230. The rules just demonstrated for the calculus of radicals, 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. This, however, has been proved on the supposition that, when the powers of the same degree of two expressions are equal, the expres sions themselves are also equal. Now, this last proposition, which is true for absolute numbers, is not always true for algebraic expressions; for it is easily shown that the same number can have more than one square root, cube root, fourth root, &c.

Let us denote the algebraic value of the square root of a by x, and the arithmetical value of it by p; we have the equations x = ± p.

x2 = a, and x2= p2, whence

Hence we see, that the square of p, (which is the root of a), wiil give a, whether its sign be + or

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

x3 = a, and x3 = p3.

The last 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 x3 - p3 is divisible by — p (Art. 61), which gives the exact quotient, x2 + px + p2, the above equation can be transformed into

(x − p) (x2 + px + p2) = 0.

Now, every value of x which will satisfy this equation, will satisfy the first equation. But this equation can be satisfied by supposing

Hence, the cube root of a, admits of three different algebraic

x4 = p1,

in which p denotes the arithmetical value of √a. This equation can be put under the form

x2 + p2 = 0, whence x = ± √√ − p2 = ± p√ - 1 We therefore obtain four different algebraic expressions for the fourth root of a.

As another example, resolve the equation

which may be satisfied by making either of the factors equal to

zero.

But

3 p30, gives

x = p, and x = p

G

And if in the equation x3 + p3 = 0, we make pp', it becomes x3 — p3 = 0, from which we deduce

Therefore, the value of x in the equation x6 p60, and con

sequently, the 6th root of a, admits of six values. If we make

We may then conclude from analogy, that in every equation of the form am — a = 0, or xm — pm = 0, x is susceptible of m different values; that is, the mth root of a number admits of m different algebraic values.

231. If, in the preceding equations, and the results corresponding to them, we suppose, as a particular case, a = 1, whence p = 1, we shall obtain the second, third, fourth, &c. roots of unity. Thus + 1 and - 1 are the two square roots of unity, because the equation 2-10, gives x = ± 1.

are the three cube roots of unity, or the roots of x3 — 1 = 0; and

are the four fourth roots of unity, or the roots of x1 — 1 = 0.

232. 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 modifica tions are more particularly necessary when applied to imaginary expressions, and are a consequence of what has been said in

Now, √2 is equal to ± a (Art. 139); 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

√=ax √=a=(√ — a)2 = — a.

Again, let it be required to form the product

- a × √-b.

By the rule of Art. 228, we shall have

√-ax√-b= √+ab.

abp (Art. 230), p being the arithmetical value

Now,

of the square root of ab; but the true result is

so long as both the radicals √ -a and with the sign +.

α.

For, √a√a.√-1; and √-6=√6.√1, hence,

√-a × √ = b = √a. √ = 1× √ b× √√−1 = √ ab (√ − 1)2

By similar methods we find the different powers of √1 to be as follows:

ཇ.

:

=1× √=1=(√ — 1)2 = — 1.

3.

√ − 1)2 = (√√ — 1)2 . (√√ − 1)2 = − 1 × −1 = +1.

. —

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

-

by the which, from the rule, would be +ab, and consequently, would give the four values (Art. 231),

+√ab, √ab, +√ab. √-1, Vab.√1.

But, √ − 1 × √ √ − 1 = (√ — 1)2 = (√ √ √ − 1)2 = √ −1;

We will apply the preceding calculus to the verification of the expression

considered as a root of the equation x3 − 1 = 0; that is, as one of the cube roots of 1 (Art. 231).

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

.

-, may be verified in the same

manner. It should be remarked, that either of the imaginary roots is the square of the other; a fact which may be easily verified.

Theory of Exponents.

233. In extracting the nth root of a quantity am, we have seen that when m is a multiple of n, we should divide the exponent m by n the index of the root. When m is not divisible by n, the operation of extracting the root is indicated by indicating the division of the two exponents. Thus,

a notation founded on the rule for the exponents, in the extraction of the roots of monomials. In such expressions, the numerator indicates the power to which the quantity is to be raised, and the denominator, the root to be extracted.

3

Therefore, Vaa; and a = a3.

If it is required to divide am by a", in which m and n are positive whole numbers, we know that the exponent of the divisor should be subtracted from the exponent of the dividend, and we have