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twelve) compose the given number. If then the number be divided by 12, the remainder will be the number of single units less than 12; and the quotient will be the exact number of twelves. If now the number of twelves be divided by 12, the remainder will be the number of single twelves, less than 12; and the quotient will be the number of twelves of twelves. In the same manner the nunber of twelves of twelves, &c. less than twelve may be found. The same may be said of any other radix.

Hence the Rule for the expression of an integral number in any scale. If the number be a fraction less than 1, it is to be expressed (using the same radix as before ;) as a number of twelfths, of twelfths of twelfths, &c. less than twelve. Now numerator and denominator of a fraction may be both multiplied by 12, without altering the value: let this then be done, and let the multiplied numerator be divided by the original denominator; the integral quotient, which is of course less than 12, will be the number of twelfths in the fraction. In like manner the fractional part of the quotient may be converted into twelfths, which will be the twelfths of twelfths in the given fraction. So the other numbers may be found. The same may be said of

any radix.

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The same Prop. exhibited algebraically:-Let N be the given number, and let do, Q1, Q2, az, &c. an be the digits in order from right to left, which represent the number. Then N = an g + an-1gb-1+

tag r2 taartao N

do = an ph-1 + an - 17 ? +

+dgr tait

do Since the digits are all less than r, a is less than 1, or ao is the remainder

T
after dividing N by r. Similarly if N be the integral quotient,

N
= an 9.0 + an- ] g -3+

aj

taat or ay is the remainder after dividing N byr. Similarly it may be shown that the successive digits are the remainders arising from successive divisions by r. If N is a fraction, and a 1, A2, &c. be the successive digits,

d2 a3 then N=- + +

+ &c.
r2

02
.. Nxr=ait-t-+ &c.

22 or the first digit is the integral part of N Xr. Similarly if N, be the fractional part of N Xr.

a2 ყვ
N=-+-+&c.

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or a, is the integral part of N Xr. Similarly it may be shown that the successive digits may be found by successive multiplications of the fractions.

B

B

A

E

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Prop. 93.-The number of superficial units in the area of a

rectangle is the product of the numbers of lireal units of

the same kind in the length and breadth. Let A B, A C, be sides of a rectangle, and A let A B contain 6, A C 4, units of the same kind. Divide A B into 6, and AC into 4 equal parts, and through the points of bisection draw lines parallel to A C, A B, thus dividing the rectangle into equal squares, each being a superficial unit. Then it is seen that C the number of these squares is 24 or 6 X 4. Therefore, if the numbers of units in the sides be integers, the Prop. is true.

Next let A B contain 3} units: A C 24 units of the same kind : produce A B to E making A E = 4 times A B, and produce A C to F making A F=twice A C. Construct a figure similar to the former, on A E, A F, as sides of a rectangle. Then it is seen that the rectangle A G is composed of 8 rectangles, each equal to BC. But A E contains 4 X 34, or 13 units, and A F contains 2 X 2), or 5 units: therefore A G contains 13X 5, or 65 superficial units. Hence BC being one eighth of A G, contains 62 superficial units. But 65 = 4 x Š = 31 X 24. Therefore the Prop. is true when the numbers of units are fractions.

Cor. 1. Hence a square whose side contains a units. contains a X a superficial units. Therefore a square whose side is 12 in. or 1 ft., contains 144 square inches ;

i.e. 1 square foot is equal to 144 square inches. In a similar manner the other parts of superficial measure may be proved.

Cor. 2. Hence a rectangle 12 in. by I in. containing 12 square inches, is the 12th part of a square foot, and if a be a number of feet in the length of a rectangle, 6 the number of inches in the breadth, the area of the rectangle is a X 12 X b square inches = a b twelfths of a square foot. Therefore a number of feet multiplied by a number of inches gives the number of superficial primes in a rectangle, whose sides contain these numbers of feet and inches. In a similar manner it may be shown that a number of inches multiplied by a number of twelfths of inches, gives the number of superficial thirds, each being the twelfth of a square inch.

Cor. 3. Hence the method of finding the area of a rectangle, whose sides are given as a number of feet, inches, &c. by Cross Multiplication, is evident. For the area of a rectangle contained by any two lines is the sum of the areas of the rectangles contained by each part of the one, and each part

А

B

с

D

F

G

of the other. Thus the rectangle conlained by AD and A E, is equal to the sum of the rectangles contained by A F, and each of A B, BC, CD; and by FG, and each of the same parts; and by GE and each of the same parts. If A B, BC, CD; AF, el FG, GE; be lengths of a certain number of feet, inches, seconds, the areas of these rectangles may be found as above, and added together. Prop. 94.— The number of solid units in a rectangular

parallelopiped is the product of the numbers of lineal units of the same kind in the length, breadth, and thick

ness.

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Let A B, A C, AD, be the three edges of a rectangular parallelopiped ; let A B, AC, A D, contain respectively 6, 3, and 4 units. Divide A B into 6 parts in F &c. through F &c. draw planes pa

Fig. 1. rallel to the side ACED of the solid ; then the

A F solid is divided into 6 solids, each equal to A H, whose base is ACED, and height 1 unit; therefore the whole solid contains 6 times as many solid units as A H. Again, if AC be divided into 3 equal parts, and planes be drawn through the points of section parallel to A K, A H will be E H divided into 3 equal solids, whose base is A K, and height 1 unit. Therefore the whole solid will be divided into 6 X 3 of these solids. Lastly, if A D be divided into 4 equal parts, and planes be drawn through the points of bisection parallel to A G, the solid with base A K will be divided into 4 equal solids, the base of (each being 1 square unit, and the height of each 1 unit; each is therefore 1 solid or cubic unit. Therefore the whole solid contains 6 X 3 X 4 cubic units. Whence the truth of the Prop. when the numbers of units in the dimensions are integers. Fig. 2.

