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By resolving both fractions as above into their prime factors, and striking out [or marking] all those that occur in both terms of the complete fraction, and are not found in the imperfect one, the required denominator appears to be 2·2·2·5=80. By following a similar process for No. 8, the required numerator would also be found to be 80.

But processes like these may be rendered much more simple by cancelling while transferring the compound fraction to the slate, reserving, of course, such factors as are found in the imperfect fraction. Thus,

No. 7.

16 4.1.6.4
1.10.8.6

1.10.8.6

No. 8.

16—4• 1 •6·4

and prove.

giving 80 as before for the required deficient term, when the two superfluous 6s are struck out. 9. Supply the denominator] 18 14.27

in

10. Supply the numerator in

16.6

and prove.

16.6

18 14.27'

Prime factors

2.3.3 2.7 X3.3.
2·2·2·2×2.$

2.2.2.2× 2.3

Prime factors

2.3.3 2.7 X3·3.3

Here, cancelling the marked 3s, and dividing by the super

fluous 7, gives 44 as the required deficient term.

11. Supply the denominator]

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2•2•2•2× Prime factors for No. 11, 2-2-2-3_2-2.2-2× 5 ×_2•7

2•3•3 ×5•7 ×2•2•2•2

Here, as there is a 3 in the given term of the imperfect fraction, and none in the corresponding term of the perfect one, it must be supplied in both terms of the latter; that is, the fraction must be multiplied by 3, giving, after proper cancellation, as marked above, 216 for the required deficient terms. Or the operation may be more simply performed by cancelling whilst transferring to the slate, as under:

24 8.1.7
9.7.8

By striking out the two 7s, and inserting 3, as before, we have again 216 as the deficient term.

No. 12 does not differ from the above, except in the reversal of the terms.

13. Supply the denominator | 14. Supply the numerator in

15

in

and prove.

25.34.28 15-16.14.17'

and prove.

16.14.17 25·34•28' Here, as the corresponding terms of the equivalent fractions have no common factor, no advantage would result from resolving them into their prime factors. But, as 17 and 34, and also 14 and 28, admit of cancellation, and as the two 2s thence arising may be cancelled with 16, the complete fraction thus becomes, giving 933 for the deficient term.

15. Supply the denominator

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16. Supply the numerator in

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Here, as no factor is common to the equivalent fractions, and the perfect fraction admits of no cancellation, the problem can only be solved as in Example 1 of this Case, namely, by placing 7 in both terms of the perfect fraction; that is, by multiplying by 7; and then dividing by the factors 4.2.13, or their product, 104.

Suggestive Questions.-How can both terms of the perfect fraction be multiplied by 7? How can the other factors of the numerator on the left, and of the denominator on the right (4.2.13), be removed, thus leaving 7 as sole numerator [or denominator]? Will this solve the question?

17. Supply denominators in 18. Supply the numerators each of the following pairs of in each of the following pairs 56 16.7 equivalent fractions: of equivalent fractions: 48.3

(solely by division of perfect 48.3

56

[blocks in formation]

(solely by division of per

15

[blocks in formation]

and multiplication);

[blocks in formation]

15

90 11

(division and multiplication);

(change mixed num- (division and multiplication);

ber to improper fraction, and 90

to eighths.) Prove as be- and mixed number to improper fraction.) Prove as before.

fore.

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[blocks in formation]

Multiply by 5, and divide by 3.4.6, or their product. Why?

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24. Change 24 of 1 of 2 of to an equivalent fraction, with 16 as numerator, by inspection, and prove.

25. Change 21 of to an 26. Change of 14 to an equivalent fraction, with 56 as equivalent fraction, with 56 as denominator, by inspection, and numerator, by inspection, and prove. prove.

27. Change of 10304 to 28. Change 16 of 1104 to an equivalent fraction, with 4 an equivalent fraction, with 4 as denominator, ascertaining as numerator, ascertaining the the sole divisor by inspection, sole divisor by inspection, and and prove.

prove.

