Prop. 69.-The cube of any number is equal to the sum of the cubes of any two parts into which it may be divided, together with three times the sum of the products of the square of each into the other. The cube of a number is obtained by multiplying the square by the number. Now the square is equal to the sum of the squares of the two parts together with twice the product of the parts. If this be now multiplied by one of the parts, the result will be the cube of the first part, twice the product of the square of the first into the second, and the product of the first into the square of the second. If it be multiplied by the other part, the result will be the product of the square of the first into the second, twice the product of the first into the square of the second, and the cube of the second. Hence adding these results, the cube of the number is equal to the sum of the cubes of the two parts, together with three times the sum of the products of the square of each part into the other. Thus the cube of 16 (122 + 2 × 4 × 12 + 42) × (12 + 4) =128+2×4X 122+42× 12+122×4+2×42×12+43 123 +3 × 4 × 122 + 3 × 42 × 12 +43 Algebraically expressed the Prop. will stand thus: (a+b)3 = a3 + 3 a2 b + 3 a b2 +b3. Prop. 70.-A power of a fraction is the fraction formed by raising the numerator and denominator to the required power. For the product of fractions is obtained by multiplying the numerators for a new numerator, and the denominators for a new denominator. When therefore all the numerators are the same, and all the denominators, the numerator and denominator of the power will be the power of the numerator and denominator of the original fraction. Cor. 1. A root of a fraction is the fraction formed by extracting the root of the numerator and denominator. For if this fraction be raised to the power indicated by the degree of the root, it will become equal to the given fraction. Cor. 2. Hence the square and cube root of decimals may be found by extracting the root of the number, considered as integral, and marking off as decimals one-half, or one-third as many figures, as there are decimals in the given number, which must therefore be some multiple of 2 or 3. Prop. 71.-To prove the Rule for pointing in extraction of the Square Root. square of 100 is 10000 the square of 1000 is 1000000, and so on, it appears that the square root of a number of 1 or 2 digits will consist of 1 digit ; of 3 or 4 digits will consist of 2 digits; of 5 or 6 digits will consist of 3 digits, and so on. Hence if every alternate figure be marked by a point, the number of points will be the number of digits in the square root. Again, since the square of a decimal contains twice the number of decimal places, which the decimal contains, therefore if a point be placed over every alternate figure in a given decimal, the number of points will shew the number of decimal places in the square root; but there must always be an even number of decimals in the given number, that the extraction of the square root may be possible. It is of no consequence with which figure the pointing is commenced, but it is usual to begin with the units' figure, and to point every alternate figure right and left. of 4 and not more than 6 digits consists of 2 digits; of 7 and not more than 9 digits consists of 3 digits, and so on. Hence if every third figure be marked by a point, the number of points will shew the number of digits in the cube root. Again, since the cube of a decimal contains three times as many decimal places as the decimal contains, therefore every decimal, which is a cube, must have a number of decimal places divisible by 3, and if a point be placed over every third figure, the number of points will shew the number of decimal places in the root. It is usual to commence the pointing with the units' figure, and to continue it right and left, pointing every third figure. Prop. 73.-To prove and explain the Rule for the extraction of the Square Root. Let N be the given number of 3 or 4 digits, whose nearest square root therefore will contain 2 digits, which call a and b. Then the integral part of the root will be denoted by 10 a+b. Let the number be pointed by the Rule for pointing, thus being divided into periods from point to point, the figures as far as the first point counting from the left being reckoned as the first period. Let c be the greatest number, whose square does not exceed the first period in N,which is the number of hundreds in N; then (c+1)2 is greater than the number of hundreds in N, and 100 (c + 1)2 or 2 2 {10 (c+1)} is greater than N. Hence a cannot be greater than c; for if it could be equal to e+ 1, then since {10 (c+1)} is greater than N, the square of a part of the root would be greater than the given number, which is the square 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 (b being a number of units less than 10) be less than 10 c, and therefore (10 a + b)2, which is the greatest square in N, would be less than (10 c)2, i.e. would be less than the greatest square number of hundreds in N, which is manifestly impossible. Hence a is equal to c, or the first digit in the root is the greatest number whose square 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 N (10 a+b+ x)2 = (10 a)2 + 2 × 10 a (b + x) + (b + x)2 fraction, the integral quotient arising from the division of N-(10 a)2 by 2 X 10 a 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 this expression may be very nearly equal to 6, or the quotient may be larger by 5, than the second digit. But from the second form given of the expression b it appears that the error decreases or increases, as and therefore the a ratio of the quotient to a, decreases or increases. And the error is larger as x, and therefore the remainder from the division, is larger. Also if a be greater than 4, or b be less than 4, the error cannot exceed 1, i.e. if the first digit be greater than 4, or the quotient less than 5. But practice will soon make the detection of the error easy. 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 |