Suppose then the second digit found; we have now to see whether the number obtained be the exact or nearest root. In order to do this, we must evidently subtract the square of the root from the given number. But since (10 a+b)2 = 100 a2 + 2 × 10 a × b + b2, and in forming the first remainder, we subtracted 100 a2, therefore we need only to subtract 2 × 10 a × b+b2 from this remainder. Now 2 X 10 a × b + b2 = (2 × 10 a + b) × b; if therefore we add to twice the first part of the root the second part, and multiply by this latter, the subtrahend will be formed. If there be no remainder, the number found will be the exact root: but if there be a remainder, either there is no exact root, or the second digit in the number found is too small. To determine whether this latter be the case, we observe that the addition of to any number increases its square by 1 more than twice the number; therefore the remainder must not be greater than twice the root found, if the second digit be correct. Now let the given number contain 5 or 6 digits, and therefore its root contain 3 digits. Let the number be pointed as by the rule: then it may be shewn as before that the whole number of tens in the root is the greatest number, whose square does not exceed the whole number of hundreds in the given number, that is the first two periods. Hence we have to find the nearest square root of the number composed of the first two periods, which, containing two digits, may be found as already explained. Let a stand for this number, a of course is less than 100. Then the whole root will be expressed by 10 a+b+x, b being the number of units, x a fraction less than 1. In precisely the same way as before it may be shewn that the digit b may be found by dividing the remainder, after the subtraction of 100 a2, by 2 X 10 a. Only in this case, as a is not less than 10, the (b + x)2 greatest value of the expressions x + 2 X 10 a is less than ; and therefore the error in the quotient cannot exceed 1. Having found the digit b, we have to determine as before, whether the exact or nearest root has been obtained, by subtracting the square of the root found from the given number; or by subtracting from the last remainder the product of the sum of twice the first part of the root and the second part by the second part. The criterion by which is known whether the last digit be large enough is the same as before, viz. the remainder must not be greater than twice the root. In the same way it may be shewn how to extract the square root of any number whatever. And the process is seen to be that described in the ordinary rule, unnecessary ciphers being omitted. If the given number be a decimal, the number of decimal places must be made even (Prop. 71), and the extraction of the square root of the number considered as integral being effected, the root of the decimal is found by marking off as decimals half as many as there are in the given number. For the root of a fraction is obtained by extracting the root of numerator and denominator. Prop. 74.-If the Square Root of a number contain 2 n + 1 digits, and n+1 of them have been found by the ordinary Rule, the remaining n may be found by dividing the remainder by the corresponding trial-divisor. For if a and b be the two parts of the root, the one a containing n+1 significant digits, followed by n ciphers, and the other b containing ʼn digits, then the remainder after a has been found, and its square subtracted is 2ab+b2, which being divided by the trial-divisor 2 a, gives a quotient b2 b2 2 a be a proper Now b containing n 62 b+ differing from b by the quantity 2 a' If then Za fraction, b will be correctly found by this division. digits is less than 10; and a containing in all 2 n + 1 digits is not less b2 b2 2 a than 102n; therefore is less than 10 2n or than . Hence 2 a being a proper fraction, the division of the remainder by the trial-divisor will give b correctly. Prop. 75.-To prove and explain the Rule for the extrac tion of the Cube Root of a number. Let N be the given number of 4, 5, or 6 digits, whose nearest cube root therefore contains 2 digits, which call a and b. Let the number be pointed according to the Rule. Let c be the greatest number, whose cube does not exceed the first period in N,which is the number of thousands in N; then (c+1)3 is greater than the number of thousands in N, and 1000 (c+1)3 is greater than N. Hence a cannot be greater than c; for if it could be equal to c +1, then, since 1000 (c+1)3 is greater than N, the cube of a part of the root would be greater than the given number, which is the cube of the whole root. Nor can a be less than c; for then 10 a would be less than 10 c, and 10 a + b would also be less than 10 c, and therefore (10 a+b)3, which is the greatest cube in N, would be less than 1000 c3, i.e, would be less than the greatest cube number of thousands in N, which is manifestly impossible. Hence a is equal to c, or the first digit in the root is the greatest number, whose cube does not exceed the first period. We have now to find b. Let x be the difference between 10 a + b, and the complete root of N; x is of course a fraction. Then, a+b+x}=(10a)3+3(10a)2(b+x)+3(10a) (b+x)2+(b+x)3 N = {10 tient arising from the division of N - (10 a)3 by 3 (10 a)2, will be the second digit. But as x may be very nearly equal to 1, and b may be as large as 9, while a may be as small as 1, the value of the above expression may be very nearly as large as 143; or there may be an excess of 14 in the quotient above the second digit, and even if a be 9, there may be an excess of 2. But as the ratio is diminished or increased, the possible b a error is diminished or increased. And if b be less than 3, the error cannot exceed 1. No Rule however can be formed for determining the exact amount of error in every particular case. Suppose then the second digit found; we have now to see whether the number obtained be the exact, or nearest, cube root. To do this, we must subtract the cube of the root from the given number. But since (10 a + b)3 = ( 10 a)3 + 3 (10 a)2 b + 3 (10 a)b2 +b3, and in forming the first remainder