground of a greater length than breadth, whose angles are right angles, and in a triple proportion, Before you at­tempt the reso­lution of such questions, you must draw their figures. and their squares taken together, are quintu­ple their sum. The sides, the diame­ter, and the capacity or superficies of that piece of ground is demand­ed, I put for the least side 1R, there­fore [Page 360]seeing they are in a triple proportion, the other side shall be 3R, their squares shall be 1Q and 9Q, which added together make 10Q, which ought to be quintuple, the sum of the numbers. Now the sum of the numbers is 4R, and its quin­tuple 20 R, and by consequence, see here an e­quation between 10Q and 20R, which are two collateral characters, and therefore dividing 20 by 10, which is the number of the greatest chara­cter, you shall finde 2 for the least side, therefore the greater side shall be 6. Therefore the superfi­cies shall be 12, and the diameter ℛQ40.

Question II.

There is an equilateral triangle, whose super­ficies is ℛQ243. The side and perpendicular is demanded, supposing that the perpendicular of an equilateral triangle doth alwayes divide the side into two equal parts, I put for the half of the side divided, 1R; therefore the side shall be 2R. Now seeing that in every equilateral triangle; the square of the side is equal to the square of the perpendicular joyned to the square of the half of the side, which is 1Q of the square of the whole side, which is 4. I shall have 3Q for the square of the perpendicular, and so ℛQ3Q shall be perpendicular, which multiplied by the half of the side, which is 1R (reducing it first to its square, because of the radical signe which is in the number multiplied) you shal have ℛQ3QQ for the superficies of the triangle: therefore there will be an equation between ℛQ243 and ℛQ3QQ, and therefore there will be also an equation [Page 361]between their squares, which are 243 and 3QQ, and the whole being divided by 3, the number of the greatest character, there will be yet an equa­tion between 81 and 1QQ. I extract therefore the squared square root of 81, and have 3 for the half of the side, 6 for the side, ℛQ27 for the per­pendicular, and ℛQ243 for the superficies of the triangle.

Question III.

There is a Semicircle, whose Diameter is divi­ded according to mean & extream reason, on wch there is raised a perpendicular produced to the circumference, and the lesser line which is drawn from the extremity of the diameter, to this point of the circumference, is ℛQ20−2, the quan­tity of the diameter, and of its parts, and of this perpendicular is demanded. To resolve this que­stion, it is presupposed that the greater part of the diameter shall be equal to the given line, as may with facility be Geometrically demonstrated. That being done, I put for the lesser part of the diameter 1R, therefore seeing that the other part is given ℛQ20−2, the whole diameter shall be ℛQ20−2+1R, which multiplied by 1R, giveth for the product ℛQ20−2R+1Q, equal to the square of the given quantity, which is 24−ℛQ320, and by due transpositi­on you shall finde 1Q equal to 24−ℛQ320+2R−ℛQ20, from which the square root ought to be extracted, taking exact notice that the particles which have the Cossick Characters may hold place with the number of roots. I consider [Page 362]therefore in this term of the equation, the num­ber of roots which is 2−ℛ Q 20, whereof I take the half, which is 1−ℛ Q 5, to whose square 6−ℛ Q 20, the absolute number 24−ℛ Q 320, ought to be added. and the sum will be 30−ℛ Q 500, whose square root ought to be extract­ed as from Apotomes, as hath been shewn in the last Section of the fourth Chapter, that root is 5−ℛ Q 5, which added to the half of the num­ber of roots 1−ℛ Q 5 gives for the sum 6−ℛ Q 20, which is the value of 1 root, that is to say, of the lesser part of the diameter; and therefore if you adde it to the greater part, you shall have 4 for the quantity of the whole diameter, from whence the perpendicular is easily known, provi­ded the rules of Geometry be in any reasonable manner understood.

Sect. 5. Questions resolved by the second Roots.

Question I.

THree men have amongst them a sum of mo­ney: The first saith to the second, If you deli­ver me the half of your money, I shall have 100 Crowns: The second saith to the third, If you deliver me ⅓ of your money, I shall have 100 Crowns: The third saith to the first, If you deli­ver me ¼ of your money, I shal have 100 Crowns. I demand how much money each one hath.

