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
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Prove by contradiction watch
- Thread Starter
- 12-04-2006 13:04
- 12-04-2006 13:18
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.Last edited by mikesgt2; 12-04-2006 at 13:19.