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Can this be proven?

  1. Jun 28, 2012 #1
    I haven't had much practice with proving things, so I'm not sure of exactly what would go into this, but can you prove that x is irrational in the following equation?

    x^x = 2
  2. jcsd
  3. Jun 28, 2012 #2


    Staff: Mentor

    start with the assumption that sqrt(2) is rational hence sqrt(2) = a / b where a and b are integers (definition of rational number) and that there are no common factors between a and b.

    from there we get a^2 / b^2 = 2 or a^2 = 2 * b^2 from which we conclude that a is even.

    since a is even then we can say a = 2 *c and we plug it in to conclude what about b?
  4. Jun 28, 2012 #3
    Well since there's no common factors between a and b I guess that means b would be odd. Meaning 2 would have to equal an even number divided by an odd number (since any even number squared is even and any odd number squared is odd (by specific examples)), which I guess is possible (30/15)

    Am I on the right track with this?

    Anyways, I appreciate the practice with proofs, but the number I was questioning about was not sqrt(2), it was x such that x^x = 2. (Not x such that x^2 = 2)
  5. Jun 28, 2012 #4


    Staff: Mentor

    its the same thing x^2 = 2 --> x=sqrt(2) the question is: is x or the sqrt(2) rational?

    to continue with the proof if you plug 2 *c in for a you can conclude that b is even too.

    Isn't that a contradiction?
  6. Jun 28, 2012 #5
    Again he is not asking about the square root of two.

    He is asking about x^x = 2

    which x = 1.5596104....
  7. Jun 28, 2012 #6


    Staff: Mentor

    got it. I misread x ^ x to be x * x = 2. okay.

    So going back the sqrt(2) proof... I think a proof by contradiction would still apply.

    assume x = a/b with a and b having no common factors

    a^a = 2^b * b^a from which we conclude a is even

    using similar logic and some added caveats about a, b and c we should be able to conclude that b is even and that x is then irrational.
  8. Jun 28, 2012 #7
    I cheated and checked this out in Wolfram Alpha. It's intimately related to the Lambert W function. I'm pretty sure the question of when x^x is rational is pretty difficult.

    This link is a spoiler.


    If you go to the "solutions" panel and click on Exact forms, you'll get a closed form expression for the solutions. I don't think it's even possible to tell if they're rational. So even with looking up the exact solution to the equation, it's still difficult to answer the question of whether it's rational.

    This link on the Lambert function is worth looking at. It's basically 18th- and 19th-century math, along with some contemporary computer visualizations. Imagine these old mathematicians like Lambert and Euler doing all this work without the benefit of any computational assistance. No calculators, computers, not even mechanical adding machines. But somehow they visualized the equations underlyling these pictures.

  9. Jun 28, 2012 #8


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    Not strictly relevant to the OP's question, but once you prove that x is irrational, you can immediately prove it's transcendental as well.

    Assume to the contrary that x is algebraic. By Gelfond-Schneider, x^x has to be transcendental. But x^x = 2, by definition. This is a contradiction, hence x is not algebraic, and therefore transcendental.
  10. Jun 29, 2012 #9
    How did you calculate this?

    From [itex]a^a = 2^b b^a[/itex] can we really conclude that [itex]a[/itex] is even?

    What if [itex]b[/itex] is a negative integer?
    Is [itex]a[/itex] necessarily even if [itex]a^a[/itex] is even?

    I'll put more thought into this, just wanted to post to keep the topic alive for the time being

    Thanks :)
  11. Jun 29, 2012 #10
    It's pretty easy, actually. Suppose that x is rational and x^x = 2. Then x = 2^(1/x) is a rational power of 2 and therefore an algebraic integer. But x is rational, so x must actually be an ordinary integer, and it's easy to see that no integer can satisfy that equation.
  12. Jun 29, 2012 #11


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    But it's not hard from first principles either. The solution must have x > 1, so we can assume x = a/b where integers a and b are positive and coprime.
    As jedishfru observed, aa = 2b * ba, from which we conclude a is even and b is odd. Suppose a = 2kc, c odd.
    (2kc)(2kc) = 2b * b(2kc)
    Comparing powers of 2:
    k2kc = b
    Since k >= 1, b is even.
  13. Jun 29, 2012 #12
    I cheated. I also use Wolfram Alpha.
  14. Jun 29, 2012 #13


    Staff: Mentor

    Just to be clear my post was designed to hint at the solution not provide a complete answer due to forum rules.
  15. Jun 29, 2012 #14
    I see how x is a rational power of 2, but I don't see why that means that it is an algebraic integer (Considering polynomials are defined as only having integer exponents). I also don't see why if it is an algebraic integer, that means that it is an ordinary integer. I can think of many rational powers of two that are rational, but not integers :S

    I thought that rule was just for homework help? I wouldn't worry about giving me the definite answer anyhow, I'm just as concerned about the proof itself.

    That's what I figured, just wanted to know for sure haha
    Last edited: Jun 29, 2012
  16. Jun 29, 2012 #15
    2^(p/q) is a root of the monic polynomial X^q - 2^p.

    It is a well-known theorem that every algebraic integer in the rational numbers is an integer. The proof is as follows: Suppose that x is a rational algebraic integer. By the definition of algebraic integer, we can find a monic polynomial f(X) with integer coefficients such that f(x) = 0. Let f be such a polynomial of minimal degree. If f(X) factors as g(X)h(X) in the polynomial ring Z[X], then the product of the leading coefficients of g and h must be 1, and therefore the leading coefficients themselves are either both 1 or both -1. Replacing g(X) and h(X) with -g(X) and -h(X) if necessary, we may assume that both g and h are monic. Since the factorization is proper, neither g(X) nor h(X) is equal to 1, so both of them have positive degree, and hence both of them have degree strictly less than the degree of f. But since g(x)h(x) = f(x) = 0, we have either g(x)=0 or h(x)=0, in either case contradicting the minimality of f. Therefore f(X) is irreducible in the polynomial ring Z[X]. By [url="http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29]Gauss's lemma[/url], f must be irreducible in Q[X] as well, and so is the minimal polynomial of x. But since x is rational, the minimal polynomial of x is X - x, so X - x has integer coefficients and thus x is an integer.

    And none of them are algebraic integers.
  17. Jun 30, 2012 #16


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    This can be proven in a much more elementary fashion using the Rational Root Theorem.
  18. Jun 30, 2012 #17
    Ah, so it can. Thanks for mentioning that, I always thought such a basic fact should have a more elementary proof. :smile:
  19. Jul 4, 2012 #18
    A very quick method to prove that 2 is irrational : use RRT (Rational Root Theorem) (see http://en.wikipedia.org/wiki/Rational_root_theorem) and apply to p(x) = (x-n)^2 +1 with n=2 (your number).

    The same method can be used to prove that the square root of n^2+1 is irrational for n > 0.
    Last edited: Jul 4, 2012
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