I put for the first mans money 1 R of Crowns, and for the second mans money 1 A, and for the third mans money 1 B: therefore the first which [Page 363]hath 1 R with ½ of the second mans money shall have 1 R+½ A equal to 100, and by conse­quence ½ A shall be equal to 100−1 R, and multiplying the whole by 2, 1 A, shall be equal to 200−2 R. I begin again therefore the opera­tion, and in stead of 1 A, I put for my second number 200−2 R. Now the question requires that the second man, (with ⅓ of the third mans) shall have 100; therefore there will be an equati­on between 200−2 R+⅓ B, and between 100, adde to both parts of the equation 2 R, and take away 200, there will yet remain an equation be­tween ⅓ B and 2 R−100, and multiplying the whole by 3, you will have 1 B equal to 6 R−300: That being found, I begin again the work, and in stead of 1 B, I put for the third number 6 R−300▪ which added to ¼ of the first mans, makes 25/4 R−300, which ought to be equal to 100, therefore if you adde on both parts 300, there will be an equation between 25/4 R and 400, there­fore if divide 400 by 25/4, you shall finde 64 for one root; therefore the second, which had 200−2 R shall have 200−128, that is to say 72, and the third shall have 84. Those three numbers do perfectly satisfie all the conditions of the que­stion.

Question II.

Two men divide between themselves three hun­dred Crowns in such sort, as that the second mans money divided by that of the first mans, makes [...]/ [...]. It is demanded, how much each of them hath. I put for the first mans money 1 R, and for that [Page 364]of the second 1 A, there is therefore an equation between 1 R+1 A, and 300, and therefore 1 A is equal to 300−1 R, therefore 300−1 R / 1 R is equal to 3/2, and by consequence in crosse multiplying these two fractions, I shall have 600−2 R equal to 3 R, therefore 600 shall be also equal to 5 R, and dividing 600 by 5, I shall find 120 for the first mans money, so shall the other have 180.

Question III.

To find two numbers whose product may be 10, and the sum of their squares 29; I put for the first 1 R, and for the other 1 A, the product is 1 R A, equal to 10; therefore in dividing the whole by 1 R, there will be an equation between 1 A and 10/1 R, and therefore I begin again the ope­ration, and put for the first 1 R, and for the se­cond 10/1 R, their squares are 1 Q+100/19 equal to 29, and after the reductions and extractions of the roots, I find 5 and 2 for my sought numbers.

Sect. VI. Questions resolved indefinitely.

THat question is said to be resolved indefinite­ly, in which the numbers are demonstrated in such Algebraical terms, as do satisfie all the con­ditions of the question proposed.

Question I.

To divide 12 into four numbers Arithmeti­cally and continually proportional, I presuppose that when there is four numbers in Arithmetical proportion, the sum of the extreams is alwayes [Page 365]equal to the sum of the means: whence it fol­lows, that in our question the sum of the ex­treams shall be 6, and the sum of the means also 6. I put for the second 1 R, therefore the third shall be 6−1 R, their difference is 2 R−6, presuppo­sing 1 R to be the greater number of the two; if therefore I adde this difference to 1 R, I shall have 3 R−6, and if I take it from 6−1 R, I shall have 12−3 R, and therefore the four num­bers in continued Arithmetical proportion, shall be 3 R−6 | 1 R | 6−1 R | 12−3 R | and the question is indefinitely resolved, in pursuit of which you may take such a number as you please for the value of 1 R, provided neverthelesse that you do admit of fained numbers lesse than no­thing: However, if you will have no other num­bers than what are real, you ought to take the value of 1 R, beneath 4 and above 2, which will be easily understood by a little experience.

Take for example 5/2 for the value of 1 R, there­fore the first number, which is 3 R−6, shall be 15/2−6, that is to say 3/2, the second shall be 5/2, the third 7/2, and the fourth 9/2, which added toge­ther make 12, and are in continued Arithmetical proportion. And so you may take an infinite number of others.

Question II.

A Vintner hath three sorts of wine, the first is worth 4, the second 6, and the third 10 pence the pinte; of these three sorts of wine he desires to fill a vessel which conteins 80 pintes, which may be worth 8 pence the pint; I demand how many [Page 366]pintes he ought to take of each sort. You ought here to consider, that the number 80 must be di­vided into three such numbers, as the first multi­plied by 4, the second by 6, and the third by 10, the sums of the three products added together may make 640 (because that all the wine which shal be in the vessel to be filled wil cost 640 pence, seeing that if one pint be worth 8 pence, 80 pintes will be worth 640 pence) I put therefore for the third number 1 R, which multiplied by 10, makes 10 R, which being taken from 640, doth leave for the residue 640−10 R, which is a number conteining the first 4 times, and the second 6 times. On the contrary, seeing that the third hath been put 1 R, therefore 80−1 R shall make the sum of the first and the second, which multi­plied by 4, will give 320−4 R, which being sub­stracted from 640−10 R, will leave 320−6 R double to the second number, and therefore the second shall be 160−3 R. In like manner the same sum 80−1 R multiplied by 6, will pro­duce 480−6 R; therefore if you take away 640−10 R, there will remain 4 R−1 [...]0, double to the first, and therefore the first shall be 2 R−80. See here the question resolved indefinitely, the first number is 2 R−80, the second 160−3 R, and the third 1 R. The terms between which you ought to take the value of 1 R are 53⅓, and 40; if therefore you take 46 for the value of 1 R, you shall have 46 pints of wine of 10 pence the pinte, 22 of 6, and 12 of that of 4 pence the pinte.

