from the product. If the excess of 9's in their product be equal to the excess of 9's in the total product, the work is right. 3580000 7 excess in the product. This method of proof is liable to the same objections here as in addition. Multiplication may also be proved by dividing the product by the multiplier. If the work be right, the quotient will be equal to the multiplicand. USE OF MULTIPLICATION. The learner can hardly need to be told, that operations in adding together the several sums are rendered very easy by this rule; while by addition it would be very long and tireIt facilitates the transaction of business in a very some. high degree. SECTION VI. SIMPLE DIVISION. Simple Division teaches how to divide a given number into a certain number of equal parts. The number to be divided is called the Dividend. The number by which we divide, or which shows into how many parts the number is to be divided, is called the Divisor. The number which shows how many each of the parts contains, is called the Quotient. Lacroix remarks, that "the manner of decomposing one number by another, in order to know how many times the last is contained in the first, is called Division, because it serves to divide, or portion out, a given number into equal parts, of which the number or value is given. "For instance, if we wish to divide the number 64 into 4 equal parts, our object is to ascertain how many times 4 is contained in 64; that is, what number, multiplied by 4, will give a product of 64." There are several ways in which a question of this kind may be solved. 1. We may solve it by addition; that is, we may add the number 4 to itself continually till the number 64 is produced: thus, 4+4+4+4+4+4+4+4+4+4+4+4+4 +4+4+4=64; then, by counting the number of times which 4 is written, we ascertain how many times this number is contained in 64, which we find to be 16 times. The same result may be obtained by subtracting 4 from 64 continually till nothing remains; thus, 64-4-50-4-56 —4—52—4—48—4—44—4—40—4—36-4-32-4-284-24-4-20-4-16-4-12—4—8—4—4—4—0. Here again, if we count the number of times which 4 has been taken from 64, we shall find it 16, as before. But it will be seen at once, that it would be very inconvenient in practice, to make use of repeated additions or subtractions for finding how many times one number is contained in another, especially if one of these numbers is much larger than the other. It then becomes necessary to have recourse to an abbreviation of this method. From the example already given, it may be seen, that the quotient, or number of times which 4 is contained in 64, is a number which, if multiplied by the divisor, 4, will make just 64; thus 16+4=64. If the dividend do not exceed 144, and the divisor do not exceed 12, the quotient may be found at once from the table given in Multiplication; since that table contains the product of all factors which do not exceed 12. If it were asked, for instance, how many times 8 is contained in 56, it would be necessary to go down the eighth column to the line in which 56 is found; the figure 7 at the beginning of this line, shows the second factor of the number 56, or how many times 8 is contained in this number. "We shall see from this table, that there are numbers which cannot be exactly divided by others. For instance, as the seventh line, which contains all the multiples of 7 below 84, has not 40 in it, it follows, that 40 is not divisible by 7." (Lacroix.) *The product of any number, multiplied by 2, 3, 4, 5, &c., is called a multiple of that number. A multiple is that which can be measured, or divided by another number without a remainder. When the number to be divided, or the dividend, is larger than 144, we must seek the quotient by some other means than by the table. Let it be required to divide 4375 dollars equally among 25 men; how many dollars will each receive? It will be borne in mind, that our object is to ascertain how many times 25 are contained in 4375; or in other words, to find a number, which, multiplied by 25, shall produce 4375; or, to divide 4375 into 25 equal parts. That we may better understand the operation, we will write the thousands, hundreds, &c. of the dividend, independently of each other. 25) 4000(100 25 1500 25) 1800(70 1750 50 70 5 It is evident, in this case, that the first figure of the quotient cannot be thousands, because 25 is not contained in 4, but if we call the 4 thousands 40 hundreds, we perceive, that the divisor, 25, is contained in this number, which is only a part of the dividend, less than 200 times, and more than 100 times; we accordingly take 100 times 25 or 2500 from this part of the dividend, and the remainder is 1500, with which we join the 300, making 1800. As 1800 contains 25 less than 100 times, it is plain, that the next quotient figure must be tens; and in 1800, 25 is contained more than 70 and less than 80 times; hence we take 70 times 25, or 1759, from this part of the dividend, and 50 remains. With this remainder we join the 7 (tens) and 5 (units,) which make up the last portion of the dividend, 125 (units,) in which 25 is contained 5 times, making the quotient 175; for 100+70+5=175. 