of times the dividend contains the divisor, is called the Quotient. If there is anything left after the operation is performed, it is called the Remainder. Sometimes there is a remainder, and sometimes none: when there is any, it is less than the divisor, and of the same denomination as the dividend. SIMPLE DIVISION, Is the dividing of one number by another, without regard to their values; as 20, divided by 5, produces 4 in the quotient; that is, 5 is contained 4 times in 20. Note. It will be proper for the scholar, before proceeding to perform any operations in division, to learn the multiplication table in an inverted order, as follows: Say 2 in 4, twice; 2 in 6, three times; 2 in 8, four times, &c.: then, taking the next column, say 3 in 6, twice; 3 in 9, three times; 3 in 12, four times, &c.; and in like manner proceed through the whole table. 1. Place the divisor at the left hand of the dividend, with à curve line between them, and draw a line under the dividend. 2. See how many of the left hand figures of the dividend must be taken to make a number that will contain the divisor; that is, a number as large at least as the divisor; and the number expressed by those figures, call the dividual.* 3. Find how many times the dividual contains the divisor, and also what remains over: The number of times the divisor is contained in the dividual will be the first figure of the quotient; which figure set down below the right hand figure of the dividual; and the remainder, (if any,) suppose to be prefixed to the next figure in the dividend, for a new dividual; but if there is no remainder, then take the said next figure of the dividend for a new dividual. 4. Find another quotient figure, as before, and so proceed, until the work is done. *The dividual is a partial dividend, or so much of the dividend as is taken to be divided at a time, and which produces one quotient figure. Note 1.-When any dividual after the first, is less than the divisor, place a cipher in the quotient, and suppose the dividual to be joined to the next figure of the dividend, for a new dividual; after which proceed as before. Note 2.-If, after the division is performed, there is a remainder left, it may be placed a little to the right hand of the quotient, as in the 1st, 2d, and 3d examples; or, if you wish to make the quotient complete, then draw a short horizontal line from the quotient towards the right, and place the remainder above, and the given divisor below this line; which will form a fraction that will express the true value of the remainder. See examples 7th, 8th, &c. To prove Division. Multiply the quotient by the divisor; to the product add the remainder, if any; and the result will be equal to the dividend, if the work be right.* In performing the work of this example, I proceed thus: I first set down the given numbers as directed in the Rule ; then I examine the dividend, and find that the two left hand figures are the fewest that will contain the divisor; and therefore I take these two figures, which make the number 37, for the first dividual. This number contains the divisor 9 times, and 1 remains over; for 9 times 4 is * Multiplication and Division being the reverse of each other, the reason of this method of proof is evident. There are also several other meth. ods sometimes used for proving division; two of which may be briefly explained as follows: If the remainder be subtracted from the dividend, the difference will be the product of the divisor and quotient, if the work be right. Therfore, if this difference be found, the work may then be proved by either of the two last methods laid down for proving Multiplication. 36, and 1 makes 37. The 9 I place underneath, for the first quotient figure; and the 1 remainder, I suppose to be prefixed to the next figure of the dividend, viz. 4; and then I have 14, for a new dividual. This dividual contains the divisor 3 times, and 2 remains over; for 3 times 4 is 12, and 2 makes 14. The 3 I place underneath, and the 2 which remains, I suppose to be prefixed to the next figure of the dividend, which makes 20, for a new dividual. This dividual contains the divisor just 5 times; for 5 times 4 is 20. The 5 I set down; and as there is no remainder, I take the next figure of the dividend, viz. 2, for a new dividual; but as this is less than the divisor, I set down a cipher below, and then suppose the 2 in the dividend to be joined to the next figure, which makes 27, for a dividual. This dividual contains the divisor 6 times, and 3 remains over. The 6 I set down, and as there are no more figures in the dividend, I set down the remainder a little to the right hand of the quotient, which finishes the work. Then, to prove the work, I multiply the quotient by the divisor, and add the remainder to the product; and as the result is equal to the dividend, I conclude the work is right. I perform the work of example 2d thus: I say 5 in 47, 9 times, and 2 over-I set down 9: then, 5 in 20, 4 times I set down 4: then, 5 in 1, no times, but 1 remains -I set down 0: then, 5 in 15, 3 times-I set down 3: then, 5 in 2, 0 times, but 2 remains-I set down 0, and place the 2 remainder near the quotient, at the right hand, which finishes the work. 11. If 758967 be divided by 11, what will the quotient Ans. 68997. be? 12. What is the quotient of 4765862 divided by 11? 13. What is the quotient of 1072687 divided by 12? Ans. 89390,7 14. What is the quotient of 9876543210 divided by 12? Ans. 8230452671% When the divisor exceeds 12. RULE.* 1. After having set down the dividend, mark off a curve line on each side of it, and place the divisor on the left hand. *This is a general rule for the division of whole numbers. The method by the rule for Case first, is a contraction of the process by the gen eral rule; the subtrahends in the former case being made out and subtracted mentally. In performing division by either of these rules, we resolve the dividend into parts, and find the number of times the divisor is contained in each of those parts. Each dividual is just so much of the dividend, as is necessary to be divided in order to obtain one quotient figure; and the only thing which remains to be proved, is, that the several figures of the quotient, taken as one number, according to the order in which they are placed, are the true quotient of the dividend by the divisor; which may be thus demonstrated. Dem.-The local or complete value of the first part of the dividend, or first dividual, is, by the nature of notation, 10, or 100, or 1000, &c. times the simple value of what it is taken in the operation; according as there are 1, 2, or 3, &c. more figures in the dividend; and consequently the true value of the quotient figure belonging to that part of the dividend, is also 10, 100, or 1000, &c. times its simple value; but the local or complete value of the quotient figure, belonging to that part of the dividend, found by the rule, is 10, 100, or 1000, &c. times its simple value; for there are as many more quotient figures as there are figures remaining in f 2. Take as many of the left hand figures of the dividend as may be necessary, for a dividual, as in Case first. Find how many times the dividual contains the divisor, and place : the given dividend therefore, the first quotient figure, taken in its complete value from the place it stands in, is the true quotient of the divisor in the complete value of the first part of the dividend. For the same reason, all the rest of the figures of the quotient, taken according to their places, are, each, the true quotient of the divisor, in the complete value of the several parts of the dividend belonging to each; consequently, taking all the quotient figures in the order in which they are placed by the rule, they make one number, which is equal to the sum of the true quotients of all the several parts of the dividend; and is, therefore, the true quotient of the whole dividend by the divisor. That no obscurity may remain in the demonstration, it is illustrated by the following example. Divisor=24)85701 Dividend. 1st dividual 85000(3000 the 1st quotient. 1st subtrahend=24×3000=72000 1st remainder=13000 +700 2d dividual 13700( 500 the 2d quotient. 2d subtrahend=24×500-12000 2d remainder = 1700 -+-00 3d dividual 1700 70 the 3d quotient. 3d subtrahend=24×70=1680 3d remainder= 20 +1 4th dividual=21(0 the 4th quotient. 4th subtrahend=24x0= 0 3570 the sum of all the 4th, and last remainder-21 quotients; or, the Answer. Explanation.-It is evident that the dividend is resolved into these parts, 85000+700+00+1; for though the first part of the dividend, or the first dividual, is considered only S5, yet it is truly 85000; and therefore its quotient is not simply 3, but is really 3000, and the remainder is 13000; and so of the rest; as may be seen in the operation. When there is no remainder to a division, the quotient is the absoluta |