[Page 367]Here I intreat you to consider, A special Note. that it is impossible perfectly to under­stand the rule of Aligation, without the know­ledge of Algebra: For if you propose this questi­on to one skilful onely in Arithmetick, he will give you for the 3 sought numbers, 40, 20, and 20. And if you tell him, that of the sort of wine of 4 pence the pinte, you have but 16 pints, he will re­main astonished; whereas by your question re­solved indefinitely by Algebra, you will be able to give satisfaction to this condition in infinite kindes and manners.

Question III.

Two numbers are sought, which have 56 for the difference of their Cubes, and which added together make 6. I put for the difference of those numbers 1 R, and if from the difference of the two Cubes you take the Cube of the difference of the sides, dividing this residue by the triple of the difference of the sides, you have for the quo­tient the product of the sides, it followes that if from 56 you take 1 C, and that the residual 56−1 C be divided by the difference of the sides, which is 1 R, the quotient 56−1C / 1R shall be triple, the product of the sides; and therefore if you divide this quotient by 3, you will have for the product of the sides 56−1C / 3R, which is equivalent to 56/3R−19/3, therefore if you divide 6, (which is the sum of your two sought numbers) into two parts, whose product may be 36/3R−59/3, you will have the que­stion resolved indefinitely. Now to attein this, I [Page 368]have given you a Canon in the second question of the third Section, the half of the sum 6 is 3, whose square is 9, from which if you take the product found, there will remain 9+19/3−56/3R, whose square root added, and taken from the half of the sum, gives for the resolution 3+ℛ (9+19/3R−56/3R) which shall be the greatest sought num­ber, and 3−ℛ Q (9+19/3−56/3R) which shall be the least. Now these two numbers resolves the question indefinitely, in such sort, that if you take two for the value of the root, you will finde that your two sought numbers shall be 4 and 2, and every other number (taken for the value of 1 R) above 2, will resolve the question.

Appendix. Questions in Algebra, most of which require the Rule of Three in their Operation.

Question. I.

A Merchant receives in exchange for 568 Crowns four kindes of money; of the first 7 makes one Crown, of the second 18, of the third 21; and of the fourth 28 make one Crown. Moreover, he received of each sort of money a like number. I demand, How much he received of each kinde of money?

Put 1 R for the quantity of each sorts of money, and then constitute the Rule of Three after this manner.

  • If 7 Pieces of mony be worth Crown 1 what is money 1 R I an­swer 1/7 R Crowns.
  • If 18 Pieces of mony be worth Crown 1 what is money 1 R I an­swer 1/12 R Crowns.
  • If 21 Pieces of mony be worth Crown 1 what is money 1 R I an­swer 1/21 R Crowns.
  • If 8 Pieces of mony be worth Crown 1 what is money 1 R I an­swer 1/28 R Crowns.

For if 7 pieces of money make 1 Crown, 1 R of money will make 1/7 R of Crowns, and so of the rest. Now those Fractions described in the fourth place, do make together 7 [...]/2 [...]2 R of Crowns, equal to the number 568 Crowns. Divide there­fore [Page 370]568 by 71/252, and you have for the value of 1 R 2016, and so many pieces of each kind of mo­ney he received: which thus I prove.

  • If 7 Pieces are worth 1 Crown what Pieces 2016 Answer 288 Crowns.
  • If 18 Pieces are worth 1 Crown what Pieces 2016 Answer 112 Crowns.
  • If 21 Pieces are worth 1 Crown what Pieces 2016 Answer 96 Crowns.
  • If 28 Pieces are worth 1 Crown what Pieces 2016 Answer 72 Crowns.

For 2016 pieces of the first kind of money do make 288 Crowns, and as many of the second kind make 112, as many of third kind make 96, and of the fourth kind 72, which added together do make 568 Crowns.

Question II.

A Certain man hath two measures of wine, the one worth 12 Crowns, the other 15. Now he desires of both these wines to fill another equal measure, whose worth may be 13 Crowns. I demand what part of each of those wines he must take, to fill the other to be worth that price?

Put 1 R for the part of the measure of the worst wine, and for the part of the measure of the best wine 1−1 R, then work by the rule of 3 thus.

 Meas.Cro. Crowns
Worst wine1121 R?makes 12 R
Best wine.1151−1 R?makes 15−15 R

For if one measure of the worst wine be worth 12 Crowns, 1 R of one measure of the same wine will be worth 12 R of Crowns, & if one measure [Page 371]of the best wine be worth 15 Crowns 1=1 R of a measure of the same wine will be worth 15−15 R. Therefore 1 R measure of the worst wine and 1−1 R measure of the best will be worth 15−3 R Crowns, which ought to be equal to to 13 Crowns. Adde therefore 3 R to each part of the equation, and the equation will be between 3 R+13 and 15; take away therefore 13 from both sides, and the equation will be between 3 R and 2. Divide therefore 2 by 3, and you shall have 2/3 for the value of 1 R, and so much ought to be taken of the measure of the worst wine, and ½ part of the measure of the best wine, which thus I prove by the Rule of Three.

