their notes and periods which is directly contrary unto that of whole numbers for in the one to wit in whole numbers we reckon unites tens hundreds thousands and so forward from the right hand to the left but in Decimal fractions we reckon tenths of unity hundreds of unity thousands from unity and so forward from the left hand to the right as doth very clearly appear by the following Table 9 6 3 4 7 6 5 8 3 Fifth place Fourth place Third place Second place First place First place Second place Third place Fourth place In this Table you may observe that the places of Decimal parts are separated from the place of unity in whole numbers by a line drawn between them so that the number on the left hand of that line expresseth 9 6 3 4 7 integers or unities and the number on the right hand of the line 6 5 8 3 expresseth the parts of an unite or an integer divided into 10000 equal parts in like manner this number 7 6 doth signifie 7 integers and eight tenth parts of an integer and this number 3 4 7 6 5 doth signifie 347 integers or unities and 65 decimal parts of an integer CHAP. XVI Of the Reduction of vulgar fractions into decimal fractions 1. HAving shewed the notation of Decimal fractions I come now unto their numeration the which as in vulgar fractions is accidental or essential 2. Accidental numeration is the Reduction of vulgar fractions into Decimals and this may be performed by the fourteenth rule of the twelfth Chapter where is shewed how fractions of diverse denominations may be reduced into fractions of any denomination desired namely by multiplying the numerator given by the denominator required and dividing the product by the denominator given now then to apply this rule to decimal fractions let it be required to reduce this vulgar fraction 5 â— of a pound into a fraction whose denominator shall be 1000 multiply 5 the numerator given by 1000 the denominator desired and the product is 5000 which being divided by 8 the denominator given the quotient is 625 the numerator answering to 1000 the denominator desired and so the vulgar fraction â— â— is reduced into the decimal fraction 625 1000. 3. Upon this ground the known or accustomary parts of money weight measure time motion c. may be reduced unto decimals and first for money if you desire to know what decimal fraction is equal to one shilling or 1 20 of a pound assigning the denominator of that decimal to be 100 the numerator thereto by the last rule will be found to be 5 so that 5 100 or .05 is a decimal fraction equal in value to one shilling or 1 20 part of a pound sterling In like manner if the Decimal answering to 9 â—â— of 1 â—â— of a pound were required that is the decimal of 9 pence the compounded fraction 9 â—â— of 1 â—0 being reduced to a single fraction is 9 240 now then if you assign the denominator of that decimal to be 1000 the numerator thereto by the last rule will be found to be 375 so that 375 10000 or .0375 is a decimal fraction equal in value to 9 pence or 9 240 of a pound sterling Or if the decimal answering to ¾ of 1 12 of 1 20 of a pound were required that is the decimal of 3 farthings the compounded fraction ¾ of 1 12 of 1 20 being reduced into a single fraction is 3 960 now then if you assign the denominator of that decimal to be 10.000.000 the numerator thereto by the last rule will be found to be 31250 so that 31250 1000000 or .0031250 is a decimal fraction equal in value to â— 960 of a pound sterling And hence it appears that to find the decimal answering to any number of shillings you must multiply the shillings propounded by the denominator 10000 c. and divide the product by 20 the number of shillings in a pound sterling And to find the decimal answering to any number of pence you must multiply the pence propounded by the denominator 10000 c. and divide the product by 240 the number of pence in a pound sterling And to find the decimal answering to any number of farthings you must multiply the number of farthings propounded by the denominator 100000 and divide the product by 960 the number of farthings in a pound sterling The decimal answering to any known parts of money weight measure c. may be otherwise found in this manner suppose a pound sterling to be divided into 1.0000000 equal parts then doth .1.0000000 represent one integer or 20 shillings and the half thereof 0.50000000 is the decimal of 10 shillings 0.500000 is the decimal of one shilling and the half thereof 0.0250000 is the decimal of six pence and the half thereof is the decimal of 3 pence to wit 0.01250000 and a third of 0.01250000 to wit 0.00416667 is the decimal of a penny the half of the decimal of a penny viz. 0.00208333 is the decimal of half one penny and the half of that to wit 0.00104166 is the decimal of