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Thread: Proving (URGENT)

  1. #1
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    Proving (URGENT)

    Help! Please!

    Prove that if n is an element of Z(the integers) and log2n is rational, then log2n is an integer.
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  2. #2
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    Consider this:
    $\displaystyle log_2 (n)^k = k log_2 (n)$

    When n = 2, $\displaystyle log_2 (2) = 1$

    If n is not a power of 2, then
    $\displaystyle log_2 n = \frac{ln (n)}{ln (2)} $

    is ln 2 rational?
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  3. #3
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    Quote Originally Posted by vincentngtf View Post
    Help! Please!

    Prove that if n is an element of Z(the integers) and log2n is rational, then log2n is an integer.
    (We can assume that n>1).
    Suppose that $\displaystyle \log_2 n = p/q$ where $\displaystyle p/q$ is a positive rational number. This means $\displaystyle 2^{p/q} = n\implies \left( 2^{p/q} \right)^q = n^q\implies 2^p = n^q$. Now the fundamental theorem of arithmetic allows us to factorize $\displaystyle n$, the thing is that all factors much be $\displaystyle 2$'s because the LHS is made out of two, thus, $\displaystyle n=2^m$. This means, $\displaystyle 2^p = 2^{mq}$. Thus, $\displaystyle p=mq$ which means $\displaystyle q|p$ which tells us that $\displaystyle p/q$ is an integer.
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