 Meas.Cro. Crowns
Worst wine112⅔?worth 8
Best wine115⅔?5
    13 Cr.

For ⅔ of a measure of the worst wine is worth 8 Crowns, and ⅓ of the measure of the best wine is worth 5 Crowns, which added together make three.

Question III.

I Have a measure of wine worth ten Crowns; How much water must I mix with one mea­sure, that a mixed (like) measure may be worth seven Crowns.

Put for the measure of water 1 R, then frame the question by the rule of Three thus.

[Page 372]

Meas.Meas. Meas. 
Wine,waterCrownsmixtCrowns
1+1 R101?worth 10/1+1 R

For if a measure of wine, together with 1 R of a measure of water, be worth 10 Crowns, one mea­sure of wine and water mingled together will be worth 10/1+1 R & so the equation will be between 10/1+1 R and 7, which by crosse multiplication is reduced to 10 and 7+7 R, take away 7 from each part of the equation, and it will be between 3 and 7 R; divide 3 by 7, and you have for 1 R, 3/7, and so much water ought to be mingled with a measure of wine, that a measure of the mixture may be worth seven Crowns, and is thus proved.

Measure of wine & waterCrownsMeasureCrowns
1 3/7101?worth 7

For if 1 of wine and 3/7 of water be worth 10 Crowns, 1 of that mixture is worth 7 Crowns.

Question IV.

THere are in a certain vessel 20 measures of wine, of which each of them are worth 12 Crowns the measure: Now this vessel is filled up with water, and then one measure of this mix­ture is worth 10 Crowns, I demand the content of the vessel.

[Page 373]Put for the measures 20+1 R, then work by the rule of Three thus.

MeasuresMeasures Meas. 
of wineof waterCro.mixtCrowns
20+1 R2401?worth 240/20+1 R

For if one measure of wine be worth twelve Crowns, a measure of wine 20+1 R measure of water together will be worth 240 Crowns: therefore one measure mixt will be worth 240/20+1 R Crowns, so the equation is found be­tween 240/20+1 R and 10, which by crosse multi­plication is reduced to 240 and 200+10 R; take away 200 from both parts, and there will remain an equation between 40 and 10 R: Divide 40 by 10, and you have 4 for the price of the root, and so many measures of water were put into the ves­sel, and therefore the whole vessel conteins 24 measures, thus proved: 24 measures wine and water worth 240 Crowns, 1 measure? worth 10 Crowns.

Question V.

TWo Letter-carriers belonging to two cities di­stant each from other 140 leagues set forth to­wards one another, at one and the same time; the one travels eight leagues a day, the other six. I demand on what day they shall meet together.

[Page 374]Put 1 R for the day, then work by the rule of Three thus.

DayesLeaguesDayesLeagues
181 R?8 R.
161 R?6 R.

The first therefore in 1 R of dayes shall travel 8 R of leagues, and the latter 6 R, and both of them together will have measured 14 R of leagues, that is 140 leagues. There is therefore an equation between 14 R and 140. Divide then 140 by 14, and you shall have 10 for the value of 1 R. So the tenth day finished, they met together, which thus I prove.

DayesLeaguesDayesLeagues
1810?80
1610?60

For the first in 10 dayes went 80 leagues, and the latter went 60 leagues, which both together make 140, the distance of the two cities from one another.

Question VI.

A Certain Merchant bought a quantity of wooll, and another quantity of wax, which cost him together 124 Crowns. Now 100 pound weight of wooll cost him 7 Crowns, and 100 pounds of wax cost him 14 Crowns; but the quantity of wooll that he bought was double to the quantity of wax I demand how many pounds of each sort the Merchant bought?

[Page 375]Put for the wax 1 R, and for the wooll 2 R of pounds, then work by the rule of Three, thus.

PoundsCrownsPoundsCrowns
100 wooll72 R wooll?14/1 [...]0 R
100 wax141 R wax?14/106 R

Therefore there will be an equation between 28/100 R of Crowns, and 124 Crowns. Divide there­fore 124 by 28/100, and you shall have for the value of 1 R 442 6/7, and so many pounds of wax he bought, and of wooll 885 5/7, which is thus proved.

100 pounds,wooll7 Crow.885 poun. 5/762 Cr.
100 pounds,wax14 Crow.442 poun. 6/762 Cr.

And so for wooll and wax together he expen­ded 124 Crowns.

Question VII.

