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Thread: Proof uniqueness of the nth root

  1. #1
    Junior Member
    Nov 2012

    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:

    x \in R, x \ge 0, n \in N \setminus \left\{ 0\right\}  \Rightarrow \exists ! y \in R: y \ge 0 \wedge y^n=x

    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.
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  2. #2
    MHF Contributor

    Mar 2011

    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 xn - 1 has n roots in the field of complex numbers

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

    lemma 3: if a > 0 is a real number, the polynomial xn - a has at least one real root.

    proof: since xn - a is a real polynomial, it's complex roots occur in conjugate pairs: that is if zn = a, then (zn)* = 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 xn - a.

    on the other hand, if n is even, then for some positive integer N, we have N2 - a > 0, so that xn - a is positive on (-∞,N) and (N,∞), since 02 - a = -a < 0, by the intermediate value theorem, xn - 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 = xn/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:

    xn - a = (y - √a)(y + √a). thus xn -a has a solution if xn/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 x2 - 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 xn - a, and let w = cos(2π/n) + i sin(2π/n). then {u,uw,uw2,...,uwn-1} are the n roots of xn - 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 uwk is a real root of xn - 1. then:

    (uwk)* = uwk that is:

    u*(wk)* = uwk, hence:

    (wk)* = wk, so wk is real. the numbers wk 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 uwn/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 xn - a is u = uw0.

    if u < 0, then since n is odd, n = 2t +1, so un = (u2t)(u) = (ut)2(u) < 0, since (ut)2 > 0.

    hence un - a < 0, contradicting that u is a root of xn - a. hence u > 0.

    (i've left some of the details out but this is the general idea).
    Thanks from wilhelm
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  3. #3
    MHF Contributor
    Dec 2012
    Athens, OH, USA

    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.
    Thanks from wilhelm
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