figures or digits; thus, for example, if we had to square 25 ; put a 20 and b=5, and we shall find (a+b)2=(25)2=625. 305. Before we proceed to the investigation of these Rules, it will be necessary to explain the nature of the common arithmetical notatión. It is very well known that the value of the figures in the common arithmetical scale increases in a tenfold proportion from the right to the left; a number, therefore, may be expressed by the addition of the units, tens, hundreds, &c. of which it consists; thus the number 4371 may be expressed in the following manner, viz. 4000+300+70+1, or by 4X1000+3× 100+7×10+1; also, in decimal arithmetic, each figure is supposed to be multiplied by that power of 10, positve or negative, which is expressed by its distance from the figure before the point: thus, 672.53=6× 102+7× 10'+2×10°+5x101+3x10-26 X 100+7×10+2×1+ 50 3 53 15 3 十 =672+ + =672 100 100 Hence, if the digits 100 of a number be represented by a, b, c, d, e, &c. beginning from the left-hand; then, A number of figures may be expressed by 10a+b. 3 figures by 100a10b+c. By the digits of a number are meant the figures which compose it, considered independently of the value which they possess in the arithmetical scale. Thus the digits of the number 537 are simply the numbers 5, 3 and 7; whereas the 5, considered with respect to its place, in the numeration scale, means 500, and the 3 means 30. 306. Let a number of three figures, (viz. 100a+10b+c) be squared, and its root extracted according to the rule in (Art. 299), and the operation stands thus ; I. 10000a2+2000ab+100b2+200ac+20bc+c2 10000a2 200a10b)2000ab+100b2 2000ab+100b2 (100a +10b+c 200a +20b+c) 200ac+20be+c2 200ac +20bc+c2 and the operation is transformed into the following one; 40000+12000+900+400+60+1(200+30+1 III. But it is evident that this operation would not be affected by collecting the several numbers which stand in the same line into one sum, and leaving out the ciphers which are to be subtracted in the operation. Let this be done; and let two figures be brought down at a time after the square of the first figure in the root has been subtracted; then the operation may be exhibited in the manner annexed; from which it appears, that the square root of 53361 is 231. 307. To explain the division of the given number into periods consisting of two figures each, by placing a dot over every second figure beginning with the units, as exhibited in the foregoing operation. It must be observed, that, since the square root of 100; is 10; of 10000 is 100; of 1000000 is 1000; &c. &c. it follows, that the square root of a number less than 100 must consist of one figure; of a number between 100 and 10000, of two figures; of a number between 10000 and 1000000, of three figures ; &c. &c., and consequently the number of these dots will show the number of figures contained in the square root of the given number. Frem hence it follows, that the first figure of the root will be the greatest square root contained in the first of those periods reckoning from the left. Thus, in the case of 53361 (whose square root is a num ber consisting of three figures); since the square of the figure standing in the hundred's place cannot be found either in the last period (61), or in the last but one (33), it must be found in the first period (5); consequently the first figure of the root will be the square root of the greatest square number contained in 5; and this number is 4, the first figure of the root will be 2. The remainder of the operation will be readily understood by comparing the steps of it with the several steps of the process for finding the square root of (a+b+c)2 (Art. 299); for, having subtracted 4 from (5), there remains 1 ; bring down the next two figures (33), and the dividend is 133; double the first figure of the root (2), and place the result 4 in the divisor; 4 is contained in 13 three times ; 3 is therefore the second figure of the root; place this both in the divisor and quotient, and the former is 43; multiply by 3, and subtract 129, the remainder is 4; to which bring down the next two figures (61), which gives 461 for a dividend. Lastly, double the last figure of the former divisor, and it becomes 46; place this in the next divisor, and since 4 is contained in 4 once, 1 is the third figure of the root; place 1 therefore both in the divisor and quotient; multiply and subtract as before, and nothing remains. 308. The method of extracting the cube root of numbers may be understood by comparing the process for extracting the cube root of (a+b+c)3, (Art. 300), with the following operations, in which is deduced the cube root of the number 13997521. 13997521(200+40+1 a3=(200)3=8000000 1st remainder 5997521 3a2=3X (200) divisor, ..3a2b=3(200) X40=4800000 Omitting the superfluous ciphers, and bringing down three, figures at a time, the operation will stand thus: 309. These operations may be explained in the following manner; I. Since the cube root of 1000 is 10, of 1000000 is 100, &c.; it follows, that the cube root of a number less than 1000 will consist of one figure; of a number between 1000 and 1000000 of two figures, &c. &c.; if, therefore, the given number be divided into periods, each consisting of three figures, by placing a dot over every third figure beginning with the units, the number of those dots will show the number of figures of which the cube root consists; and for the reason assigned in the preceding Article, (respecting the first figure of the square root), the first figure of the root will be the cube root of the greatest cube number contained in the first period. 11. Having pointed the number, we find that its cube root consists of three figures. The first figure is the cube root of the greatest cube number contained in 13; this being 2, the value of this figure is 200, or a=200, conseqently a3= 8000000; subtract this number from 13997521, and the remainder is 5997521. Find the value of 3a2, and divide this latter number by it, and it gives 40 for the value of b, the second number of the root; put this in the quotient, and then calculate the value of 3ab+3ab2+63 and subtract it, and there remains 173521. Find now the value of 3×(a+b)3, and divide 173521 by it, and it gives 1 for the value of c, the third member of the root; put this in the quotient, and then calculate the amount of 3(a+b)3c+3(a+b) c2+c3, which subtract, and nothing remains. III. In reviewing the first of these two operations, it is evident that six ciphers might have been rejected in the value of a3, and three in the value of 3a2b+3ab2+b3, without affecting the substance of the operation; having therefore simplified the process as in the second operation, we are furnished with the following rule, for extracting the cube root of numbers. RULE. 310. Point off every third figure, beginning with the units; find the greatest cube number contained in the first period, and place the cube root of it in the quotient. Subtract its cube from the first period, and bring down the next three figures; divide the number thus brought down by 300 times the square of the first figure of the root, and it will give the second figure; add 300 times the square of the first figure, |