A Certain man buyes a number of Ells of Vel­vet, which he selleth again; he buyes 5 Ells for 7 crowns, and sels 7 Ells for 11 crowns, and gained on the bargain 100 crowns. I demand how many Ells of Velvet he bought and sold in all.

Put for the quantity of Ells 1 R, then work by the rule of Three thus.

EllsCrownsEllsCrowns
571 R?7/5 R
7111 R?1 1/7 R

[Page 376]You see if 7/5R of crowns which he laid out be substracted from 1 1/7R of crowns which he recei­ceived, there will remain 6/35R of crowns for the profit. Therefore the equation will be between 6/35R and 100 R, divide 100 by 6/35, and you have 583⅓ for the value of 1 R, and so many Ells he bought and sold. Thus proved.

EllsCrownsEllsCrowns
57583⅓?816⅔
711583⅓?916⅔

By which you may perceive there is 100 crowns gotten.

Question VIII.

Certain man buyes a number of Ells of Vel­vet, paying 11 crowns for 7 Ells: Now he sels the whole again after the rate of 5 Ells for 7 crowns, and lost 100 crowns by the bargain. I demand how many Ells he bought and sold in all.

Put for the number of Ells 1 R, and then work by the rule of Three thus.

EllsCrownsEllsCrowns
7111R?1 1/7R.
571R?7/5R.

Then if 7/5R of crowns which he received, be substracted from 1 1/7R of crowns which he expen­ded, the losse will happen to be 6/35R of crowns. So the equation will be between 6/35R of crowns, [Page 377]and 100 crowns. Divide therefore 100 by 6/35, and you shall have for the value of 1 R, 583⅓ Ells, and so many Ells were bought and sold, which thus I prove.

EllsCrowns  
711583⅓?916⅔
57583⅓?816⅔

Where you see he lost 100 crowns by the bar­gain.

Question IX.

A Man buyes 100 pounds of wax for 17 crowns. I demand, how many pounds he must sell for one crown, that so on 102 crowns he may gain 18 crowns.

Put 1R for the number of pounds, then work by the rule of Three thus framed.

CrownsPoundsCrownsPounds.
17100102600
11 R102+18?120 R.

For 102 crowns do give 600 pounds, and if for one crown there be given 1R of pounds there will be 1 [...]0R of pounds given for 102+18, which said 120R are equal to 600 pounds, the quantity of the wax sold. Therefore the equati­on shall be between 120R of pounds and 600 pounds. Divide therefore 600 by 120, and so you have for the value of 1 R, 5 pounds, and so many pounds are sold for one crown; so as that in 600 pounds 18 crowns may be gained on 102, thus proved. [Page 378]

PoundsCrownsPoundsCrowns
10017600?102
51600?120

Where in the first example 600 pound made 102, in this it makes 120, that is 102+18.

Question X.

A Man buyes 100 pounds of wax for 17 crowns, in disposing of which he loseth 18 crowns, on 102 crowns. I demand how many pounds he sold for one crown?

Put for the number of pounds 1 R, then work by the rule of Three, thus constituted.

CrownsPoundsCrownsPounds
17100102?600
11 R102−18?84 R.

So there will be equation between 84R of pounds and 600 pounds. Divide therefore 600 by 84, and the value of 1R will be 7 pounds and 1/7 and so many pounds, he sold for one crown, and lost 18 crowns on 102 by the bargain, which thus I prove.

PoundsCrownsPoundsCrowns
10017600?102
7 1/71600?84

Where you see he laid out 102 crowns for 600 pounds, and in stead thereof received but 84 crowns, that is 102−18 crowns.

Question XI.

A Certain man agrees with a servant for 12 moneths service, for ten crowns and a coat; but at the end of seven moneths, he gives him the coat and two crowns. I demand then at what rate he esteemed the coat.

Put for the price of the coat 1R of crowns, and say by the rule of three. If 12 moneths require 1R+10 crowns, how much will one moneth require? and you shall find that it will require 1R+10/12 of crowns: again say, If 7 moneths require 1R+2 crowns, how much will one moneth require, and you shall finde it to be 1R+2/7 as here under appears.

MonethsCrownsMonethCrowns
121R+101?1R+10/12
71R+21?1R+2/7

Therefore there will be an equation between 1R+10/12 and 1R+2/7 seeing both are the re­ward of one moneth, which equation by crosse multiplication is reduced to 7R+70 and 12R+24, take away 24 from both parts, and the e­quation will be between 7R+46 and 12R. [Page 380]Again, take away 7 from both parts, then it will be 46, equal to 5R; divide 46 by 5 and the price of a root, and so of the coat is 9⅕ crowns, as ap­pears hereunder. The reward of 12 moneths is 19⅕, and of 7, 11⅕.

MonethsCrownsMonethsCrowns
1219⅕7?11⅕

Question XII.

