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cal sign; therefore we have (√2a)2= √2a.

6

Again, squaring √36, we have √36-√√√36; hence....

3

( √36)2 = √3b.

3b=

Consequently, when the index of the radical is divisible by the exponent of the power, perform this division, leaving the quantity under the radical as it was.

To extract the root of a radical, multiply the index of the radical by the index of the root to be extracted, leaving the quantity under the sign as it was.

Thus,

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This rule is nothing more than the principle of No. 161, enunciated in an inverse order.

of

When the quantity under the radical is a perfect power the degree of the root to be extracted, the result can be reduced.

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are equal to Va. (161.)

5

m

9a2

√ a =

5

5

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m

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Va; because both expressions

176. The rules just demonstrated for the calculus of radicals, principally depend upon the fact that the nth root of the product of several factors is equal to the product of the nth roots of these factors; and the demonstration of this principle depends upon this: When the powers (of the same degree) of two expressions are equal, the expressions are also equal. Now this last proposition, which is true for absolute numbers, is not always true for algebraic expressions.

To prove this, we will show that the same number can have more than one square root, cube root, fourth root, &c.

For, denote the general expression of the square root of a by

x, and the arithmetical value of it by p; we have the equation x2=a, or x2=p2, whence xp. Hence we see that the square of p (which is the root of a) will give a, whether its sign

be + or

In the second place, let x be the general expression of the cube root of a, and p the numerical value of this root; we have the equation

a3=a, or x3=p3.

This equation is satisfied by making x=p.

Observing that the equation x3=p3 can be put under the form. x3-p30, and that the expression 3-p3 is divisible (31) by x-p, which gives the exact quotient, x2+px+p2, the above equation can be transformed into

(x−p) (x2 +px+p2)=0,

which can be verified by supposing x-p=0, whence x=p; or by supposing

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Hence, the cube root of a admits of three different algebraic values, viz.

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Again, resolve the equation =p', in which p denotes the

arithmetical value of Va. This equation can be put under the form x1-p1=0. Now this expression reduces to (19) . . . (x2—p2) (x2 +p2). Hence the equation reduces to

(x2—p2) (x2 +p2)=0, and can be satisfied by supposing x2-p2-0, whence x=p; or by supposing x2+p2=0, whence a=V-p2 =±p√ −1.

We therefore obtain four different algebraic expressions for the fourth root of a.

For another example, resolve the equation which can be put under the form

Now x-p" reduces to

therefore the equation becomes

x°=p®, x®-p=0.

3

(x3-p3) (x3 +p3),

(x3—p3) (x3+p3)=0,

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3

And if in the equation x3+p3=0, we make p=-p', it becomes x3-p'3=0, from which we deduce x=p', and

3.

3

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Therefore the equation x-p=0, and consequently the 6th root of a admits of six values, p, ap, a'p, -p, -αp —α'p, by making

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We may then conclude from analogy, that every equation of the form a-a=0, or x-p=0, is susceptible of m different values, that is, the mth root of a number admits of m different algebraic values:

177. If in the preceding equations and the results corresponding to them, we suppose as a particular case a=1, whence p=1, we will obtain the second, third, fourth, &c. roots of unity. Thus +1 and -1 are the two square roots of unity, because the equation 2—1=0, gives x=±1.

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cube roots of unity, or the roots of x3-1=0.

+1, −1, + √−1, −√—1, are the four fourth roots of unity, or the roots of x-1=0.

178. It results from the preceding analysis, that the rules for the calculus of radicals which are exact when applied to absolute numbers, are susceptible of some modifications, when applied to expressions or symbols which are purely algebraic; these are more particularly necessary when applied to imaginary expressions, and are a consequence of what has been said in No. 176.

For example, the product of a by the rule of No. 174,

would be

√=ax √ = a = √ +a2.

Now, Va is equal to ±a (No. 176.); there is, then, apparently, an uncertainty as to the sign with which a should be affected. Nevertheless, the true answer is -a; for, in order to square √m, it is only necessary to suppress the radical; but the √ ax √ a reduces to (a), and is therefore equal to

-a.

Again, let it be required to form the product √—a× √—b, by the rule of No. 174, we will have

√=ax v=b=√ +ab.

Now, Vab±p (No. 176), p being the arithmetical value of the square root of ab; but I say that the true result should be —p or —√ab, so long as both the radicals √—a and √— are considered to be affected with the sign +.

For,

√=a=√a. √—1 and √—b=√b. √-1; hence √—a× √—b= √a. √—1 × √—b× √=1= √ab( √—1)2

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Upon this principle we find the different powers of √-1 to be, as follows:

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Again, let it be proposed to determine the product of a

by the which, from the rule, will be
—b
quently will give the four values (No. 176).

+ab, and conse

+ Vab, -Vab, + Vab. √-1, Vab. √1.

To determine the true product, observe that

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tion x-1=0, that is, as the cube root of 1. (See No. 177.) From the formula

(a+b)3=a3 +3a2b+3ab2 +b3,

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(−1) 3 + 3(−1)2. √—3 +3(—1). ( √—3)2 + ( √ —3)

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§ IV. Theory of Exponents. Of Series.

179. In extracting the nth root of a quantity a", we have seen that when m is a multiple of n, we should divide the exponent m by the index of the root n; but when m is not divisible by n, in which case the root cannot be extracted algebraically, it has been agreed to indicate this operation by indicating the division of the two exponents.

n'

ገዜ

Hence vama, from a convention founded upon the rule for the exponents, in the extraction of the roots of monomials. Therefore, va2=a3; √a^=a3.

3

2

4

In like manner, suppose it is required to divide am by a". We know that the exponent of the divisor should be subtracted from the exponent of the dividend, when m>n, which gives . . .

am

an

am-n. But when m<n, in which case the division cannot be effected algebraically, it has been agreed to subtract the exponent of the divisor from that of the dividend. Let p be the absolute difference between n and m; then will n=m+p,

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Therefore the expression a is the symbol of a division which it has been impossible to perform; and its true value is

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