Proof uniqueness of the nth root

Hi. I'm not sure if I put it in the correct section. I hope I did.

I was wondering if there is a relatively short proof of the following:

Which means that each number x has only one nth root. There is a proof in Walter Rudin's book but it is really long, so this is why I'm asking , if you know any simpler ones.

Re: Proof uniqueness of the nth root

this is the only proof i know:

lemma 1: every polynomial f(x) in F[x], for a field F has at most n roots.

proof: fields are euclidean domains with d function deg(f). the result follows by induction on the degree of a polynomial and the division algorithm. (note: a purely analytic argument using the continuity and differentiability of polynomials and rolle's theorem and the intermediate value theorem is also possible).

lemma 2: the polynomial x^{n} - 1 has n roots in the field of complex numbers

proof: by demoivre's theorem, the complex numbers z_{k} = cos(2πk/n) + i sin(2πk/n) for k = 0,1,2,...,n-1 are all roots of x^{n} - 1. by lemma 1, these are all the roots of x^{n} - 1.

lemma 3: if a > 0 is a real number, the polynomial x^{n} - a has at least one real root.

proof: since x^{n} - a is a real polynomial, it's complex roots occur in conjugate pairs: that is if z^{n} = a, then (z^{n})* = a* = a so z and z* are both n-th (complex) roots of a.

if n is odd, we must have z = z* for some n-the root z of a, which is therefore a real root of x^{n} - a.

on the other hand, if n is even, then for some positive integer N, we have N^{2} - a > 0, so that x^{n} - a is positive on (-∞,N) and (N,∞), since 0^{2} - a = -a < 0, by the intermediate value theorem, x^{n} - a must have a root in (N,0) and another root in (0,N).

(alternatively: if you are willing to accept the theorem as true for n = 2, then if n is even let y = x^{n/2}, and let √a be the square root (the unique positive one that exists in the case of n = 2, by assumption), so we have:

x^{n} - a = (y - √a)(y + √a). thus x^{n} -a has a solution if x^{n/2} - √a has a solution, if n/2 is odd, we are done. if not, repeat. if n is a power of 2, eventually we will reach the case where we have x^{2} - b

(where b is some square root of a square root of a..(depending on how many times we have to do this)..square root of a)). if n has any other factor, eventually we'll factor out all the powers of 2, and terminate with an odd case).

lemma 4: let u be a real root of x^{n} - a, and let w = cos(2π/n) + i sin(2π/n). then {u,uw,uw^{2},...,uw^{n-1}} are the n roots of x^{n} - a.

proof: follows by lemma 1, lemma 2 and lemma 3 (lemma 3 for the existence of u, lemma 1 to show these are all the possible roots, lemma 2 to show these actually are roots).

now all we have to do is show we can take u > 0, and that there is always exactly one such u.

suppose uw^{k} is a real root of x^{n} - 1. then:

(uw^{k})* = uw^{k} that is:

u*(w^{k})* = uw^{k}, hence:

(w^{k})* = w^{k}, so w^{k} is real. the numbers w^{k} all lie on the unit circle |z| = 1, which intersects the real line in just two places: 1 (= (1,0) = 1+i0) and -1 (= (-1,0) = -1+i0).

thus 2πk/n = π or 2πk/n = 0. in the first case, we have k = n/2, and since the k's are integers, this can only happen when n is even. in this case we get two roots: u and uw^{n/2} = -u.

exactly one of these is positive. if n is odd, then there is no solution to 2πk/n = π for any integer k = 0,1,2,...,n, so the only real root of x^{n} - a is u = uw^{0}.

if u < 0, then since n is odd, n = 2t +1, so u^{n} = (u^{2t})(u) = (u^{t})^{2}(u) < 0, since (u^{t})^{2} > 0.

hence u^{n} - a < 0, contradicting that u is a root of x^{n} - a. hence u > 0.

(i've left some of the details out but this is the general idea).

Re: Proof uniqueness of the nth root

Think about elementary calculus. Given a positive, there is exactly one x > 0 with x^n = a: (if a = 0, it's "easy")

First, since the function x^n "clearly" approaches infinity as x approaches infinity, the intermediate value theorem guarantees that there is at least one x > 0 with x^n = a -- 0^n < a < b^n for some b.

Next the function x^n is increasing for x > 0 since the derivative is positive. So the function is one to one; i.e., if x_1^n = x_2^n, then x_1 = x_2.