a farthing and by either of these waies may a Table be made to shew the decimal of any number of shillings pence and farthings 5. But if any dislike this because the decimals of some parts of a pound cannot be exactly found though it may be thus found so infinitely near that all questions may be resolved without considerable though not without sensible errour yet to satisfie the curiosity of such as shall make this objection let them make the third part of a pound to be the integer that is six shillings and eight pence so may they have the decimal of any number of shillings pence and farthings exactly as by the following operation doth appear Let 6s 8d be divided into 1.000000 Then shall the decimal of 8d be 0.100000 The decimal of 4d shall be 0.050000 The decimal of 2d 0.025000 The decimal of 1 penny 0.012500 The decimal of 2 farthings 0.006250 The decimal of 1 farthing 0.003125 6. The like may be done for weights and measures but as for weights there will be no great advantage by turning them into decimals because all questions concerning them may be more conveniently resolved without them and as for measures especially long measures there needs no reduction of them for the foot yard or chain for measuring of land being divided first into ten equal parts and then each of those into ten other equal parts they will be more useful than as they are now usually divided 7. And in Astronomical fractions the Reverend Learned and the famous Oughtred in the 1 Chap. of his key to the Mathematicks doth tell us that this Decimal Logistica or accounting is much easier and nearer than that Sexagenary commonly used which he plainly perceived saith he whatsoever he was that first reduced the Canon of Sines from a Semidiameter of 60 parts unto 1 with circles or cyphers annexed and
new fraction will be 84 â—0 as before 5 Example If 4¼ of a Yard of Cloth in length being 1 Yard and half broad will make a Cloak how much Plush that is ¾ of a Yard in breadth will serve to line that Cloak Here the fourth term required must be greater than the second term given and therefore the lesser extream must be the Divisor and the less extream being in the third place plainly sheweth that this question doth belong to the Rule of Three inverse and the numbers must stand as here you see breadth length breadth 1 1â— 4¼ ¾ or 3 â— 17 4 ¾ And ¾ being the Divisor 4 being multiplied first by 17 and then by 3 the last product is 204 and 3 being multiplied by 4 and by 2 the last product is 24 and so the new fraction or quotient sought is 204 24 which being reduced is 8½ yards 6. Example in Decimal fractions If 7 ounces 3 penny weight and 12 grains be worth 21l 11s 6d what is the value of 1½ ounce The Decimal of 3 penny weight is 0.15 And the Decimal of 12 ounces is 0.025 And the sum of these 2 Decimals is 0.515 Wherefore the first number in the Rule of Three is 7.175 Again the Decimal of 10s 6d is 0525 Wherefore the second number is 21.515 And the third number is 1.5 Now to resolve this question the 3 given numbers will stand thus 7.175 − 21.525 − 1.5 Lastly multiplying the second by the third the product is 32.2875 which being divided by 7.175 the quotient is 4.5 that is 4 pounds and ten shillings which is the price of an ounce and half of that Gold CHAP. XXIII Of the Rules of Practice 1. WHen the Rule of Three direct hath 1 or an Integer for the first term it is commonly called a Rule of practice not only for the speedy but the practical resolution of such questions for whereas other questions belonging to this Rule do for the most part require Multiplication and Division these in the ordinary way are resolved by Multiplication and whereas Multiplication is generally much easier than Division these questions are in an accustomary or practical way for the most part performed by Division and yet done with more ease than they can be by Multiplication nay where Multiplication and Division are both used the resolution is yet easier than it is or can be by Multiplication only 2. The questions to be resolved by these abbreviations of the Rule of Three or practical operations do for the most part come under one or other of these five cases The price of 1 or an Integer doth consist either 1. Of a certain number of shillings under twenty 2. Of a certain number of pounds and shillings 3. Of a certain number of pence under twelve 4. Of a certain number of shillings and pence 5. Of a certain number of pounds shillings and pence with the parts of a penny 3. Now for the resolution of such questions which may or do fall under any of these cases it is necessary that the Aliquot parts of a pound and shilling be first known that is how often the shillings of pence which are the given price of one Integer are conteined