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  1. #1
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    log proof

    prove log base 2 of 3 is irrational
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by teramaries View Post
    prove log base 2 of 3 is irrational
    $\displaystyle \log_23=\frac ab \implies 2^{a/b}=3\implies 2^a=3^b $

    Now look at the prime factorization of both sides and see this forces $\displaystyle a=b=0 $ which is absurd in our case.
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  3. #3
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    Quote Originally Posted by teramaries View Post
    prove log base 2 of 3 is irrational
    Use proof by contradiction.

    Assume $\displaystyle \log_2 3 = \frac{p}{q}$ where p and q are co-prime positive whole numbers

    $\displaystyle \Rightarrow 3^q = 2^p$.

    Now consider the last digits of $\displaystyle 3^q$ and $\displaystyle 2^p$ to get the required contradiction.
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    A slight variation of previous proofs . . .


    Assume $\displaystyle \log_23 = \frac{p}{q}$ where $\displaystyle p$ and $\displaystyle q$ are coprime positive integers.

    Then: .$\displaystyle 2^{\frac{p}{q}} \:=\:3 \quad\Rightarrow\quad 2^p \:=\:3^q$


    Since $\displaystyle p$ is a positive integer, $\displaystyle 2^p$ is even.

    Since $\displaystyle q$ is a positive integer, $\displaystyle 3^q$ is odd.

    Hence: .$\displaystyle 2^p \:\ne\:3^q$

    . . Q.E.D.
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