we subtracted (10 a)3, therefore we need only to subtract 3 (10 a)2 b+ 3 (10 a) b2 + b3 from this remainder. Now this expression may be put into the form {3 (10 a)2 + {3 (10 a) +b } b} b, so that if to the divisor 3 (10 a)2 there be added, the product by the second part of the root of the number, formed by adding the second part to three times the first part, and this whole sum be multiplied by the second part, the subtrahend will be formed. If there be no remainder, the number found will be the exact root, but if there be a remainder, either there is no exact root, or the second digit in that found is too small. To determine whether this latter be the case, we observe that the addition of 1 to any number increases its cube by 1 more than three times the product of the original, by the increased number; therefore the remainder must not be greater than this product, if the second digit be correct. Now let the given number contain more than 6 and not more than 9 digits; its nearest cube root therefore will contain 3 digits. Let the number be pointed according to the Rule; then it may be shewn as before that the whole number of tens in the root is the greatest number, whose cube does not exceed the whole number of thousands in the given number, that is the first two periods. Hence we have to find the nearest cube root of the number composed of the first two periods, which, containing two digits, may be found as already explained. Let a stand for this number, a of course being less than 100. Then the whole root will be expressed by 10 a+b+x, b being the number of units, a a fraction less than 1. In precisely the same way as before it may be shewn, that the digit b may be found by dividing the remainder, after the subtraction of (10 a)3 by 3 (10 a)2. Only in this case, as a is not less than 10, the greatest possible excess of this quotient over b will be 2. Having found the digit b, we have to determine, as before, whether the exact or nearest root has been obtained, by subtracting the cube of the root found from the given number; or by subtracting from the last remainder a subtrahend formed in the same manner as the former one. The method of forming the second trial-divisor without the labour of squaring the root may be shewn thus. Let a, b, be the first two digits in the root, then the number of hundreds in the next trial-divisor will be 3 (10 a + b)2, or 3 (10 a)2 +2 {3 (10 a+6} b+b2, or is the sum of the last complete divisor, and the quantity {3 (10 a) + b}b increased by the square of the last figure in the root. In the same way it may be shewn how to extract the cube root of any number whatever. And the process is seen to be that described in the Rule, unnecessary ciphers being omitted. If the given number be a decimal, the number of decimal places must be made some multiple of three, (Prop. 72), and the cube root of the number, considered as integral, being effected, the root of the decimal is found by marking off as decimals one-third as many as there are in the given number. Prop. 76.-If the cube root of a number contain 2 n + 2 digits, and n + 2 have been found by the ordinary Rule, the remaining n may be found by dividing the remainder by the corresponding trial-divisor. For if a, b, be the two parts of the root, the one a containing n +2 significant digits, followed by n ciphers, and the other 6 containing n digits, then the remainder, after a has been found, and its cube subtracted, will be 3a2b+3a b2 + b3, which, being divided by the trial-divisor 3 a2, gives b2 63 a quotient b+· b3 + " b2 differing from b by the quantity + a 3 a2 a 3a2 If this quantity be less than 1, b will be correctly found by the division. Now b containing n digits is less than 10", and a containing in all 2 n + 2 digits or 100 1 1 +. 10 100 Hence b is accurately found by dividing the remainder by the trial-divisor. Prop. 77.-In any Arithmetic Series the sum of any two terms, equidistant from the extremes is always the same; and, when the number of terms is odd, twice the middle term is equal to the sum of the extremes. For every Arithmetic series, whose terms increase by a common difference, may by inverting the order of the terms, be written as a decreasing series, and any term of the increasing series will be as much greater than the first term as the corresponding term of the decreasing series is less than its first term: that is if a and be the first and last terms of the series, and b and c two terms equidistant from a and b, the excess of b above a will be equal to the defect of c from 1, or b-al-c: hence b+c=a+l. Also if m be the middle term, then m — a — 1 — m, or 2 m = a + 1. Prop. 78.-To find the sum of an Arithmetic Series. Let the order of the terms of the series be inverted, and another series formed; and let the corresponding terms of these two series be added together, the result will (Prop. 77) be a series of terms, each equal to the sum of the first and last terms of the original series, and the number being the same as in that series. Also the sum of these terms, being the sum of two identical series, is twice the sum of one of them. Hence twice the sum of the series is equal to the sum of the first and last terms multiplied by the number of terms: or the sum of an Arithmetic series is the sum of the first and last terms multiplied by half the number of terms. Prop. 79.-To find any required term of an Arithmetic Series. Since every term is greater than the preceding or following by the common difference, therefore the second differs from the first by once the common difference, the third differs from the first by twice the difference, and so on. So it appears that to form any term from the first and the common difference, we must increase or diminish the first term by the product of the common difference multiplied by a number less by one than the number of the term. Cor. 1. Hence the last term is equal to the first term increased or diminished by the product of the common difference multiplied by a number less by one than the number of terms. Cor. 2. Hence if the difference between the first and last terms of an |