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the fraction to its equal #, which is now in its lowest terms, for no greater number than 1 will again measure them. Note 1. — Any fraction may evidently be reduced to its lowest terms by a single division, if we use the greatest common divisor of the two terms. Thus, we may divide by 32, which we found ("I 62) to be the greatest common divisor of 128 and 160. 32) #3–# Ans.

Hence, To reduce a fraction to its lowest terms,

Divide both terms of the fraction by any number which will divide them both without a remainder,and the quotients thence arising in the same manner, and so on, till it appears that no number greater than 1 will again divide them.

Note 2. — A number ending with a cipher is divisible by 10. If the two right hand figures are divisible by 4, the whole number is also.

A number is divisible by 2 when it ends with an even number, and by 5 when it ends with 5, or 0.

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Note 3. — Let the following examples be wrought by both methods; by several divisors, and also by finding the greatest common divisor.

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5. Reduce for to its lowest terms. $. 6. Reduce # to its lowest terms. Ans. #. 7. Reduce for to its lowest terms. Ans. }}}. 8. Reduce #### to its lowest terms. Ans. }.

Note. —The reducing of a compound fraction to a simple one will be considered in the multiplication of fractions, where it properly belongs. The reducing of fractions to a common denominator will be presented in connection with addition and subtraction of fractions, in which operations only it is necessary. The reducing of complex to simple fractions will be considered after the pupil shall be made aequainted with the division of fractions, a knowledge of which is indispensable to understanding the operation.

Questions. – T 67. Give the illustration with the half apple. Reverse the operation. What follows 2 What is the process mentioned, and what is it called 2 When is a fraction in its lowest terms? Explain Ex. 1. How can a fraction be reduced by a single division? Rule. Give the note by which you determine by what number you divide.

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Solution.— The adding together of É, $ and # of an orange is the same as the adding of 2 oranges, 4 oranges, and 1 orange, which would make 7 oranges. The 8 is called the common denominator, as it is common to the several fractions; and we write over it the sum of the numerators, to express the answer. Ans. #.

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Solution.—so of a dollar is one dime, or ten cent piece. . The operation, then, is to subtract 3 ten cent pieces from 7 ten cent pieces,

which will leave 4 ten cent pieces, or, Ans. Ho.

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T 69. 1. 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?

Solution. — The fractions 3 and of an orange can no more be added than 3 oranges and 1 apple, which would make neither 4 oranges nor 4 apples, as they are of different kinds, (‘s 12.) But if 1 orange made 2 apples, the 3 oranges would make 6 apples, and the 1 apple being added we should have 7 apples. Now 3 does make just 3, and consequently # make $, to which if # be added we shall have the Ans. #.

The denominator, 4, of the fraction #, is a factor of 8, the denominator of the fraction #. And if each term of the fraction # be multiplied by 2, the remaining factor of 8, it will be reduced to 8" (š,) without altering its value. (T67.) Hence, if the denominator of one fraction be a factor of the denominator of another fraction, and both its terms be multi

Qnestions. – s 68. Like what, is the process of adding eighths? What is the 8 called, and why What is the tenth of a dollar?

plied by the remaining factor, it will be reduced to the same denominator with the latterfraction, without altering its value. (T 5S.) For example:

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Solution. — The factors of 12 are 3 and 4, the latter of which is the denominator of #, and multiplying both terms of 3 by 3, the other factor, we have so, a fraction of the same value as #, but having a different denominator. Ans. To.

3. A man has or of a barrel of sugar in one cask, $ in another, and # in another; how much in all ?

Solution. — The denominator 6, of the second fraction, is a factor of 12, the denominator of the first; and if both terms of # be multiplied by the other factor, 2, it will become #. Also 4, the denominator of the third fraction, is a factor of 12, and if both its terms be multiplied by 3, the other factor, it will be #. And so-Ho-H or =##=1}} barrels. Ans.

