DECIMAL FRACTIONS 



1734 



DECIMAL FRACTIONS 



Multiplicand and multiplier change: 



424X142 =60208 



424X14.2 = 6020.8 

 42.4X14.2 = 602.08 

 42.4X1.42 = 60.208 

 4.24X1.42 = 6.0208 



From this it is seen that the product varies 

 in value as the multiplicand and multiplier 

 together vary, that for each division of either 

 multiplicand and multiplier by 10 the product 

 is divided by 10; put in the common language 

 of arithmetic, "The product contains as many 

 decimal places as the multiplicand and the 

 multiplier together." (See Fig. 6.) 



If (5) nX.08=.0056, n=.07. Why? Because 

 the number of decimal places in n added to 

 the number in .08 makes 4, the number in 

 .0056, the product. 

 Express the same problems in division: 



(2) 6.6-:-7 = n (n stands for the quotient) 



(3) .56-=-7 = n 



(4) .056^.7=n 



(5) .0056^. 07 = n 



One sees how many decimal places there 

 are in n each time by seeing what number of 

 decimal places must be added to the number 

 of decimal places in the divisor to make the 

 number of decimal places in the dividend (or, 



.4F-U 



'.7ft 



.4X.6-.32 



FIG. 6 



7X2 = 1.4 



The formal rule is: Multiply as in whole 

 numbers; point off in the product as many 

 decimal places as there are decimal places in 

 the multiplicand and the multiplier. 



Division of Decimals. This follows directly 

 from multiplication; with two points clear in 

 mind division of decimals presents no new diffi- 

 culty. These points are: 



(1) A dividend is the product of the divisor 

 and quotient. 



(2) A product contains as many decimal places 

 as the multiplicand and multiplier together con- 

 tain. That Is, (1) 7X8 = 56, and (2) 7Xn = 5.6. 



Number (2) above says, "What number mul- 

 tiplied by 7 gives the product 5'.6?" From the 

 work we have done in multiplication of deci- 

 mals we know that the missing number must 

 have one decimal place because the product 

 has one; now the other term, 7, has none, and 

 therefore n=.8. In this statement (3), 7Xn= 

 .56, the product has two decimal places, and 

 the given number, 7, has none ; so n must have 

 two, and n=.08. 



In (4), .7Xn=.056, the product has three 

 decimal places and the given number, .7, has 

 one decimal place; therefore n must have two 

 decimal places and n=.08. 



in other words, he sees the dividend as a prod- 

 uct). This attack upon division of decimals 

 makes it a very simple piece of work. The 

 dividend is a product; therefore, it contains 

 as many decimal places as the two terms, 

 namely, divisor and quotient, that produce it. 

 Expressed as a rule: 



Divide as in whole numbers, and point off as 

 many decimal places in the quotient as added to 

 the number of decimal places in the divisor will 

 give the number of decimal places in the dividend. 



One special case must be considered under 

 this that often gives much trouble; namely, 

 where the dividend does not have so many 

 decimal places as the divisor. Since the divi- 

 dend is the product of divisor and quotient, it 

 must have at least as many places as one of 

 those terms. For example, put into multipli- 

 cation form, 15-.75=n; 15=.75Xn. When 

 .75 and n were used to give a product, that 

 product must have had at least two decimal 

 places; therefore 15 in its present form could 

 not have come from such multiplication; it 

 must have been 15.00 or have had more deci- 

 mal zeros. The division problem now is: 



15.00-^-. 75 = n t>roof: .75 



n = 20 2 



15.00 