A Certain Citizen agrees with a slothful servant for 30 dayes, that every day he wrought he would give him 7 Groats; but for every day that he idled, and wrought not, he was to allow his master five Groats: when the 30 dayes were past, it happens that the servant was to receive nothing from his master, nor the master from the servant. I demand then, how many dayes he laboured, and how many he idled?

Put 1R for the dayes of labour, and 30−1R for the dayes of idlenesse, and then frame the rule of Three thus.

DayGroatsDayesGroats
Labour 171R?7R
Idleness 1530−1R?150−5R

Now seeing that his work and play came to one reckoning, there will be an equation between 7R and 150−5R, adde 5R to each part, and the equation will be between 12R and 150. Divide therefore 150 by 12, and you have for the value of 1R, 12½, and so many dayes he laboured, and 17 dayes ½ he idled, which is thus proved.

[Page 381]

DayGroatsDayesGroats
Labour 1712½87½
Idleness 1517½87½

Where you see the reward is the same with the mulct.

Question XIII.

ONe sels 20 pound weight, part Saffron and part Ginger, for 45 crowns; but he sold 1 pound of Saffron for 3 crowns, and 1 pound of Ginger for ½ a crown. The question is, how many pounds of each sort he sold?

Put for the Saffron 1R of pounds, and for the Ginger 20−1R of pounds, then by the rule of Three work thus.

PoundsCrownsPoundsCrowns
Saffron 131R?3R
Ginger. 1½20−1R?20−1R / 2

Therefore the equation is between the sum of 3R of crowns, and 20−1R / 2 of crowns added to­gether, and 45 crowns, now that sum is 10+5/2R of crowns (for 20−1R / 2 is equal to 10−½R, to which if you adde 3R of crowns, it makes the sum 10+5/2R) therefore the equation shall be between 10+5/2R of crowns, and 45 crowns. Take away 10 from both parts, and it will be be­tween 5/2R and 35. Divide therefore 35 by 5/2, and [Page 382]you shall have 14 for the value of 1 R. And so many pounds of Saffron were sold, and 6 pound Ginger, which thus I prove.

PoundCrownsPoundsCrowns
Saffron 1314?42
Ginger 1½6?3

Where you see that the price of 14 pound Saf­fron, and 6 pound Ginger added make 45 crowns.

Question XIV.

A Certain Trades-man hath 2 sorts of coyn, in number 560 pieces, worth 160 crowns; a cer­tain part thereof is worth each piece ⅓ of a crown, and each piece of the rest ¼ of a crown. I demand the number of the first and latter sort of money?

Put 1R for the first, and 560−1R for the latter, and then constitute the rule of Three after this manner.

MoneyCrownMoney 
11R?⅓R
1¼560−1R?560−1R / 4

The fourth number found is equal to 160 crowns, and the sum of their numbers makes 140+1/12R (for 560−1R / 4 is equal to 140+¼R, to which if you adde ⅓R, makes the summe 140 +1/12R.) There is therefore an equation between 140+1/12R, and 160 crowns. Take away 140 [Page 383]from both parts, and then the equation is between 1/12 and 20. Divide therefore 20 by 1/12, and you have 240 for the value of 1 R, and so much mo­ney there was of the first sort, of which each piece was worth ⅓ of a crown of the latter kinde 320, each worth ¼ of a crown. Thus proved.

Money MoneyCrowns
1240?80
1¼320?80

Where you see the numbers in the fourth place make 160 crowns.

Question XV.

IN the Army of the Emperour, the number of the Infantery were octuple to the number of the Cavallery, among them there is distributed 392000 crowns, so as that every Foot Souldier had 5 crowns, and every Horseman 16. The question is, Of how many Horsemen the Army consisted, and of how many Footmen.

Put 1R for the Horsmen, and 8R for the Foot, according to the condition of the question, and then constitute the rule of Three thus:

 CrownsHorsmenCrowns
Horse 1161R16R
Foot 15Foot 8R40R
   56R

Therefore 56R of crowns shall be equal to 392000 crowns: wherefore divide 392000 by 56, and you shall have for 1R 7000 for the number [Page 384]of Horsmen; therefore the Foot shall be 56000, eight times as many, and so there will be distri­buted to the Horsmen 112000 crowns, and to the Foot 280000, which together make 392000 crowns.

Question XVI.

A Man hath a certain sum of money in a purse, which a stander by judgeth to be 600 crowns, whose errour he thus corrects. If to what I have in this purse, there be added ½ ⅓ and ¼, and from the sum there be substracted 1/12 part of my mo­ney, then I should have 600 crowns. The que­stion is, How many crowns he had in the purse!

