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    (Original post by nuodai)
    Proof by contradiction is a method of proof whereby you assume the conclusion is false, and then show this assumption leads to something which can't be true (e.g. 1=0 or "2 is odd").

    A number is rational if it is in the form \dfrac{p}{q}, where p,q are integers (q \ne 0).

    Piecing this together, we want to show that \log_2 3 is irrational; i.e. that it can't be written in the form \dfrac{p}{q} for any integers p,q. So, we start our proof by assuming that there exist integers p,q (q nonzero) such that \log_2 3 = \dfrac{p}{q}.

    By the definition of logarithms, this gives 3 = 2^{p/q}, and raising both sides to the power of q gives 3^q=2^p. This can only happen if p=q=0, but we can't have q=0 so this can't be true, so the assumption can't be true, so it must be false; hence the proposition is true.
    Why is it that only q can not equal 0? Shouldn't p and q can not equal 0?
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    (Original post by Punit21)
    Why is it that only q can not equal 0? Shouldn't p and q can not equal 0?
    q is in the denominator of the fraction, but p isn't, so the restriction applies only to q
 
 
 
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Updated: January 24, 2013
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