same principles hold; in fact, the measured quantity, $4 × 12, is identical with $12 ×. In this conception of quantity (money value) we have nothing to do with yards, and either form of the measurement may be taken. In fact, $X 12 is of $1+ of $1+ of $1, and so on 12 times that is, of $12. The multiplicand is always a unit of measure; the multiplier always shows how this unit is treated to make up the measured whole. It is purely an operation. In this example the denominator shows how the unit $12 is to be dealt with in order to yield the derived unit of measure it is to be divided into four parts, and the derived unit thus found is to be taken three times. As already shown, from the nature of the fraction it denotes three times one fourth of the multiplicand, or one fourth of three times the multiplicand—that is, $123, or $ 12×3. 4 (3) The explanation usually given of the process is in harmony with this. This explanation considers the multiplicand as a case of pure division; that is, is one fourth of 3, and to multiply a quantity by is to take one fourth of 3 times the quantity. In fact, in all operations with fractions the idea of division, as well as of multiplication, is present; a factor and a divisor are always elements in the problem. (4) The method to be followed when both factors are of fractional form involves nothing different from the other two cases. The price of yard of cloth at $2 a yard is to be found. The result is indicated by $; that is, as before, 4 times a certain quantity is to be divided by 9, or of the quantity is to be multiplied by 4. In the first case, $X 4 is either $1 × 3 × 4 = $1 × 12 = $3; or $4×4 × 3 = $1 × 3 = $3; and of this is $3, or $3. It may be observed that we may change the multiplicand into an equivalent fraction with a unit of measure determined by a multiple of the denominators. In $, for example, we have $33 × 1 = $3% × 4 = $} = $3. The complete process is seen to be × = }}. But since numerators are always factors of a dividend, and denominators factors of a divisor, common factors may be divided out. In $3 X 4, for instance, the value of the quantity is the same whether we take 4 times the number of units (= $12), or make the units 4 times as large ($). This is nothing but the application of fundamental principles (see page 226) of multiplication and division. If we have to divide 210 by 21, we may proceed thus: 210 3 × 7 × 5 = × 7×5 = 5. Again, 32340 385 = X 6 1 X 84 = 3 × 7 4. Division of Fractions.—(1) When the divisor is an integer and the dividend a fraction. Paid $ for 5 yards of calico, what was the price per yard? One yard will cost one fifth of $, or $3. At $5 a yard, how much lace can be bought for $? The answer is indicated in $7÷$5; the quantities must have the same unit of measure, and the expression is equivalent to $7$10750; hence, yard of lace can be bought. 10 (2) When the divisor is a fraction, and the dividend an integer. At $a yard, how many yards of dress goods can be bought for $6 ? The number of yards is given in $6÷$, where, again, the quantities must be reduced to the same unit of measure: $6 ÷ $4 = $24 ÷ 4 = 24 ÷ 3 = 8; hence, 8 yards can be bought. Paid $6 for yard of velvet, what was the price per yard? of The cost is given in $6÷, which means that 3 times the quantity sought is $6, and therefore it is Or, by the law of commutation, $8 X $6×4 ÷ 3 = $8. = $4 × 8 = $6; and $6÷ $4 $8, as before. = (3) When both divisor and dividend are fractions. What quantity of cloth at $ a yard can be bought for $? The quantity is given in $4 ÷ $33, where again the quantities must be expressed in terms of a common unit of measure: there results $18÷$316 ÷ 3 = 51, which is the number of yards. = If 5 yards of calico cost $, what is the price per yard? We have $÷that is, one third of 16 times some quantity = $; 16 times the quantity the quantity is $1 × 3 × 16 = $3. = $ × 3; The Inverted Divisor.-It is obvious in all these cases that practically the divisor has been inverted and then treated as a factor with the dividend to get the quotient. It must be clear, too, that this is simply reducing the quantities to be compared to the same unit of measure. When $12 is to be divided by $-i. e., when their ratio is to be found-they must be expressed in the same unit of measure. The divisor is measured off in fifths of a dollar; the dividend, then, must be expressed in fifths of a dollar-that is, it becomes 5 X 12, or 60. The question is now changed to one of common division: $12÷÷÷60÷415. Similarly, in $1}÷ $1 = 88 = 18, the divisor is expressed in fifths of a dollar; the dividend $1 must be expressed in fifths; how is this done? By multiplying $13 by 5, which gives the number of fifths in $13, namely, $ig; for if $12 is 60 fifths, of $12 must be fifths. The unit of measure is now the same, and we have fg (fifths) ÷ 4 (fifths) 1. "Inverting the divisor," then, makes the problem one of ordinary division by expressing the quantities in the same number measure. = Though formal proofs of rules are in general too abstract to begin with, yet after the pupil has freely used and learned the nature of the processes involved in concrete examples, he will quite readily comprehend the more abstract proof, and even the general demonstration. Take a few instances: 1. To prove that the product of two fractions has for its numerator the product of the numerators of the given fractions, and for its denominator the product of their denominators: (1) Prove & × 5 = 45 X515; but this product is 9 times. too great, and therefore the required product is of 15 = 15. (2) × 8=3; for & X 8 = 3 × ×8; and 95; multiply these equals; .:.8×× 93 X 5; divide by 8 X 9; tors, etc. 8×9 α (3) Generally, let and be any two fractions. b= a; (because, from the nature of number, b=1); similarly, d = c; multiply these equals; .'. xx bd = a × c; divide these equals by bd; axc; that is, the product of the numer bx ators of the given fractions is the numerator of the required fraction, and the product of their denominators its denominator. 2. To prove the rule for division of fractions, "invert the divisor and proceed as in multiplication." (1); divide by 5 and there results; but it is required to divide not by 5 but by 1 of 5; the required quotient must therefore be 9 times 13, which is 3×9 8 × 5 (2) X 8 = 3; multiply these equals by 9; X that is, (1) Similarly, × 9= 5; multiply these equals by 8; ...X9X8 = 5 X 8. Divide (1) by (2): (2) Similarly, a general proof may be given, as in multiplication. |