and the quotient is; and is multiplied by 2, by multiplying the numerator, 3, by 2 (220), and the product is equal to the part of the mill sold. Hence,

To multiply a fraction by a fraction, or to change a compound fraction to a single one.

RULE.-Multiply the numerators together for a new numerator, and the denominators together for a new denominator.

QUESTIONS FOR PRACTICE (56).

2. A man owning of a farm, sold of his share; what part of the farm did he sell? Ans. -3. What part of a foot is of a foot?

of

Ans.

4. What part of a mile is of of a mile?

Ans.

5. Change of of of to a single fraction. Ans. 1032

of

6. Multiply

by 27.

225. DIVISION BY FRACTIONS.

1. In 6 dollars, how many times of a dollar?

Here we wish to divide $6 into parts, each of which shall be of a dollar, or in other words, divide 6 by Now in order to find how many times in 6, we reduce 6 to 4ths, by multiplying it by 4, the denominator of the fraction, thus: 4 times 6 are 24; 6 dollars, then, are 24 fourths, or quarters of a dollar; and dividing 24 fourths by 3 fourths (the numerator), the quo tient, 8, is evidently the number of times of a dollar may be had in 24 or 6 dollars. Hence,

226. To divide a whole number by a fraction. RULE.-Multiply the number to be divided by the denomina. tor of the fraction, and divide the product by the numerator.

QUESTIONS FOR PRACTICE.

2. In 7 shillings, how many

times

of a shilling?

Ans. 28.

3. In 17 bushels of wheat, how many times of a bushel?

Ans. 85. 4. In 1 gallon of wine, how many times of a gallon?

Ans. 17. 17 times. 5. In 5 eagles, how many of a dollar? Ans. 200.

6. In a pound of tobacco, how many quids, each weighof an ounce ?

ing

7.

Ans. 384—1014. How many are 7÷? 8÷4? 2÷32?

NOTE. Here it will be seen that division by a fraction, gives a quotient larger than the dividend.

227. DIVISION OF ONE FRACTIONAL QUANTITY BY AN

OTHER.

ANALYSIS.

1. If of a bushel of wheat cost of a dollar, what is

that per bushel?

To find the cost per bushel, we must divide the price by the quantity (154), that is, we must divide by 2. But to divide a number by a fraction, we multiply it by the denominator, and divide the product by the 3X4 12 numerator (226); hence, we must multiply by 4, as

5 5

is divided by 3, by multiplying the denominator, 5, by 3, as,
of a dollar then is the price of one bushel. Hence,

(121);

228. To divide a fraction by a fraction.

(220), and

12 12 5X3 15.

RULE.-Multiply the numerator of the dividend by the denominator of the divisor for a new numerator, and the denominator of the dividend by the numerator of the divisor, for a new denominator.

NOTE.-In practice, it will be most convenient to invert the divisor, and then proceed as in Art. 224.

1?

QUESTIONS FOR PRACTICE.

2. In 7 how many times Ans. 45.

3. In 22 how many times Ans. 118-1.

22?

4. At of a dollar a bushel for oats, how many can I of a dollar?

buy for

Ans. 3 bush.

229. ALTERATION

5. If of a yard cost of a dollar, what is that a yard? Ans. 16-$1.773. of a piece of cloth

6. If

be worth

of of an eagle, what is the whole piece worth? Ans. 12 eag.

IN THE TERMS OF A FRACTION

WITHOUT ALTERING ITS VALUE.

ANALYSIS.

A fraction is multiplied by multiplying its numerator, and divided by multiplying its denominator (219); hence if we multiply both the terms of a fraction at the same time by any number, we both multiply and divide the fraction by the same number, and therefore do not alter its value. Again, a fraction is divided by dividing its numerator, and multiplied by dividing its denominator (219); hence if we divide both the terms of a fraction at the same time by any number, we both divide and multiply the fraction by the same number, and therefore do not alter its value. Hence,

230. To enlarge the terms of a fraction

RULE.-Multiply both the terms of the fraction by the number which denotes how many times the terms are to be enlarged.

231. To diminish the terms of a fraction

RULE.-Divide

both the terms of the fraction by such a number as will divide each without a remainder.

QUESTIONS FOR PRACTICE.

1. What is the expression 1. What is the expression for, in terms which are 10 for 18, in terms 10 times less? times as large?-for, the-for f, the terms being diterms being increased 9 minished 9 times?

times?

232. OF THE GREATEST COMMON DIVISOR OF TWO

NUMBERS.

ANALYSIS.

1. If the two terms of a fraction be 8 and 38, what is the greatest number that will divide them both without a remainder?