Put 1R for the number of crowns. If the parts ½R, ⅓R, and ¼R, which together make 1 1/12R, be added to 1R, the whole makes 2 1/12R; take a­way 1/12R, and there will remain 2R equal to 600. Divide therefore 600 by 2, and you have 300 for the value of 1R, and so much money was in the purse. For if you adde the ½, ⅓, and ¼, to wit 150, 100, and 75, it will make the sum of 625, and taking away 1/12, to wit 25, the rest will be the number 600, which resolves the question.

Question XVII.

Certain Traveller goes 9 miles a day, ano­ther Traveller after the tenth day past, be­gins his journey from the same place, and goes every day 14 miles. I demand in how many dayes the latter will overtake the first.

[Page 385]Put 1R for the number of dayes, therefore the first over and above 90 miles, which he hath gone in ten dayes, goeth in 1R of dayes, besides 9R of miles, seeing that every day he goeth nine miles. But the latter going 14 miles each day, goeth in 1R of dayes 14R of miles, and because the first of necessity in 1R of dayes goes as many miles together with 90, which he went in ten dayes as the latter went in 1R of dayes; sith that then they are to meet together, the equation shall be between 9R+90 and 14R. Take away 90 from both parts, and it will be between 90 and 5R. Divide therefore 90 by 5, and 1R makes 18. Therefore in 18 dayes they shall come together. For the first in 18 dayes went 162 miles, which added to 90, he made the first 10 dayes, makes 252 miles, which the latter went in 18 dayes.

Question XVIII.

A Traveller goes nine miles a day, another Traveller after the end of ten dayes begins the same journey. I demand, how many miles a day the latter ought to travel, that so in 18 dayes he may overtake the first.

Put 1R for the miles. Therefore in 18 dayes he will have travelled 18R of miles. Seeing therefore that the first travelling every day nine miles, went in 18 dayes 162 miles, and adding thereto 90, which he went the first ten dayes be­fore the second set forth, it is manifest that he tra­velled 252 miles. Therefore the equation is be­tween [Page 386]18R and 252. Divide 252 by 18, and you have 14 for the value of 1R, and so many miles the latter ought to travel, to overtake the first in 18 dayes.

Question XIX.

A Certain man dying, made his will and testa­ment, leaving 3000 crowns to be distributed between his Wife, Son, and two Daughters, on this condition, that the portion of the Son might be double to that of the Mother, and the portion of the Mother double also to the portion of each of the Daughters. The question is, how much each ones portion was?

Put for the portion of one of the Daughters 1R, for the Mothers portion 2R, and for the Sons portion 4R. So there will be an equation between 8R and 3000 crowns.

  • The portion of one Daughter 1R that is 375 Cro.
  • The portion of th'other Daughter 1R that is 375 Cro.
  • The portion of the Mother 2R that is 750 Cro.
  • The portion of the Son 4R that is 1500 Cro.

Divide 3000 by 8, so the value of 1R will be 375, the portion of one of the Daughters, and therefore the Mothers portion will be 750, & the Sons 1500.

Question XX.

A Certain man receiveth of a Merchant, a quan­tity of Saffron for 10 crowns; and again he receives of the same man 24 pounds more of [Page 387]Saffron, at length he returns to him 30 pounds thereof again, and the merchant computing the price of the Saffron, restoreth to him 14 crowns. I demand the price of a pound of that Saffron.

Here you see 10 crowns +24 pounds to be the whole debt which the buyer owed to the mer­chant, & in like manner 30 pounds −14 crowns. Therefore there will be an equation between 10 crowns +24 pounds, and 30 pounds −14 crowns. Adde 14 crowns on both parts, and the equation will be between 24 crowns +24 pounds and 30 pounds, take away 24 pounds from both parts, and it will be between 24 and 6 pounds. Divide 24 by 6, and you have 4 for the root, and so many crowns one pound of Saffron is worth, which I thus prove. 2½ pounds are bought for 10 crowns, and so the buyer received of the merchant 26½, which were worth 106 crowns. If therefore to the merchant there be re­stored 30 pounds, the merchant oweth to the buyer 3½, seeing he received onely 26½, but 3½ pounds are worth 14 crowns, which the mer­chant rendered to the buyer.

Question XXI.

TWo men enter into fellowship in trade, now the second brings with him double the money that the first brings, and 5 crowns over and above. They gain by their traffick 960 crowns, of which the first takes to himself 300 crowns, and the se­cond 660. I demand how much each put in bank.

[Page 388]Put for the first 1R, and for the second 2R +5, the sum of both together is 3R +5, which have gained 960 crowns. Then work by the rule of three thus:

3R+59601R?960R / 3R+5

You shall finde that the first which brought in 1R, hath gained 960R / 3R+5 which number is equal to 300 crowns which he received. This equation by crosse multiplying will be reduced to 960R, equal to 900R+1500. Take away therefore 900R from both parts, and there will Remain 60R equal to 1500, divide 1500 by 60, and you have 25 for one R, and so much the first put in bank, therefore the second put in 55 crowns, and is thus proved. For both put in 80 crowns.