Next let AB, AC, A D, contain respectively 21, 33, 41 units; produce them to E, F, G; A E being 2 A B, A F being 3 AC, AG being 4 AD; and describe a rectangular parallelopiped on A E, AF, AG as edges. If A E, A F, AG, be divided into 2, 3, and 4, equal parts, and planes be drawn through the points of section parallel to the sides of the solid, the solid will be divided into a number of solids, each equal to that contained by A B, AC, AD, in number

K K

Ін

D

A А

F

1

th

2 X3 X4 part that of the larger. But the larger contains a number of cubic units equal to 5 X 10 X 17, since A E, A F, AG, contain 5, 10, and 17 lineal

5 X 10 x 17 units, therefore the smaller contains

cubic units, or s x 10

2 X 3 X 4 x 17, or 2 X 33, X 4+ cubic units. Hence the Prop. is true when the dimensions are fractional.

Cor. 1. Hence a cube, whose edge contains a units, contains a Xa xa cubic units ; therefore a cube, whose edge is 12 inches, or 1 foot, contains 12 X 12 X 12 or 1728 cubic inches; i.e. I cubic foot is equal to 1728 cubic inches. In the same way the other parts of solid measure may be proved.

Cor. 2. Hence a rectangular parallelopiped, whose edges are 12 in. 12 in. I in. containing 144 cubic inches, is one-twelfth of a cubic foot: and if a, 6 be numbers of feet in two edges, and c a number of inches in the third, the content of the parallelopiped is a X 12 X 6 X 12 X c cubic inches, or axbXcX 144 cubic inches, or a X6 X c twelfths of a cubic foot. But a X b is the number of square feet in a rectangle whose sides contain a, and 6 lineal feet: therefore a number of square feet in a rectangle multiplied by a number of lineal inches gives the number of cubic primes in the solid, whose base is the rectangle, and height the inches. Also 6 X c is the number of superficial primes in a rectangle whose sides are b feet and c inches; therefore a number of superficial primes in a rectangle multiplied by a number of ļineal feet gives the number of cubic primes in a solid whose base is the rectangle, and height the feet. In the ordinary mode of speaking, square feet multiplied by lineal inches give cubic primes; and superficial primes multiplied by lineal feet give cubic primes. Similarly it may be shown that superficial primes multiplied by lineal inches, or superficial seconds by lineal feet, give cubic seconds.

Cor. 3. Hence the method of finding the solid content of a rectangular parallelopiped by Cross Multiplication is evident, the dimensions being given in feet, inches, &c. For the content of any parallelopiped (as that in fig. 2) is evidently the sum of the contents of those whose base is E F, and heights AD, DH, A K, KG. And the contents of these are equal to the sum of the contents of others whose bases are parts of E F, and heights as before. Therefore if these contents be determined, (as they may be by multiplying the square units in the bases by the lineal units in the heights,) and be added together, the sum will be the content of the parallelopiped.

APPENDIX.

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1.-- To explain the Rules of Reduction. The rul for converting numbers of one denomination another appear axiomatical, when an exact nnmber of one is contained in one of another. Thus, since £1 is equivalent to 20s. it is evident that any sum of money will be equivalent to 20 times as many shillings as pounds; and therefore that, if pounds are to be reduced to shillings, we must multiply by 20, and if shillings are to be converted into pounds, we must divide by 20. And similar reasoning may be used in all other cases of the same kind. But if no exact number of the one denomination be contained in one of the other, the question is evidently one of finding how many times one of the new denomination is contained in the given quantity, or of finding the ratio of the given quantity to the unit of the given denomination. The explanation therefore of the rule is by Prop. 45. 2.- To prove the Rule for conversion of shillings, pence, and

farthings, into decimals of a pound.
Since 2s. = 1)£. =.£.

ls. 2£. = .050£.
6d. is. = .025£.
1}d. = of 6d. = .00625£.

{d = 1 of 1 d. = .00104 £. therefore for every pair of shillings 1 must be placed in the first place of decimals; and for an odd shilling 50 in the 2nd and 3rd places : also for every farthing besides the shillings there must be l in the 3rd place, and, since for 6d. there must be 25 in the 2nd and 3rd places, 1 extra must be added for 6d. Also it appears that for every farthing above the last sixpence, 4 must be put iu the 4th and 5th places, which for 6 farthings or 11d. would become 24, but, since 25 appears in the 4th and 5th places for every 14d. 1 extra must be added for every 6 farthings. Lastly for every farthing above the last six there will appear in the 6th and following places the decimal figures equivalent to the vulgar fraction }; therefore the figures in these places will be found by converting the fraction, whose denominator is 6, and numerator the number of farthings above the last six, into a decimal.

Conversely, the approximate value of a decimal of a pound may evidently be obtained by this Rule :- Take a pair of shillings for every one, in the 1st place, and an odd shilling for 50 (if there be 50) in the 2nd and 3rd places.

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