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    Can anyone help me do this exam question... I have never met these sort of proofs ever!

    Prove by contradiction that log23 is irrational.

    thanks in advance

    Suppose it is rational, then there exist integers m, n >= 1 with

    log_2(3) = m/n

    But then 3 = 2^(m/n) so 3^n = 2^m. This is impossible, as the LHS is odd while the RHS is even. We have deduced a contradiction, hence our original assumption must be false. Therefore log_2(3) is irrational.
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Updated: April 12, 2006
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