RULE OF THREE, OR PROPORTION IN VULGAR FRACTIONS.

1. Prepare the first and third terms as directed in Addition of Vulgar Fractions; if more than three terms are given, prepare all but the middle term as above.

2. If the middle term is a compound or a mixt fraction, reduce it to a single fraction, and to its lowest term.

3. In stating, use the numerators only, and proceed with them as with whole numbers, the result will be the numerator of the quotient, or answer, whose denominator is the same as that to which the middle term was reduced.

NOTE. These examples may be proved by varying their order.

EXAMPLES.

1. If of a yard of cloth cost of a dollar, how much willa yard cost?

2. If of a ton of hay cost $16,

of a ton cost?

by Case X, will be

" of T7

a dollar,

for the answer.

3. If of a hogshead of wine cost $30,

cost?

how much will Ans. $18. what will Ans. $5. what is it a Ans. $40. 5. If of a yard of painting cost of a dollar, what is it a yard? 8 Ans. 41300 or 45 cents.

4. If of a hogshead of wine cost 85, hogshead?

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4 fo

=

6. If 1 yard of paint cost 45 cents, what will of a yard cost? Ans. 40 cents, of a dollar. NOTE. The 5 and 6 examples prove each other by varying their order.

7. If of a dollar be worth 5 shillings, how much of the same currency is 32 dollars worth?u Ans. 7 of a shilling, which reduced to its proper quantity is 9 15.

7

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EXTRACTION OF THE SQUARE ROOT.

The square root of any number is that number which being multiplied by itself will produce the number of which it is called the square root, and the number so produced is called its square.

The square root of a given number is found as follows: viz.

Begin with the unit figure, and place a point over it, then proceed towards the left, pointing every second figure, as is done with the example below; each two figures are called a period.

If there are decimals, proceed towards the right from the unit figure, pointing every second figure : should a place be wanting in the decimals, supply it with a cipher.

The figure under the left hand point and that to the left of it (if any) is the first resolvend.

Find such a figure as, being involved to the second power,t will come the nearest to this resolvend, and not exceed it.‡

Place the number thus found under the resolvend, and subtract it therefrom; and place the figure, so involved, in the quotient, or root; to the remainder bring down two figures, or the next period, placing them at the right of it, for a new resolvend; draw a crooked stroke or line at the left of this new resolvend.

Double the quotient, and place it for a divisor at the left of the new resolvend, leaving room for more figures at the right of this divisor.

See how often this divisor is contained in the resolvend, excluding its right hand figure, allowing for the

By some called involving the number to the second power, &c. thus 4 x 4=16, the square of 4, or 4 involved to the sec ond power; also 4x4x4-64, the cube of 4; or 4 involved to the third power.

† Multiplied by itself.

This is called a subtrahend.

1

increase in multiplying the divisor, with the figure thus found annexed to it, by said figure.

Place this figure in the quotient, and also at the right of the divisor.

Multiply the divisor (thus increased) by this last figure in the quotient, the product will be the subtrahend to place under the resolvend, to be subtracted therefrom.

To the remainder bring down another period, and proceed as before, till the work is completed.

NOTE. The number of points over the given square, shows how many figures there will be in the root.

1

These examples are proved by multiplying the root by itself, and if any remain take it in, or add it to the product, as in proving division, the result will be equal to the first given number.

In this example I find that 1 is the highest number to place in the root or quotient, as the left hand period is 1, which I place both in the quotient, and under that period, and subtract it from the period, find none remains; I then bring down the next period 44, and double the quotient for a divisor, to find a second figure

of the root; this divisor 2 I can have twice in 4, (omitting the right hand figure in the resolvend) I set 2 in the quotient or root, and also at the right of the divisor, then I multiply the divisor thus increased, by this last figure in the root, and set the product under the resolvend, for a subtrahend, and subtract it therefrom; and as there are no more periods to bring down, the work is done; and as there is no remainder, the root is a perfect one.

NOTE 1. The figure found by dividing all but the right hand figure in the resolvend, will sometimes prove too great on account of the increase, by multiplying it with the other figures in the divisor; in which case a less number must be taken, so as the product may not exceed the resolvend.

NOTE 2. When there is a remainder after extracting the square root from the last resolvend, and a more exact root is required, annex two ciphers to the remainder, and proceed as before; if there be another remainder, annex two ciphers to it, and proceed on to as many places of figures as you please.

NOTE 3. All figures found in the root after annéxing ciphers, will be decimals, ni timu 29sig

5. Required the square root of 1729,