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Math Help - Rational number proof

  1. #1
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    Rational number proof

    I'm supposed to prove that log(m)/log(n) is rational if and only if there is some integer k such that m and n (which are integers) are powers of k.

    It's an "if and only if" proof, so there are two parts:

    a) Show that the existence of said integer k implies that log(m)/log(n) is rational. I did this, but I don't understand how to do the second part:

    b) Show that log(m)/log(n) being rational implies that there is some integer k such that m and n are powers of k.

    If log(m)/log(n) is rational, it can be represented as a/b where a and b are integers and b is not 0. log(m)/log(n) = a/b implies that m^b = n^a.
    But I don't know where to go from there.
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  2. #2
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    Re: Rational number proof

    Quote Originally Posted by euphony View Post
    I'm supposed to prove that log(m)/log(n) is rational if and only if there is some integer k such that m and n (which are integers) are powers of k.

    It's an "if and only if" proof, so there are two parts:

    a) Show that the existence of said integer k implies that log(m)/log(n) is rational. I did this, but I don't understand how to do the second part:

    b) Show that log(m)/log(n) being rational implies that there is some integer k such that m and n are powers of k.

    If log(m)/log(n) is rational, it can be represented as a/b where a and b are integers and b is not 0.
    And, a/b is reduced to lowest terms. That is, a and b have no factors in common.

    log(m)/log(n) = a/b implies that m^b = n^a.
    But I don't know where to go from there.
    m^b= n^a is the same as saying m= n^(a/b). Since m is an integer, so is n^(a/b) and since a and b have no common factors, so is n^(1/b). Let k= n^(1/b).
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  3. #3
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    Re: Rational number proof

    and since a and b have no common factors, so is n^(1/b)
    I don't quite understand this.
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  4. #4
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    Re: Rational number proof

    Quote Originally Posted by euphony View Post
    I don't quite understand this.
    You you understand why m=n^{\frac{a}{b}}~?

    If so, m is an integer, therefore n^{\frac{a}{b}} is also an integer.

    But because \frac{a}{b}} is reduced form that means n^{\frac{1}{b}} must an integer.

    Let k=n^{\frac{1}{b}} so m=k^a~\&~n=k^b.
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  5. #5
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    Re: Rational number proof

    I don't understand why a and b having no common factors implies that n^(1/b) has to be an integer.
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  6. #6
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    Re: Rational number proof

    Quote Originally Posted by euphony View Post
    I don't understand why a and b having no common factors implies that n^(1/b) has to be an integer.
    Oh come on. If you are going to work at this level then think at this level.

    It is a matter of standard practice. If \rho is a rational number then \exists\{a,b\}\subset\mathbb{Z} such that \text{GCD}(a,b)=1 and \rho=\frac{a}{b}.
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  7. #7
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    Re: Rational number proof

    Hi,
    Maybe what you're missing is this fact: if b and c are integers greater than 1, then the bth root of c is integral iff for any prime p with pk exactly dividing c, b divides k. From this it immediately follows that if a and b are relatively prime with na/b an integer, then n1/b is an integer.

    By the way, your original problem statement should have included the statement that m > 1, otherwise of course it's false. I think you should foster the habit of specifying the exact hypotheses of a statement.
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