Since 36 is the greatest square contained in 43, the first figure of the root is 6. We then subtract its square 36 from 43, and to the remainder 7 bring down the next period 16. Now, since the last figure 6 of the result 716, contains no part of the double product of the first figure of the tens by the second, it follows, that the second figure of the root will be obtained by dividing 71 by 12, double the first figure of the tens. This gives 5 for a quotient, which we place in the root, and at the right of the divisor 12. Then subtract the product of 125 by 5 from 716, and to the remainder bring down the next period, and the result 9149 will contain twice the product of the tens of the root multiplied by the units, plus the square of the units. If this result be then divided by twice 65, that is, by double the tens of the root, (which may always be found by adding the last figure of the divisor to itself), the quotient will be the units of the root. 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 be. ginning at the right hand, the period on the left will often contain but one figure. II. Find the greatest 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 the 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 di. visor. Seek how many times the divisor is contained in the dividend, exclusive of the right hand figure, and place the figure in the root and also at the right of the divisor. IV. Multiply the divisor, thus augmented, by the last figure of the root, 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. Ist. REMARK. 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, you 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 665, we should find 25 for the entire part of the root and a remainder of 40, which shows that 665 is not a perfect square. But is the square of 25 the greatest perfect square contained in 665? that is, is 25 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 unity. Hence, the entire part of the root cannot be augmented, unless the remainder exceed twice the root found, plus unity. But 25×2+1=51>40 the remainder: therefore, 25 is the entire part of the root. 2d. REMARK. The number of figures in the root will always be equal to the number of periods into which the given number is separated. EXAMPLES. 1. To find the square root of 7225. 2. To find the square root of 17689. 3. To find the square root of 994009. 4. To find the square root of 85678973. 5. To find the square root of 67812675. 118. The square root of a number which is not a perfect square, is called incommensurable or irrational, because its exact root can not be found in terms of the numerical unit. Thus, √2, √5, VT, are incommensurable numbers. They are also sometimes called surds. In order to prove that the root of an imperfect power cannot be expressed by exact parts of unity, we must first show that, Every number P, which will exactly divide the product AXB of two numbers, and which is prime with one of them, will divide the other. Let us suppose that P will not divide A, and that A is greater than P. Apply to A and P the process for finding the greatest common divisor, and designate the quotients which arise by Q, Q', Q" ... and the remainders R, R', R" ... respectively. If the division be continued sufficiently far, we shall obtain a remainder equal to unity, for the remainder cannot be 0, since by hypothesis A and P are prime with each other. Hence we shall have the following equations. A=PQ+R P=R Q'+R' Multiplying the first of these equations by B, and dividing by P, we have AB=BQ+BR. But, by hypothesis, AB is an entire number, and since B and are entire numbers, the product BQ is an entire number. Hence If we multiply the second of the above equations by B, and divide by P, we have hence BRQ' BR But we have already shown that is an entire number; BR is an entire number. This being the case, P must also be an entire number. If the operation be continued until the number which multiplies B becomes 1, we shall have equal to an entire number, which proves that P will divide B. Bx1 In the operations above we have supposed A>P, but if P>A we should first divide P by A. Hence, if a number P will exactly divide the product of two num. bers, and is prime with one of them, it will divide the other. We will now show that the root of an imperfect power cannot be expressed by a fractional number. Let c be an imperfect square. Then if its exact root can be expressed by a fractional number, we shall have or a C C by squaring both members. a b But if c is not a perfect power, its root will not be a whole num. ber, hence 이 will at least be an irreducible fraction, or a and b will be prime to each other. But if a is not divisible by b, axa or a3 will not be divisible by b, from what has been shown above; neither then can a2 be divisible by . Since to divide by bo is but to divide a2 twice by b. Hence, is an irreducible fraction, and therefore cannot be equal to the entire number c: therefore, we cannot assume a b or the root of an imperfect power can not be expressed by a fractional number that is rational. 1 Extraction of the square root of Fractions. 119. Since the square or 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. For example, the square root of a is equal to : for b But if neither the numerator nor the denominator is a perfect square, the root of the fraction cannot be exactly found. We can however, easily find the exact root to within less than one of the equal parts of the fraction. To effect this, multiply both terms of the fraction by the denomina. tor, which makes the denominator a perfect square without altering the value of the fraction. Then extract the square root of the perfect square nearest the value of the numerator, and place the root of the denominator under it; this fraction will be the approximate root. Thus, if it be required to extract the square root of 15 3 에, we mul. tiply both terms by 5, which gives : the square nearest 15 is 4 25 16: hence 이 is the required root, and is exact to within less 1 than 5 120. We may, by a similar method, determine approximatively the roots of whole numbers which are not perfect squares. Let it be required, for example, to determine the square root of an entire number a, nearer than the fraction 1 n : that is to say, to find a |