Doubling 3456 (thousands), we obtain 6912 (thousands), which we employ as divisor; bringing down the outstanding periods, and cutting off the last three figures,--that is, as many as there are root-figures still to be found, we obtain 5454190 (thousands), which we employ as dividend; and in the resulting quotient (789)--the remainder being disregarded--we have the last three figures of the root: 5454 required root=3456789. This is easily explained. Putting x for the last three figures of the root, we have V11949390190521=3456000+1; 11949390190521=(3456000+x)=3456000+2x 3456000 X 3 +xz; 11949390190521 – 3456000ʻ=2x3456000 xx+x+; i.e., 5454190521* =6912000 Xx+x; 5454190521-6912000=x+ zce? 6912000 As the number represented by x occupies only three places, the number represented by x cannot occupy more than 23 sic places ($ 230); so that is a proper fraction, the 6912000 denominator occupying seven places.f When, therefore, we divide 5454190521 by 6912000,-or 5454190 by 6912,-and reject the remainder, we obtain č. * From what has already been explained, it is evident that 5454190521 is the difference between 11949390190521 and 3456000%. + It is obvious that every such fraction must be proper : the denominator terminating with as many ciphers as there are figures in the square root of the numerator, and the ciphers being preceded by more than an equal number of other figures-obtained from the doubling of the known part of the required square root. a 232. As the square of a fraction is obtained when the square of the numerator is divided by the square of the denominator-so, the square root of a fraction is obtained when the square root of the numerator is divided by the square root of the denominator. Thus, (3) =5x5=5X 5._52_25 (7)=7x1 25. 64 v 49 7 &c. 2 JI II -; &c.: V253 I21 8 .; II I21 233. The following is a comparatively easy way of extracting the square root of a fraction whose terms are not square numbers, or the square roots of whose terms cannot be found by inspection : Multiply the numerator by the denominator, extract the square root of the product, and divide this root by the denominator. Thus, 5x8_V40_140 ; 8 77 i &c. а In the majority of cases, however, the square root of a fraction is most easily extracted when we begin by converting the fraction into a decimal ($ 234). 234. To extract the Square Root of a number which contains a Decimal: Regard the number as an integer—having, if necessary, first made the number of decimal places even, by annexing a cipher; and remove the decimal point in the resulting root a place to the left for every pair of decimal places in the given number. As an illustration, let it be required to extract the square root of .4. The square root of 4 being 2, the square root o •4 might, at first sight, appear to be .2. If, however, .2 were multiplied by itself, the product would be (not '4, but) '04. The given decimal can be written under any of the following forms: 4 40 400 4,000 40,000 400,000 &c. Reject10'100' 1,000 10,000' 100,000' 1,000,000' ing the fractions (4, 400, 40,000, &c.) whose denomina 10' 1,000' 100,000' tors are not square numbers, we thus have 6 V4 :6 nearly N 100 N40 IO N:4 a 1,000 N 400,000 632 .632 still more nearly N 1,000,000 1,000 &c. &c. In practice, :4 would simply be written '40, or •4000, or •400000, &c., and treated as a whole number--the first figure of the root, however, being made to occupy the first decimal place. As a second illustration, let it be required to extract the square root of 1234 56789. This number is equivalent to the 123456789 fraction As 100,000 is not a square number, we 100,000 multiply the terms of the fraction by 10: we then have V 1234567890 V1234 56789 35136 =35-136—a result more N 1,000,000 easily obtained when we annex a cipher to 123456789 (in order to make the number of decimal places even), extract the square root of 1234567890, and remove the decimal point in the root (35136) three places to the left; that is, a place for every pair of figures in •567890. Note. For every figure which we desire to have in the decimal part of the square root of a number, there must be a pair of figures—including, if necessary, one or more ciphersin the decimal part of the number itself. This is rendered obvious by what has just been explained, as well as by the fact that ($ 59) the square of a decimal occupies twice as many places as the decimal itself. 