4. What is the amount of 4, §, and #2

Solution. — As the denominators are not factors of each other, we must take some number of which each is a factor. 36 is such a number. The first denominator, 4, being a factor of 36, both terms of # may be multiplied by 9, the other factor, and we shall have jo. In like manner, both terms of É being multiplied by 6, we have . . ; and both terms of § being multiplied by 4, we have #3; then, is + #á-H #=#=1#s. Ans.

The process in the above examples is called reducing fractions to a common denominator, and is necessary when we wish to add or subtract those of different denominators. The common denominator, it will be perceived, must contain, as a factor, each of the other denominators.

It is not always manifest what number will contain all the denominators. There are two methods of finding such a

number. ~.

FIRST METHOD. T70. If several numbers are multiplied together, each

will evidently be a factor of the product. "We have, then, the following

Questions. – T 69. How can eighths and fourths be added ? When, and how, can one fraction be reduced to the denominator of another Explain the third example ; the fourth. What is the process called ! When is this necessary 2 What must the common denominator contain? What is not manifest ? How many methods of finding it?


Multiply the numerator and denominator of each fraction by the product of the other denominators.

The several new denominators will be products of the same numbers, and, therefore, will be alike; and the numerator and denominator of each fraction being multiplied by the same number, its value is not altered. See s 58.

Note. — The common denominator of two or more fractions is the common multiple of all their denominators; see "I 55.

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TV1. While we can always find a common denominator by the above rule, it will not always give us the least common denominator. In the last example, 12 as well as 24 is a common denominator of 3 and 3. Let us see how the 12 is obtained.

One number will contain another having several factors, when it contains all these factors. For example, let 18 be resolved into the factors 2 × 3 × 3, which, multiplied together, will produce it. It contains 6, the factors of which, 2 and 3, are the first and second factors of 18. It also contains

Questions. – T 70. What is the rule in the first method Whence its propriety? What is a common multiple f Explain the first example. 9, the factors of which, 3 and 3, are the second and third factors of 18. But it will not contain 8=2 × 2 × 2, for 2 is only once a factor in 18. Now 12, the factors of which are 2 × 2 × 3, will contain 4 = 2 × 2, since these factors are the first and second factors of 12. It will in like manner contain 6. And it is the least number that will contain 6 and 4, for 2 must be twice a factor, or it will not contain 4, and 3 must be a factor, or it will not contain 6. Hence, no one of these factors can be spared. But 24 = 2 × 2 × 2 × 3, has, it is seen, 2 three times as a factor; so one 2 can be omitted, and we have the factors of 12 as before. We have 2 as a factor once more than necessary, because it is a factor in both 4 and 6. Hence, when several of the denominators have the same factor we need retain it but once in the common denominator.

T72. The process of omitting the needless factors is called getting the least common denominator of several fractions, and is as follows:

2 || 4 .. 6 4 and 6 are each divided by 2; and the divisor and remainders being taken 2 .. 3 for the factors of the common denomina

2 × 2 × 3 = 12. tor, we have rejected 2 once.

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2 || 2 . 4. 6 . S. 10 Solution. — We write the denomi- nators in a line, and divide as here seen. 2 || 1 . 2. 3. 4. 5 By the first division, 2 existing as a

factor in each of the five numbers, is

1 . 1 .. 3. 2. 5 rejected four times, being retained

- - - - - - - once ; as the divisor is substituted for

2 × 2 × 3 × 2 × 5=120 the five factors 2, which we should

have had by multiplying all the num

bers together. But 2 being a factor in two of the remainders, it is: rejected once more by a second division.

Ans. 120.

2. Find the least common denominator of #, or, and 33.

Questions.—T 71. Why the necessity of a second method? When will one number contain another? What numbers will 18 contain, and why? What will it not contain why? Why will 12 contain the denominators of both # and #2 Why is it the least number that will contain them Why is a factor in 24 once more than necessary What may then be done?

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