PROOF. If the work be right, (¶ 16, "Proof,") the product of the quotient into the divisor will be equal to the dividend; thus, X. This, it will be perceived, is multiplying the price of one yard (3) by the quantity (7) to find the cost (;) and is, in fact, reversing the question, thus, If the price of 1 yard be 3 of a dollar, what will of a yard cost? Ans. of a dollar. Note. Let the pupil be required to reverse and prove the succeeding examples in the same manner.

4. How many bushels of apples, at bushel, may be bought for of a dollar?

of a dollar per Ans. 43 bushels.

5. If 4 pounds of butter serve a family 1 week, how many weeks will 367 pounds serve them?

The mixed numbers, it will be recollected, may be reduced to improper fractions.

Ans. 8

weeks.

¶ 58.

The RULE for division of fractions may now be pre

sented at one view :

I. To divide a fraction by a whole number,-Divide the numerator by the whole number, when it can be done without a remainder, and under the quotient write the denominator; otherwise, multiply the denominator by it, and over the product write the numerator.

II. To divide a whole number by a fraction,-Multiply the dividend by the denominator of the fraction, and divide the product by the numerator.

III. To divide one fraction by another, Invert the divisor, and multiply together the two upper terms for a numerator, and the two lower terms for a denominator.

Note. If either or both are mixed numbers, they may be reduced to improper actions.

EXAMPLES FOR PRACTICE.

1. If 7 lb. of sugar cost

of a dollar, what is it per pound?÷7 how much? of is how much? 2. At for of a barrel of cider, what is that per barrel?

3. If 4 pounds of tobacco cost of a dollar, what does 1 pound cost?

4. If of a yard cost $4, what is the price per yard? 5. If 14 yards cost $75, what is the price per yard?

Ans. 5

6. At 31 dollars for 10 barrels of cider, what is that per bariel?

7. How many times is contained in 746 ?

Ans. $3.

Ans. 1989.

11. Divide 43 by § of 4.

12. Divide § of 4 by 4ž.

Quot. 2.

Quot. .

ADDITION AND SUBTRACTION OF FRACTIONS.

π 59. 1. A boy gave to one of his companions of an orange, to another, to another; what part of an orange did he give to all? ++how much? Ans.. 2. A cow consumes, in 1 month, of a ton of hay; a horse, in the same time, consumes of a ton; and a pair of oxen, ; how much do they all consume? more does the horse consume than the cow?

than the horse? ++++$= how much? how much? how much?

how much the oxen

how much?

4. 26+2%+2%+1+2=how much? how much?

5. A boy, having of an applc, gave of it to his sister; what part of the apple had he left?

how much?

When the denominators of two or more fractions are alike, (as in the foregoing examples,) they are said to have a common denominator. The parts are then in the same denomina tion, and, consequently, of the same magnitude or value. It is evident, therefore, that they may be added or subtracted, by adding or subtracting their numerators, that is, the num ber of their parts, care being taken to write under the result their proper denominator. Thus, ++; } } =f.

6. A boy, having an orange, gave of it to his sister, and of it to his brother; what part of the orange did he give away?

4ths and 8ths, being parts of different magnitudes, or value, cannot be added together. We must therefore first reduce them to parts of the same magnitude, that is, to a common denominator. are 3 parts. If each of these parts be divided into 2 equal parts, that is, if we multiply both terms of the fraction by 2, (T 46,) it will be changed to; then § and Ans. of an orange.

are .

7. A man had of a hogshead of molasses in one cask, and of a hogshead in another; how much more in one cask than in the other?

Here, 3ds cannot be so divided as to become 5ths, nor can 5ths be so divided as to become 3ds; but if the 3ds be each divided into 5 equal parts, and the 5ths each into 3 equal parts, they will all become 15ths. The will become 19, and the & will become; then, taken from + leaves Ts, Ans.