29. Change 100 of 12 to 30. Change % of to an an equivalent fraction, with 4 equivalent fraction, with 4 as as denominator, ascertaining the numerator, ascertaining the sole sole divisor by inspection, and divisor by inspection, and prove. 32. Change into an equiva

prove.

31. Change into an equivalent fraction, with as nulent fraction, with as de- merator, and prove by again nominator, and prove by again resolving it into a simple fracresolving it into a simple frac- tion of lowest denomination. tion of lowest denomination.

3 33. Change and into 告

34. Change

$ and

3

5

into

equivalent simple fractions, and

equivalent simple fractions, and prove by again resolving the

prove by again resolving the first into a fraction whose nufirst into a fraction whose de-merator shall be , and the nominator shall be , and the second into one whose numesecond into one whose denomi- rator shall be 5.

nator shall be 5.

2

35. Change and

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10

into

9 36. Change and into

equivalent simple fractions, and equivalent simple fractions, and prove by again resolving the prove by again resolving the first into a fraction whose nufirst into the denomination of merator shall be g, and the §, and the second into the de- second into one whose numerator shall be 9.

nomination of 9ths.

A remarkable property of numbers is developed by the above examples, namely, that any number whatever may be expressed by a common fraction, whose numerator (or whose denominator) shall consist of any specified number, whether whole or fractional.

II.-Addition and Subtraction of Common Fractions.

Suggestive Questions.-Can numbers of different denominations be added together or subtracted? See Oral Arithmetic, Chap. III., Sect. IV., p. 91. What previous operation is necessary?

Exercises for the Slate or Black-board.

1. Find the sum and the difference of 3 and.

Ans. Sum; Diff. . 2. Find the sum of the five following fractions, and prove the operation by subtracting from it the sum of the last four: 4, 3, 2, 3, 7%. What fraction will be left?

6

3. Add 4, 15, 14, 20, 25, and prove by subtracting the sum of the first four.

4. Add 73, 35, 9, 54, and prove by subtracting the sum of the last three. [Change the fractional parts of these numbers to the same denomination, add them, carrying what integers they may contain to the given integers.]

5. What is the sum and difference of

33

21

and
62 74

39

Ans. Sum, 38; Diff., 1258.

6. Add ,,, and, and prove the operation by giving

these 4 fractions a decimal form, and changing their sum into a common fraction, which, of course, will be equivalent to the sum of the four common fractions.

7. Add 3, 8, and 7, and prove by the same process as in the last example.

III.-Multiplication and Division of Common Fractions. [See Oral Arithmetic, Chap. III., Sect. V., p. 93.]

Exercises for the Slate or Black-board.

35

1. Multiply by 4, by 7, by 3, and by 7, all by division; or, which is the same thing, by cancelling what would otherwise be equal factors in both terms of the product. Prove the operations by performing them by multiplication, and bringing each fraction to its lowest denomination.

2. Divide by, or, which is the same thing, change

into a simple fraction, or integer.

3. Divide by ; by ; by 7.

Ans. 2.

Remark, that in the above example, and in all other cases of division where the numerators (or where the denominators) are alike in the divisor and dividend, the quotient is formed by removing the like term in the dividend, and putting the unlike term of the divisor in its place. Thus, į÷÷=}; 3. Why is this so? Perform the above example in the usual method, and see.

33

4. Divide by 17; 17 by 29; 29 by 23; 95 by 4; each at a glance.

5.32 4

5'32 8.15

5. Divide by 15, by cancelling what would otherwise be equal factors in the quotient. Solution. ÷35=1= 8153; or, omitting the superfluous steps, 3.5= 6. Divide by 2 by inspection, first cancelling equal factors in the numerators, and equal factors in the denominators. Why? Ans..

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7. Divide by ; by 18; & by ; by; by cancelling as in the preceding example. Prove by reproducing the dividend as in division of integers; that is, by considering the divisor and quotient as factors of the dividend.

8. Change of of of of of, by cancellation, into a simple fraction, by inspection, and prove by dividing the

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