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two being given. Evidently the amount of coin varies directly as the sum which is to be exchanged for it; therefore the solution of these questions is obtained by a Simple Proportion, which, when the first term is unity, may be solved by ordinary Multiplication, or by Practice.

2nd. Let it be required to determine the course of exchange between two places, the courses between each of the two and a third being known. Thus let it be given that a coins of the 1st place are equal to b coins of the 3rd, and c coins of the 3rd equal to d coins of the 2nd, then it is required to find how many coins of the 2nd are equivalent to a coins of the 1st. This is of course the same number as are equivalent to b coins of the 2nd; so that the question is:-If d coins of a given place are equivalent to c coins of another, how many of the former are equivalent to b of the latter? Which is a question precisely similar to the 1st case, and the solution therefore is obtained in the same way.

3rd. Let it be required to find the course of exchange between two places, which are connected with several other places by having given the courses between the 1st and 3rd, the 3rd and 4th, the 4th and 5th, &c. the last and the 2nd. Thus let it be given that a, coins of the 1st are equivalent to b, coins of the 3rd place; a, of the 3rd equivalent to be of the 4th ; ag of the 4th to bz of the 5th ; 24 of the 5th to be of the 2nd; then it is required to find how many coins of the 2nd are equivalent to a, coins of the Ist. The solution of this question might be effected by a series of Simple Arbitrations by last case: thus the course between the 1st and 4th places might be found from the known courses between the 1st and 3rd, and the 3rd and 4th: then the course between the 1st and the 5th might be found in the same way; and then that between the 1st and 2nd. Thus:-.

Let the course between the 1st and 4th be x, that between the 1st and 5th be y; that between the 1st and 2nd be z: then the following proportions will give x, y, %.

bi

b2 ba

y

64 Therefore, compounding these proportions, we have:a, Xas Xa4

bi

b, Xbs X14 bi Xb, Xbs Xb

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a, Xag Ха4 The same result would be obtained by writing a series of equations expressiug the given courses of exchange, (thus

a, coins of Ist kind = b, coins of 3rd kind
a2

3rd kind = b ... 4th kind
a3 ......

4th kind = bg 5th kind

5th kind = 64 04

2nd kind)

......

and multiplying the numbers on the right hand side, and dividing the product by the product of all but the 1st on the left hand side. Whence has arisen the Rule called the Chain Rule.

Prop. 90.- To explain the Rules for questions in Barter.

1st. The value of goods given in exchange is to be equal to that of the goods received ; therefore the quantity of the goods exchanged must be the quantity which might be purchased for the value of the goods received. Hence thc first Rule.

But evidently questions in Barter are questions in Exchanges, in which, instead of different coins, we have different articles concerned. Hence the solution may be effected by the Chain Rule.

2nd. If it be given that a certain quantity of one article is exchanged for a certain quantity of another, whose rate of cost is known, and it be required to find the quality, or the cost of an unit, of the former, evidently the value of the number of units in the former quantity is equal to the value of the latter, which being determined, the value of one unit is obtained by division by the number of units.

3rd. If the cash price of one article is altered in a bartering transaction, in strict justice that of the article exchanged ought to be altered in the same ratio : or the ratio of the nett to the bartering prices ought to be the same for both articles. For instance, if in bartering the rate of cost of an article be doubled, only one half as much would be received in exchange for a quantity of another article, as would have been received, if the price had been unchanged. But if the rate of cost of the latter be doubled also, then the same quantity will have to be received as if neither price had been altered. Hence the nett, or bartering price of an article may be determined when either is given, the nett and bartering prices of another being given, by a Simple Proportion.

Prop. 91.-To explain the Rule of Alligation.

The value of the mixture is the sum of the values of the ingredients ; hence if these values be added, and divided by the number of units of quantity in the mixture, the result will be the value of one unit of the mixture. Hence the Rule.

Prop. 92.To explain the method of passing from one scale

of Notation to another. To express a number in any given scale, as the duodenary, we have to find what number of collections of single units, of twelves of units, of twelves of twelves of units, &c. (the number of each kind being less than 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 nuinber 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

a

any radix.

8

The same Prop. exhibited algebraically:—Let N be the given number, and let Qo, 21, 22, 23, &c. an be the digits in order from right to left, which represent the number. Then N = An goo + anap – 1

-1+

tag g2 taartao
N
=an yn - 1 + an - 171—2 +

taartai +

do

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ao Since the digits are all less than r, is less than 1, or ao is the remainder after dividing N byr. Similarly if N be the integral quotient, Ni

ai = an poni - + an- 1 g 3 + taat or a, 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,

ai a 2 ag then N=- +

+ &c, g2

+

r

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72

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

N=-+

+ &c.

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r

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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.

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D

B

A

E

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F

G

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 34 units: A C 23 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 31, or 13 units, and AF contains 2 X 21, or 5 units: therefore A G contains 13X 5, or 65 superficial units. Hence BC being one eighth of A G, contains 66 superficial units. But 6 = 1X

34 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 lin. 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, b 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

B

с

D

F

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, 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.

G

E

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.

A E

L

MI

DK

Let A B, A C, AD, be the three edges of a rectangular parallelopiped ; let A B, AC, AD, 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 solid is divided into 6 solids, each equal to A H, whose base is ACE D, 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 equal solids, the base of seach 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 2}, 3}, 44 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, AF, AG,

VITZ be divided into 2, 3, and 4, equal parts, and planes be drawn through the points of section

c/ parallel to the sides of the solid, the solid will be divided into a number of solids, each equal to that contained by AB, AC, A D, in number

K

H

D

A

B

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