8) 38 (4

32

6)8(1

2)(3

It is evident that the greatest common divisor of 8 and 38, cannot exceed the smallest of them. We will therefore see if 8, which divides itself, and gives 1 for the quotient, will divide 38; if it will, it is manifestly the greatest common divisor sought. But dividing 38 by 8, we obtain a quotient, 4, and a remainder, 6; hence 8 is not a common divisor. Again, it is evident, that the common divisor of 8 and 38 must also divide 6, because 38-4 times 8 plus 6; hence a number which will divide 8 and 6 will also divide 8 and 38; we will therefore see if 6, which divides itself, will divide 8. But dividing 8 by 6, we have a quotient 1, and remainder 2; hence 6 is not a common divisor. Again, for the reason above stated, the common divisor of 6 and 8 must also divide the remainder, 2; and by dividing 6 by 2, we find that 2, which divides itself, divides 6 also; 2 is therefore a divisor of 6 and 8, and it has been shown that a number which will divide 6 and 8, will also divide 8 and 38. Hence 2 is the common divisor of 8 and 38, and it is evidently the greatest common divisor, since it is manifest from the method of obtaining it that 2 will divide by it, and a number will not divide by another greater than itself. Therefore,

233. To find the greatest common divisor of two numbers.

RULE.-Divide the greater number by the less, and the divisor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remains; then will the last divisor be the common divisor required.

QUESTIONS FOR PRACTICE.

2. What is the greatest common divisor of 580, 320 common divisor of 24 and 36? and 45?

Ans. 5.

NOTE. When there are more than two numbers, find the common divisor of two, then of that divisor and one of the others, and so on to the last.

6. What is the greatest common divisor of 918, 1998 and 522? Ans. 18.

REDUCTION OF FRACTIONS ΤΟ THEIR MOST

SIMPLE EXPRESSION.

ANALYSIS.

1. What is the most simple expression, or the least terms of 48?

The terms of a fraction are diminished, or made more simple, by division (230). Now, if we divide so long as we can find any number greater than 1 which will divide them both without a remainder, the fraction will evidently be diminished to the least terms which are capable of expressing it, since the two terms now contain no common factor greater than unity. Thus, 2)&z=136, 2)2+6=1,2)=34, and 2)=1, least terms. Or if we find the greatest common divisor of the two terms, 48 and 272, we may evidently reduce the fraction to its lowest terms at once by dividing the two terms by it. By Art. 233, we find the greatest common divisor to be 16, and 16), least terms as before. Hence,

235. To reduce a fraction to its least terms.

RULE.-Divide both the terms of the fraction by the greatest common divisor, and the quotient will be the fraction in its least

terms.

QUESTIONS FOR PRACTICE.

2. What are the least terms

5. Reduce

terms.

6. Reduce

terms.

to its least
Ans.
to its least

Ans..

7. Reduce 1 to its least

terms.

236.

COMMON MULTIPLES OF NUMBERS.

1. What number is a common multiple of 3, 4, 8 and 12?

3X4X8X12 1152. Ans. First, 3 times 4 are 12; 12 then is made up of 3 fours, or 4 threes; it is, therefore, divisible by 3 and 4. Again, 8 times 12 are 96; then 96 is divisible by 8, and as it is made up of 8 12s, each of which is divisible by 3 and 4, 96 is divisible by 3, 4 and 8. Again, 12 times 96 are 1152; 1152 then is divisible by 12; and as it is made up of 12 ninety-sixes, each of which is divisible by 3, 4 and 8, 1152 is divisible by 3, 4, 8 and 12; it is therefore a common multiple of these numbers (215. Def. 9).

237. 2. What is the least common multiple of 3, 4, 8 and 12?

4)3, 4, 8, 12 3')3, 1, 2, 3

1, 1, 2, 4×3×2-24

Every number will evidently divide by all its factors: our object then is to find the least number of which each of the numbers, 3, 4, 8 and 12 is a factor. Ranging the numbers in a line, and dividing such as are divisible by 4, we separate 4, 8 and 12, each into two factors, one of which, 4, is common, and the others, 1, 2 and 3 respectively. Now as the products of the divisor, multiplied by the quotients, are, severally, divisible by their respective dividends, the products of these products by the other quotients, must also be divisible by the dividends; for these products are only the dividends a certain number of times repeated. The continued product then of the divisor, 4, and the quotient 1, 2, 3, (4×1×2×3—24) is divisible by each of the dividends, 4, 8 and 12, and 24 is obviously the least number which is divisible by 4, 8 and 12, since 12 will not divide by 8, and no number greater than 12, and less than twice 12, or 24, will divide by 12. But the undivided number, 3, must also divide the number sought; we therefore bring it down with the quotients, and dividing the numbers by 3, which are divisible by it, we find that 3 is already a factor of 24, and will therefore divide 24. Thus, by dividing such of the given numbers as have a common factor by this factor, we suppress all but one of the common factors of each kind, and the continued product of the divisors, and the numbers in the last line, which include the quotients and undivided numbers, will contain the factors of all the given numbers, and may therefore be divided by each of them without a remainder; and since the same number is never taken more than once as a factor, the product is evidently the least number that can be so divided. Hence,

238. To find the least common multiple of two or more numbers.

RULE. Arrange the given numbers in a line, and divide by any number that will divide two or more of them without a remainder, setting the quotients and undivided numbers in a line below. Divide the second line as before, and so on till there are no two numbers remaining, which can be exactly divided by any number greater than unity; then will the continued product of the several divisors, and numbers in the lower dine be the multiple required.