units' place of the integers, 6; then passing one figure 7, we point the next 8; and passing the 2 we point the 6: in the same way the decimals are pointed in pairs counted from the units' place of the integers to the right hand. We next inquire what root or number will, when squared, give a product equal to or nearest under the first period on the left hand, which is here 46: this the preceding Table shows to be 6, whose square is 36, for the square of 7 being 49, would be too high: we then write the root 6 in the quotient, and the square 36 under the period 46, to be ́subtracted from it. To the remainder 10 we bring down the whole of the following period 28, and saying twice or double the figure in the quotient 6, is 12, we write 12 as a divisor to the dividend 1028, leaving room for a figure or two in the units' and tens' places. We then inquire how often this divisor 12 can be contained in the dividend 102 (for the units' place is always excluded), and finding it can be contained 8 times, we write 8 in the quotient after 6, and also in the void space of the divisor after 12, making altogether for a divisor 128. This is multiplied by the 8 of the quotient, and the product 1024 subtracted from 1028 leaves 4 for a remainder, to which bringing down the next period 76, we have a new dividend, and for a divisor, doubling the whole quotient 68, we have 136, to be taken out of 47, (excluding the 6 in the units' place, as before directed), which being impossible, we write 0 in the quotient, and also after the 136 of the divisor. Then bringing down the following period 12, which in this example consists of decimals, we have 47612 for a dividend to be divided by the same divisor increased by the nought, viz 1360: this may be done 3 times, the three being written in the quotient, and after the 0 of the divisor, which multiplied by 3 gives 40809 to be subtracted from 47612, leaving the remainder 6803, to which bringing down the last period or pair of deeimals 25, we have a new dividend 680325 to be divided by

the double of the quotient 6803, equal to 13606: this sum being contained 5 times in the dividend, we write 5 in the quotient and in the void space of the divisor, which being multiplied by 5, the product is equal to the dividend, and nothing remains. In this manner we obtain the quotient or root required: but as part of the given square is a decimal fraction consisting of four figures, forming two periods, we set off from the right hand of the quotient two figures, one for each period, separating them by the decimal point from the integral part of the root, which thus is found to be 680,25. As a proof that the operation is correct, we may square this root, or multiply it by itself, and the product, if all is right, and that there is no remainder, will be the same with the given square, the root of which was to be extracted. Had there been any remainder, it should have been added to the square of the quotient, to have made it coincide with the given square.

2d. To extract the cube root of any given quantity, we work as in the following example, where it is required to discover the root or first power of the cubical quantity 43169672,512.

3 times 1225

35

6125

3675

42875

Having written down the given cubical quantity, distinguishing by a point the decimals from the integers, we begin to divide the integers into periods, by placing a dot under or over the units place 2; then passing over two figures we place a dot under or over the fourth figure 9; and again passing over two figures, we dot the seventh figure 3: then returning to the units' place of the integers, we count off three figures to the right hand in the decimals, and place a dot over or under the third figure. By this process we discover that there will be 3 integers and 1 decimal in the root required; since there are always as many figures in the root, as there are dots or periods in the cubic dividend. We now inquire what number, when cubed or multiplied twice. successively into itself, will produce a cube equal to or the nearest under the first period on the left hand, which is 43: this upon trial, or on inspecting the foregoing Table will be found to be 3; for 3 times 3 are 9, and 3 times 9 are 27; whereas 4 times 4 are 16, and 4 times 16 are 64, which is a greater sum than 43; we therefore write in the quotient, or where the root is to appear, the 3, and its cube 27 under 43, from which subtracting it we have for a remainder 16. To this 16 we bring down the first figure of the next period, viz. 1, making 161 for a new dividend. For a divisor we take 3 times the square of the root or quotient, that is 3 times 9 are 27, which we write before 161, and finding it can be contained 5 times in that dividend, we write 5 after the 3 in the quotient, making in all 35. This quantity is then cubed, as is here shown, and the product is written under 161, but separated from it by a line; for the

new found cube is not to be subtracted from 161, but from the cube originally given at the head of the operation; care must therefore be taken to place the first figure of the product 4 under the first figure of the top line, the second figure 2 under the second figure 3 of the top line, and so on to the last figure 5 of the new cube, which will come. immediately under the 9 of the top; thus occupying two periods, corresponding to the two figures in the root or quotient. Subtracting the new cube from the top, we have for a remainder 294, to which bringing down the first figure of the succeeding period of the given cube, viz. 6, we obtain 2946 to be divided, as before, by 3 times the square of the quotient 35, which is, as shown in the example, . 3675: but this divisor being greater than the dividend 2946, it cannot be taken out of it; we therefore write a nought in the root after the 5, making the whole 350. This sum should next be cubed, and the product subtracted from the given cube; but as the addition of a nought to the factors would produce no change on the products, it is needless to take this trouble, we therefore bring down the whole period, of which the first figure 6 was already brought down, together with 5, the first figure of the following period, making the new dividend 2946725, or rather 29467, excluding the two last figures, to correspond to the two noughts; which, had the work been completed, would have made the divisor 367500: we then inquire how often this divisor can be found in the new dividend, and the quotient being 8, we write this figure in the root, and cubing the whole 350,8, the product turns out to be precisely the cubical quantity given in the question; and it consists of three places of integers and one of decimals, agreeably to the number of periods in cach portion of the given cube, as was before remarked would be the case.

Having

Having thus gone through the most necessary branches of Arithmetic, integral and fractional, it remains only to apprise the student, for his satisfaction and encouragement, that the rules laid down to govern his practice, in the various operations he has been directed to perform, are by no means either arbitrary or merely mechanical; on the contrary they are all founded in the nature of things and of numbers, and are susceptible of the most accurate and evident demonstration. To give however such demonstrations in this limited work would be impracticable, and fully to comprehend them would require a more extensive knowledge of various parts of Geometry, than can be expected to be possesed by those young persons for whose use these pages have been compiled. Opportunities for illustrating arithmetical operations will, notwithstanding, occasionally occur in the progress of the work; and such opportunities shall not be neglected.

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CHAP.