in a pound or shilling without leaving any remainer thus 4 shillings is an aliquot part of a pound or 20 shillings because 4 may be taken 5 times in 20 without either excess or defect and 3 is an aliquot part of a shilling or 12 pence for 4 times 3 doth make 12 but 7 is not the aliquot part of 20 for two times 7 is less than 20 and 3 times 7 is more In like manner 5 is not the aliquot part of 12 for 2 times 5 is but 10 and 3 times 5 is 15 the one wants of 12 and the other exceeds it but yet any number of shillings under 20 may be reduced to the aliquot parts of a pound and any number of pence under 12 may be reduced to the aliquot parts of a shilling Thus 7 shillings is one fift more 3 twenties of a pound and 5 pence is one third more one twelveth of a shilling or one fourth more one sixth Such aliquot parts of a pound or shilling as are most useful in the resolution of questions of this nature are here expressed in two Tables with which I have joyned a third shewing the product of 12 by any number not exceeding it self which in many questions will be found as useful as either of the other Aliquot parts of a pound Aliquot parts of a Shilling The product of 12 by any number less    Pence   10 0 ½ 11 â…“+â…“+¼ 2 × 12. 24 6 8 â…“ 10 ½+â…“ 3 × 12. 36 5 0 ¼ 9 ½+¼ 4 × 12. 48 4 0 â…• 8 â…“+â— â— 5 × 12. 60 3 4 â…™ 7 ¼+â…“ 6 × 12. 72 2 6 â…› 6 ½ 7 × 12. 84 2 0 1 10 5 ¼+1 â— 8 × 12. 96 1 8 1 12 4 â…“ 9 × 12. 108 1 4 1 15 3 ¼ 10 × 12. 120 1 3 1 â— 2 1 â— 11 × 12. 132 1 0 1 â— 1½ â— â— 12 × 12. 144 Shil Pence  1 1 20  These things premised I will now shew how any question coming under the aforesaid five cases may be resolved 1. Case where the price of an Integer is shillings only 4. Where the price of 1 or an Integer is two shillings the price of as many Integers as you will may be discovered by bare inspection for two shillings being the tenth of a pound the double of the first figure towards the right hand is the number of shillings required and the rest of the figures are so many pounds Example 567 yards at 2 shillings the Yard will cost 56 pounds 14 shillings for the double of 7 is 14 which I write down by it self as shillings then taking the rest of the figures towards the left hand for pounds the answer is 56l 14s 5. When the given price of 1 or an Integer is an even number of shillings greater than two multiply the number of Integers whose price is required by half the number of shillings given the double of the first figure towards the right hand in the product being set down for the shillings apart all the other figures towards the left hand shall be the pounds required Example let the price of 365 Yards be required at 14 shillings per 1 Yard if you multiply 365 by 7 which is the half of 14 the number of shillings given the product will be 2562 now the double of 2 the first figure towards the right hand is 4 the other figures are 256 pounds and so the answer is 256l 4s Here note that 4 shillings being the 5th part of a pound if that be the price of an Integer it will be all one to multiply by 2 or divide by 5 if the double of the first figure in the product towards the right hand be taken for the shillings according
36 6.7 42 6.8 48 6.9 54 7.7 49 7.8 56 7.9 63 8.8 64 8.9 72 9.9 81 8. The Multiplication of single figures being thus attained the Multiplication of compounded numbers cannot be difficult 9. In Compound Multiplication set down the given numbers as in Addition and Subtraction so 3421 being given to be multiplied by 2 place 2 the Multiplicator under 3421 the Multiplicand and having drawn a line write the double of every figure in the Multiplicand under the line because the Multiplicator is 2 in the same degree always with that figure of which it is the double Thus the double of one is 2 the double of 2 is 4 the double of 4 in the place of hundreds is 8 the double of 3 in the place of thousands is 6 and the product of 3421 multiplied by 2 is 6842. 10. When the product of any of the particular figures exceeds ten place the excess under the line as before and the last figure in the next place to it but a line lower and so the rest until the multiplication be finished as if the same number 3421 were to be multiplied by 8. every figure of the multiplicand is to be made 8 times as much as it is and subscribed in their proper places and then to be collected into one summe Thus 8 times 1 is 8 which I place under the 1. and 8 times 2 is 16 I write 6 in the place of tens a line lower and 1 in the upper line in the place of hundreds 8 times 4 is 32 I write 2 in the lower line in the place of hundreds and 3 in the upper line a place forwarder that is in the place of thousands 8 times 3 is 24 thousands I write 4 therefore in the place of thousands in the lower line and 2 a place forwarder in the upper or the lower line it matters not which because it is the last figure in the product And these two lines of numbers being added together the summe is 27368 = 3421 × 8. 