25)125(5 125 000 How many dictionaries can be purchased for 675 dollars, at 5 dollars each? If the dictionaries were only one dollar apiece, we could buy 675; if they were 2 dollars, one half as many; if 3 dollars, one third; if 4 dollars, one quarter; but as they are 5 dollars, we can buy only one fifth as many; that is, as many as the number of times which 5 is contained in 675. To make the manner of ascertaining how many times 5 is contained in 675 more plain, we may separate the number into four parts, 5 is contained in 5 once; and in 500 one hundred times; in 10 twice, in 10 times 10, i. e. 100, 20 times; in 50, 10 times; in 25, 5 times; and 100÷20+10+5=135 the answer. Thus by dividing the parts of the whole separately, and adding the several quotients together, we obtain the number of times which one number is contained in another. If it be required to divide 1656 by 3, this question may be changed into another form, viz: To find such a number, that multiplying its units, tens, hundreds, &c. by 3, the product of these units, tens, hundreds, &c. may be the dividend, 1656. This number will not have units of a higher order than hundreds; for if it had but one thousand, the product would contain at least 3, which is not the case. It appears, then, that the thousand in the dividend is a number reserved, when the hundreds of the quotient were multiplied by 3, the divisor. The figure occupying the place of hundreds in the required quotient, ought to be such, that when multiplied by 3, its product may be 16, or the greatest multiple of 3 less than 16. This restriction is necessary, on account of the reserved numbers which the other figures of the quotient may furnish when multiplied by the divisor, and which should be united to the product of the hundreds. The number which fulfils this condition is 5; but 5 hundreds multiplied by 3 gives 15 hundreds, and the dividend, 1656, contains 16 hundreds; the difference, 1 hundred, must have come then from the reserved number, arising from the multiplication of the other figures of the quotient by the divisor. If we now subtract the partial product, 15 hundreds, or 1500, from the total product 1656, the remainder 156 will contain the product of the units and tens of the quotient by the divisor; and the question will be reduced to finding a number, which, multiplied by 3, gives 156, a similar question to that which presented itself above; and the process is repeated again and again, till the dividend is exhausted. "The operation just described may be disposed of in this way: 15 15 15 06 The divisor is placed on the left hand, separated by a line, and a similar line is placed at the right of the dividend to separate the quotient from it. We then take the 16 (hundreds) on the left hand of the dividend, and dividing this number by 3, we get 5 (hundreds,) which is placed for the first figure in the quotient; then taking the product of the divisor by this figure, and subtracting it from 16 (hundreds) the partial dividend, we write the remainder 1 (hundred) below, by the side of which we place the 5 (tens) of the dividend. We divide this number by the divisor, and obtain 5 (tens) for the second figure of the Quotient; we then take the product of this figure by the divisor, and subtracting it from the partial dividend, get 0 for the remainder. We then bring down the last figure of the dividend, 6 (units;) divide this by 3, and get 2 (units) for the last figure of the quotient. "It is manifest, that if we find a partial dividend that cannot contain the divisor, it must be because the quotient has no units of the order of that dividend, and that those which it contains, arise from the products of the divisor by the units of the lower orders in the quotient. It is necessary, therefore, whenever this is the case, to put a cipher in the quotient to occupy the place of the order of units that is wanting. Let 1632 be divided by 4. 4)1632 (408 16 032 32 By this example we see, that the divisor is contained just 4 (hundred) times in the 16 (hundreds) of the div idend, without leaving any remainder. The 3 (tens) in the dividend will not contain 4; there can, therefore, be no tens in the quotient; hence we put a cipher in the place of tens in order to represent the true value of the other quotient figures." 00 If the divisor consists of several figures, it is not always easy at first to tell how many times the partial dividend will contain it. The following example and illustration will aid the learner to obviate this difficulty. |