No. 12 does not differ from the above, except in the reversal of the terms. 13. Supply the denominator 15 in and prove. 14. Supply the numerator in 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 934 for the deficient term. 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 inĮ 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 (solely by division of per 15 15 fect fraction); (division 20 12 18 27 (division and multiplication); 90 11 (change mixed num- (division and multiplication); ber to improper fraction, and 90 1 3 2 to eighths.) Prove as be- and mixed number to improper fraction.) Prove as before. fore. Multiply by 5, and divide by 3•4•6, or their product. Why? prove. 16 before. 24. Change 2 of 1% of of to an equivalent fraction, with 16 as numerator, by inspection, and prove. 26. Change of 1 to an equivalent fraction, with 56 as numerator, by inspection, and prove. 10304 prove. 27. Change of 10304 to 3 28. Change of 1 to 16 1104 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. 29. Change 100 of 12 to 30. Change 75% 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. prove. 32. Change into an equiva 5 12 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. 5 34. 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 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. 2 and into 9 36. Change and into nator shall be 5. 35. Change 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 numeIrator 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 g and 20 Ans. Sum ; Diff. z. 2. Find the sum of the five following fractions, and prove the operation by subtracting from it the sum of the last four: 6 4, 3, 1, 3, 7%. What fraction will be left? 导 3 5 6 3. Add 4, 15, 14, 20, 25, and prove by subtracting the sum of the first four. 4. Add 73, 35, 9, 5%, 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.] 33 21 5. What is the sum and difference of and 9 62 3 59 Ans. Sum, 81; 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. 3 7. Add §, %, 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. ΤΣ 8 35 49 1. Multiply by 4, 5 by 7,2 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 ; 7 by 7. Ans. 2. 6 12 77 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, д÷÷= 2. Why is this so? Perform the above example in the usual method, and see. 4. Divide by 17; 17 by 28; 27 by 28; 48 by ; each at a glance. 17`by 23 8 5. Divide by 1, by cancelling what would otherwise be equal factors in the quotient. Solution. §÷15§×??= 5.32 ; or, omitting the superfluous steps, = 香 5.32 4 8.15 3 36 8.15 6. Divide 1 by by inspection, first cancelling equal factors in the numerators, and equal factors in the denominators. Why? Ans. 1. 7. Divide by 7; by 15 ; & by ; 12 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. 6 57 T25 25 8. Change of § of 7 of % of of, by cancellation, into a simple fraction, by inspection, and prove by dividing the result by each of these several factors except ğ, employing cancellation in the division wherever practicable. 9. Multiply 6 by 43; 343 by 6; 27 by 8; 88% by It is usual to change mixed numbers into improper fractions before multiplying them. But, in some of these, as well as in many other cases, it will be found quite as easy, and much shorter, to multiply without any such change. For example, 433 27=4×61 3143×62 But in division, where the divisor is a mixed number, and the dividend is either a mixed number or an integer, it will be found most convenient to change both to the form of improper fractions. 10. Divide 3 by 4§; also 16 by 52; and prove by multiplication. RAPID AND CONCISE METHODS OF COMPUTING WITH COMMON FRACTIONS. In all treatises on arithmetic, the pupil is directed to bring fractions that are to be added or subtracted to a common denominator. But, when the fractions do not exceed two in number, these operations can frequently be performed much more rapidly and quite as correctly by bringing them to a common numerator, as will appear from the following exemplifications and exercises: I.—Addition by a Common Numerator. 1. Find the sum of 4 and 1. (†+3)=(35+33%) = 13. Suggestive Questions.-1. Compare the new numerator with the given denominators, and say what relation it bears to them. |