  • 80. 960 25? 300
  • 80. 960 55? 660

Question XXII.

THree Merchants gain together 700 crowns, which thus they distribute amongst them­selves (having regard to the summe each one brought into bank) so as that the portion of the second surmounted the portion of the first by 12 crowns, and the portion of the third surpassed that of the second by 16 crowns. I demand how much each mans portion was.

Put for the portion of the first man 1R, and then shall the portion of the second be 1R+12, and the portion of the third 1R+28. These 3 [Page 389]portions together make 3R+40, equal to 700. Take away 40 from both parts of the equation, and the equation will be between 3R and 660. Divide therefore 660 by 3, and you shall have 2 [...]0 for the value of 1R, and so much was the first mans portion, so the portion of the second shall be 232, and of the third 248, which all to­gether make 700.

Question XXIII.

A Caterer buyes a number of Fowls, so as that if he had bought the ⅓ ¼, and ⅕ of the number, and over and above 6, he would then have 100 just. I demand the number of Fowls he bought.

Put 1R for the number, whose ⅓, ¼, and ⅕ make 47/60R, which with 6, make 47/60R+6 equal to 100. Take away 6 from both parts of the equation, and the remaining equation will be between 47/60R and 94. Divide therefore 94 by 47/60, and the value of 1R will be 120, to wit, the number of Fowls that were bought.

For its ⅓ is 40, and its ¼ is 30, and its ⅓, 24; which all together with 6, make the sum 100.

Some Examples in Algebra con­cerning Squares.

To finde two numbers in a given Excesse, so as their Squares may have also a given Excesse.

LEt there be sought two numbers, whose Ex­cesse is 4, and the Excesse of their Squares 144.

Put for the lesser number 1R, and therefore the greater number must be 1R+4, whose squares are 1Q and 1Q+8R+16, their ex­cesse is 8R+16, which ought to be equal to 144. Take away 16 from both parts of the equa­tion, and the equation will be between 8R and 128. Divide therefore 128 by 8, and you shall have for the value of 1R, 16 for the lesser num­ber, the greater therefore will be 20, that the Excesse may be 4. The squares of those 2 num­bers are 256, and 400, whose difference or excesse is 144.

Two numbers being given, to finde another, with which multiplying both the given numbers, makes the grea­ter number a square, and its lesser the side of that square.

LEt the two given numbers be 200 & 5, and let the sought number be put 1R. Now 1R mul­tiplied in 200, produceth 200R, and 1R multi­plied [Page 391]in 5, makes 5R, which ought to be the side, and so multiplied in it self, ought to make a num­ber equal to 200. But 5R multiplied in it self, makes 25Q, the equation therefore is between 200R and 25Q. Divide therefore 200 by 25, and the value of 1R is 8, which multiplied in 200, make the square 1600, whose side is the num­ber 40, which is also produced by the multiplica­tion of 8 in 5.

Some Examples relating to Cubes.

To finde a number, which multiplied in it self, and the product multiplied by some given number, may pro­duce a number in a given proportion to the Cube of the found number.

LEt the given number be 20, find another num­ber, which being first multiplied in it self, and then the product multiplied by the given number 20, may produce a number in a quintuple propor­tion to the Cube made of the found number.

Put for the sought number 1R, which multi­plied in its self makes 1Q, which also multiplied by 20, makes 20Q, the Cube of 1R is 1C, to which 20Q ought to have a quintuple propor­tion; so the equation is between 20Q and 5C: Divide 20 by 5, and you have for the value of 1R, 4, the number songht. This 4 multiplied in it self makes 16, and 16 multiplied by 20, makes 320, which is quintuple to 64, the Cube of 4.

To divide a given number in two parts, so as that their Cubes may make a given sum, which shall be greater than the quarter part of the Cube described of the given number.

LEt the given number be 10, to be divided into two parts, whose Cubes may make 370, which number is greater than the ¼ part of the Cube of 10. Put for the first number compounded of 1R, and half of the given number 1R+5, and for the second 5−1R. So these two numbers do make the given number. Their Cubes are 1C+15Q+75R+125, and 125−75R+15Q−1C; their sum is 30Q+250: For +1C, and −1C, as also +75R, and −75R, do mutu­ally destroy one another: and of 15Q, and 15Q, are made 30Q; also 125, & 125, make 250. Ther­fore the equation is between 30Q+250, and 370. Take away 250 from both parts, and the equation will be between 30Q and 120. Divide 120 by 30, and you have 4 for the value of 1Q, and the value of 1R, 2. The first part put 1R+5 shall be 7; and the second put 5−1R shall be 3. The Cubes of those parts are 343 and 27, which together make 370.

FINIS.

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