11. But this may be more compendiously performed if that every figure to be set in the upper line shall be committed to memory and added to the next rank So 3421 being given to be multiplied by 8 as before for 8 times 1 I write 8 under the line in the place of unites for 8 times 2 being 16 I write 6 under the line in the place of tens and reserve 1 for the ten it exceeds to be added in the next rank the place of hundreds Then I say 8 times 4 is 32 unto which if I adde 1 which I kept in mind the whole is 33 wherefore subscribing 3 in the next rank under the line in the place of hundreds and carrying 3 in mind for the three tens that it exceeds I proceed to perform the rest of the work as you see it in the Example 12. When the Multiplicand and the Multiplicator are both compounded numbers for so many figures as are in the Multiplicator so many several products must be subscribed under the line which at last being added into one summe gives you the total product of all so 3421 being given to be multiplied by 28 the operation will stand thus For 3421 being multiplied by 8 the product is 27368 again 342 I being multiplied by 2 the product is 6842 which several products standing in their due order and added together do make 95788 the product required 13. When in the Multiplicator there are Cyphers between the significant figures you must multiply by the significant figures only neglecting the Cyphers and set each particular product in its due place according to what hath been said in the last rule and as is done in these examples following 14. When the numbers given to be multiplied do one or both of them end with Cyphers neglecting the Cyhers multiply by the significant figures only and having added together the several products according to the former directions annex as many Cyphers to the summe of all the products as were at the end of both the numbers given so shall you have the product desired as in the examples following 15. When one of the Numbers given is an unite with nothing but Cyphers annexed the Multiplication is performed by annexing so many Cyphers to the Multiplicand as there are in the Multiplicator so if 4327 were to be multiplied by 1000 the product will be 4327000. CHAP. V. Of Division 1. DIVISION is that by which we discover how often one Number is conteined in another and thereby find the Quotient of the greater Number 2. Division hath three parts the Dividend the Divisor and the Quotient The Dividend is the Number given to be divided The Divisor is the Number by which the Dividend is divided and the Quotient is the Number which the Division doth produce So if 18 were given to be divided by 3 the Number produced would be 6 for 3 is found 6 times in 18 and here 18 is the Dividend 3 the Divisor and 6 the Quotient 3. Division is either single or compound When the Divisor is a single figure and the Dividend less than 10 the answer may be given by memory or if the Dividend be the product of two single figures the answer may be had from the multiplication table if need be for finding the Divisor in the first Column and the Dividend in the second the other figure in the first Column is the Quotient sought as if 56 were to be divided by 7 in the same line where I find 7 in the first Column and 56 in the second I find 8 also which is the Quotient 4. When the Divisor is a single figure and the Dividend a Number of many places make a crooked line at each end of the Dividend that on the left hand serving for the place of the divisor that on the right for the Quotient then proceed gradually ask how often the divisor is contained in the Dividend and place that which answers the question in the Quotient then multiply the Divisor by that figure in the Quotient substract the product from the dividend setting down the remainer For example let 2916 be given to be divided by 4. placing the numbers as hath been said I ask how often 4 is conteined in 29 the answere is 7 times I therefore put 7 in the Quotient then I multiply 4 by 7 and it makes 28 which being subtracted from 29 the remainer is 1. Then drawing a stroke between the remainer and the Dividend I draw down the next figure which is 1 also and so the new Dividend is 11 then I ask how often 4 is conteined in 11 the answer is twice therefore I put 2 in the Quotient and 4 multiplied by 2 makes 8 which being deducted from 11 the remainer is 3 which I write under 11 and draw a stroak between them as before and draw down the next figure which is 6 and so the new divided is â—â— Then I ask how often 4 is conteined