number has ciphers at its right, and the significant figures do not exceed those of the smallest number, the ciphers may be omitted and this number taken for a multiplier, as with the following numbers, 4800 and 724. 724 4800 5792 2896 3475 200 Again, that you may see the nature of the work as plainly as possible, I will give another example. 3. In this example you see that the multiplicand is decompounded into 20,000, 4,000, 600, 80 and 9, each of whose products is placed separately, which shows plainly the reason for placing the products as directed. When there are ciphers in the middle of the multiplier, they may be omitted, carrying the product of the next figure two, three, &c. places to the left according to the number of ciphers, thus: 1246 204 49 84 2 5 4 1 8 4 When either of the factors is a unit with ciphers at its right, place the ciphers at the right of the other number, and it will be the product of the two numbers. Multiply 2486 by 1000. Ans. 2486000. Every removal towards the left increases the value of the figures ten times, and here 2486 is to be repeated 1000 times, which is done by removing it three places to the left. Pup. I think I now understand Multiplication sufficiently to answer any questions which may be put in this rule, and should like to apply the rule by performing a few examples. Tut. You may multiply together the following numbers. Multiply 5943267 by 42. Ans. 249617214. Ans. 2222052120. Having now learned how to multiply numbers together, so that one number may be repeated a given number of times, and the amount of these repetitions expressed in one sum, you will want to know how to bring this number into its former state, viz. to divide it into a given number of parts. And as multiplication can be performed by addition, so division can be performed by subtraction, for multiplication and division are the reverse of each other, the same as addition and subtraction are. To reverse the example given in multiplication (1. Con. II.) we should have to find how many times 12 is contained in 48; this may be done by subtraction in the following manner : 4 8 1 2 once, 36 1 2 twice, 24 1 2 three times, 1 2 12 four times. Here by subtracting 12 from 48 you decrease 48 by 12 and then, again subtract 12 from the remainder, and con tinue to subtract till you can subtract no longer. Then the number of subtractions you made will be the number of times that 48 contains 12. Thus in dividing 48 by 12 you subtract 12 four times, and 4 times 12 are 48. But if it were required to divide 288 by 12, this process would be very long and tedious. To remedy this inconvenience we have the following rule for performing SIMPLE DIVISION.* Write the dividend, and on the right and left of it draw a curved line. Place the divisor at the left of the dividend. Seek how often the divisor is contained in just so many of the left hand figures of the dividend as are necessary, and set the result on the right of the dividend, for the first figure of the quotient. Multiply the divisor by the quotient figure, and place the product under those figures of the dividend which were used for obtaining the quotient figure. Subtract this product from the dividend, and to the remainder bring down the next figure of the dividend towards the right, which makes a new dividend, which must be proceeded with as was the first; and proceed in this manner till you have brought down all the figures of the dividend. If it were required to divide 248 by 16, the work would stand thus: dividend. divisor, 16) 2 4 8 1 5 quotient. 16 8 8 8 0 8 remainder. Pup. The work here is plain, but I do not understand the reason of multiplying the divisor by the quotient figure, and subtracting as I do. I should like to have you explain every part to me that I may understand the whole. *Before learning the rule for simple division you must be informed, that the number given to be divided is called the dividend; the number by which it is divided is called the divisor; and the result or answer to the question is called the quotient ; whatever remains after division is called the remainder. Tut. I am glad that you wish to understand the reasons of the rule, and do not wish to get along merely by doing the sums without knowing any thing else about them, as scholars generally do. I will now give an example with an explanation of the whole operation. Suppose it were required to divide 64268 by 28; the numbers for the operation would stand thus: 2 4 ) 6 4 2 6 8 ( 4. Here you are in the first place to find how often 24 is contained in 64; 64 here is not merely 64, but 64,000; therefore we will suppose the dividend to be 64,000, and divide it by 24. 2 4) 6 4 000 (2 4 8 16 Here we find that 64 contains 24 twice, and 16 over; but as 64 signifies 64,000, so 16 must signify 16,000; and the 2 in the quotient does not signify 2 merely, but 2,000, and the product of 2,000 by 24 is 48,000, which 48 in the work signifies; and 48,000 subtracted from 64,000 leaves 16,000, which is expressed by the 16 in the work. We have now found that 64,000 contains 24 two thousand times, and 16,000 remainder; now had the given dividend been only 64,000, the next step would be to find how often 16,000 contains 24. But the given dividend is 64,268; therefore we have 200 to add to 16,000, because 2 in the dividend is in the place of hundreds, and is 200. We will consider the dividend to be 64,200, and the work will stand thus: 2 4) 6 4 2 0 0 (26 4 8 162 144 1 8 Here we seek how often 24 are contained in 162, and find it is 6 times; and as 162, the new dividend, is really 16,200, 6 must be 600, which multiplied by 24 gives 14,400, and this subtracted from 16,200 leaves 1800, which the remainder 18 expresses. Now had the dividend been only 64,200, the next step would be to find how often 1800 contains 24; but the dividend being 64,268 we have 60 to add to 800, for 6 in the dividend expresses 60. We will now consider the dividend to be 64,260, and the work will stand thus : 2 4) 6 4 2 6 0 ( 2 6 7 162 1 4 4 186 168 1 8 We again seek how often the divisor is contained in the dividend, which is now reduced to 1860, which 186 expresses, and find that 24 is contained in 186 seven times; but 186 is really 1860, so that 7 must be 70; 70 times 24 are 1680, which 168 expresses, and 1680 subtracted from 1860 leaves 180, which 18 expresses. Had the given dividend been but 64260, we should now have only tofind how often 180 contains 24; but the dividend is 64268, therefore we have 8 to add to 180. We will now consider the dividend as it was given, viz. 64268, and the work is as follows: 2 4) 6 4 2 6 8 ( 2 6 7 7 4 8 162 1 4 4 186 1 6 8 188 1 6 8 20 Pup. I think I now understand every part of this work completely, and should like some examples that I may render the rule familiar. |