Placing on the right of the given number, and subtracting its square, 49, from the left hand period, we find 7 for a remainder, to which we annex the next period, 82. Separating the last figure at the right from the others by a point, and dividing the number at the left by twice 7, or 14, we have 5 for a quotient figure, which we place at the right of the figure already found, and also annex it to 14. Multiplying 145 by 5, and subtracting the product from 782, we find the remainder 57. Hence, 75 is the number of tens of tens, or hundreds, of the required square root. To find the number of tens, bring down the next period and annex it to the second remainder, giving 5714, and divide 571 by double 75, or by 150. The quotient 3 annexed to 75 gives 753 for the number of tens in the root sought. We may, as before, find the number of units, which in this case will be 8. Therefore, the required square root is 7538. A similar course of reasoning may be applied to a number expressed by any number of figures. Hence, for the extraction of the square root of numbers, we have the following RULE. I. Separate the given number into periods of two figures each, beginning at the right hand: the period on the left will often contain but one figure. II. Find the greatest perfect square in the first period on the left, and place its root on the right after the manner of a quotient in division. Subtract the square of this root from the first period, and to the remainder bring down the second period for a dividend. III. Double the root already found and place it on the left for a divisor. See how many times the divisor is contained in the dividend, exclusive of the right hand figure, and place the quotient in the root and also at the right of the divisor. IV. Multiply the divisor, thus augmented, by the last figure of the root found, and subtract the product from the dividend and to the remainder bring down the next period for a new dividend. V. Double the whole root already found, for a new divisor, and continue the operation as before, until all the periods are brought down. REMARK I.-If, after all the periods are brought down, there is no remainder, the proposed number is a perfect square. But if there is a remainder, we have only found the root of the greatest perfect square contained in the given number, or the entire part of the root sought. For example, if it were required to extract the square root of 168, we should find 12 for the entire part of the root and a remainder of 24, which shows that 168 is not a perfect square. But is the square of 12 the greatest perfect square contained in 168? That is, is 12 the entire part of the root? To prove this, we will first show that, the difference between the squares of two consecutive numbers, is equal to twice the less number augmented by 1. The entire part of the root cannot be augmented by 1, unless the remainder is equal to, or exceeds twice the root found, plus 1. But, 12 × 2+1=25; and since the remainder 24 is less than 25, it follows that 12 cannot be augmented by a number as great as unity: hence, it is the entire part of the root. The principle demonstrated above, may be readily applied in finding the squares of consecutive numbers. If the numbers are large, it will be much easier to apply the above principle than to square the numbers separately. For example, if we have (651)2 = 423801, and wish to find the square of 652, we have, REMARK II. The number of places of figures in the root will always be equal to the number of periods into which the given number is separated. EXAMPLES. 1. Find the square root of 7225. 9. Find the square root of 918741672704. REMARK III.—The square root of an imperfect square, is in commensurable with 1, that is, its value cannot be expressed in exact parts of 1. a To prove this, we shall first show that if is an irreduci b A number is said to be prime when it cannot be exactly di vided by any other number, except 1. prime numbers. Thus 3, 5 and 7 are It is a fundamental principle, that every number may be resolved into prime factors, and that any number thus resolved, is equal to the continued product of all its prime factors. It often happens that some of these factors are equal to each other. For example, the number 50 = 2 × 5 × 5; and, 180 = 2 × 2 × 3 × 3 × 5. Now, from the rules for multiplication, it is evident that the square of any number is equal to the continued product of all the prime factors of that number, each taken twice. Hence, we see that, the square of a number cannot contain any prime factor which is not contained in the number itself. But, since a b' is, by hypothesis, an irreducible fraction, a and b can have no common factor: hence, it follows, from shown, that a2 and b2 cannot have a com what has just been Now, let c represent any whole number which is an imperfect square. If the square root of c can be expressed by a fraction, we shall have or a whole number equal to an irreducible fraction, which is absurd; hence, e cannot be expressed by a fraction. We conclude, therefore, that the square root of an imperfect square cannot be expressed in exact parts of 1. It may be shown, in a similar manner, that any root of an imperfect power of the degree indicated, cannot be expressed in exact parts Extraction of the Square Root of Fractions. 96. Since the second power of a fraction is obtained by squaring the numerator and denominator separately, it follows that the square root of a fraction will be equal to the square root of the numerator divided by the square root of the denominator. But if the numerator and the denominator are not both perfect squares, the root of the fraction cannot be exactly found. We can, however, easily find the root to within less than the fractional unit. Thus, if we were required to extract the square root of the Let 2 represent the greatest perfect square in ab, then will ab ab be contained between r2 and (r + 1)2, and will be con 62 |