235. To extract the Square Root of either a Fractional or a Mixed number: Convert the Fraction into a Decimal, and then proceed as already directed (234). Thus, in extracting the square root of is, we write is under the form 384615 &c. ; in extracting the square root of 2463, we write 2463 under the form 246•375; and so on. 236. To find an Approximation which shall differ from the Square Root of a number by less than a given Fractional Unit: Multiply the number by the square of the denominator of the fractional unit, and divide the denominator into the square root of the product. Thus, to find an approximation which shall differ from v 47 by less than £, we multiply 47 by 8%, and divide 8 into the square 8 root of the product: 47= 47 X 82_47X64_3008 N 475 82 84 N 3008 As v 3008 lies between V 2916 and N 3025, or 8 between 54 V 47 must lie between 54 and 55 and these two fractions differ by exactly each differs from N 47 I 82 and 55, as I co! I I by less than 8 Again : to find an approximation which shall differ from W365 by less than we multiply 365 by 1,000?, and divide 1,000' 1,000 into the square root of the product : 3655 365 X 1,000? 1,000 365,000,000 365,000,000 ;N 365= 1,0002 As w 365,000,000 lies be 1,000 tween 364,962,816 and V 365,001,025, or between 19,104 and 19,105, N 365 must lie between 19,104 19,105 i and as these 1,000 1,000 two fractions differ by exactly V 365 does not differ from 1,000 either 19·104 or 19:105 by so much as (001). 1,000 I I EXTRACTION OF THE CUBE ROOT. 237. When the integral part of a number occupies1,R2, or 3 places, the integral part of i place 4, 5, , 6 2 place the cube root of the 7, 8, 9 number will occupy IO, II, &c. &c. Thus, as every number whose integral part is expressed by 12 a n nts, or or one, two, or three figures must be less than 1,000, but not less than 1, the cube root of every such number must be less than 13/1,000, but not less than Ý T-that is, must be less than 10, but not less than 1; so that the integral part of the root will occupy one place. Again : as every number whose integral part is expressed by four, five, or six figures must be less than 1,000,000, but not less than 1,000, the cube root of every such number must be less than 31,000,000, but not less than 13/1,000—that is, must be less than 100, but not less than 10; so that the integral part of the root will occupy two places. In like manner, as every number whose integral part occupies seven, eight, or nine places must be less than 1,000,000,000, but not less than 1,000,000, the cube root of every such number must be less than 1,000,000,000, but not less than 3/1,000,000—that is, must be less than 1,000, but not less than 100; so that the integral part of the root will occupy three places. Speaking generally, therefore, we are able to say that when the integral part of a number occupies n places, the integral part of the cube root of the number will occupy n+2 ' places according as n, or n+1, or n+2 is 3 3 3 a multiple of 3. EXAMPLE I._Extract 3 185,193. This number being expressed by six figures, its cube root must (§ 237) be expressed by two-a tens' and a units' figure, In looking for the tens' figure of the root, we disregard the last three figures (193) of the given number—-knowing that (226) the cube of any number of tens is some number of thousands. As 185 lies between 125 and 216, or between 53 and 63, 185,000and therefore 185,193—must lie between 503 and 603; so that the tens' figure of the root is 5. Putting u for the units' figure of the root, we have 3 185,193=50+u ; 185,193=(50+u):= 50% +3% 50% Xu+3X 50Xu2 tu; 185,193-503=3X 50*Xu-+ 3 x 50 xu? +23 ; i.e., 60,193=3X 50% Xu+3X 50 Xu” +23. As 3X50% Xu is obviously a much larger number than 3 X 50 xu? ++, the division of 60,193 by 3x 502, or of 60,193 by 7,500, or of 601 by 75, will give the units' figure of the root, or a figure not much higher. Seeing that 75 is contained 8 times, but not 9 times, in 601, we conclude that the units' figure of the root cannot be higher than 8, which, accordingly, we try : 3 x 502 x 8=60000 3 x 50 x 82= 9600 83= 512 70112 |