¶ 60. From the very process of dividing each of the parts, that is, of increasing the denominators by multiplying them, it follows, that each denominator must be a factor of the common denominator; now, multiplying all the denominators together will evidently produce such a number.

Hence, To reduce fractions of different denominators to equivalent fractions, having a common denominator,-RULE: Multiply together all the denominators for a common denominator, and, as by this process each denominator is multiplied by all the others, so, to retain the value of each fraction, multiply each numerator by all the denominators, except its own, for a new numerator, and under it write the common denominator.

EXAMPLES FOR PRACTICE.

1. Reduce 3, and to fractions of equal value, having a common denominator.

3 X 4 X 560, the common denominator.

2 X 4 X 540, the new numerator for the first fraction. 8 × 3 × 545, the new numerator for the second fraction. 8 X 4 X 4 = 48, the new numeror for the third fraction.

The new fractions, therefore, are 48, 48, and 8. By an inspection of the operation, the pupil will perceive, that the numerator and denominator of each fraction have been multiplied by the same numbers; consequently, (T 46,) that their value has not been altered.

2. Reduce, and t to equivalent fractions, having a common denominator. Ans. 128, 128, 218, 123.

3. Reduce to equivalent fractions of a common denominator, and add together,,, and .

4. Add together

Ans. 28 +88+j8=ff=1ff, Amount.

and §.

5. What is the amount of +++++†?

Amount, 17.

Ans. 247=137 6. What are the fractions of a common denominator Ans. 1 and 2, or and t

equivalent to and ?

We have already seen, (59, ex. 7,) that the common denominator may be any number, of which each given denominator is a factor, that is, any number which may be divided by each of them without a remainder. Such a number is called a common multiple of all its common divisors, and the least number that will do this is called their least common multiple; therefore, the least common denominator of any fractions is the least common multiple of all their denominators. Though the rule already given will always find a common multiple of the given denominators, yet it will not always find their least common multiple. In the last example, 24 is evidently a common multiple of 4 and 6, for it will exactly measure both of them; but 12 will do the same, and as 12is the least number that will do this, it is the least common multiple of 4 and 6. It will therefore be convenient to have a rule for finding this least common multiple. Let the numbers be 4 and 6.

It is evident, that one number is a multiple of another, when the former contains all the factors of the latter. The

L

factors of 4 are 2 and 2, (2 × 2 = 4.) The factors of 6 are 2 and 3, (2 × 3 = 6.) Consequently, 2 × 2 × 3 = 12 contains the factors of 4, that is, 2 × 2; and also contains the factors of 6, that is. 2 × 3. 12, then, is a common multiple of 4 and 6, and it is the least common multiple, because it does not contain any factor, except those which make up the numbers 4 and 6; nor either of those repeated more than is necessary to produce 4 and 6. Hence it follows, that when any two numbers have a factor common to both, it may be once omitted; thus, 2 is a factor common both to 4 and 6, and is consequently once omitted.

¶ 61. On this principle is founded the RULE for finding the least common multiple of two or more numbers. Write down the numbers in a line, and divide them by any number that will measure two or more of them; and write the quotients and undivided numbers in a line beneath. Divide this line as before, and so on, until there are no two numbers that can be measured by the same divisor; then the continual product of all the divisors and numbers in the last line will be the least common multiple required.

Let us apply the rule to find the least common multiple of 4 and 6.

2)4.6

2

3

4 and 6 may both be measured by 2; the quotients are 2 and 3. There is no number greater than 1, which will measure 2 and 3. Therefore, 2 × 2 × 3 = 12 is the least common multiple of 4 and 6.

If the pupil examine the process, he will see that the divisor 2 is a factor common to 4 and 6, and that dividing 4 by this factor gives for a quotient its other factor, 2. In the same manner, dividing 6 gives its other factor, 3. Therefore the divisor and quotients make up all the factors of the two numbers, which, multiplied together, must give the common multiple.

7. Reduce 2, 1, and to equivalent fractions of the least cominon denominator.