How do you prove $log_2 3$ is irrational number ?

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Answer 1 · 2017-10-27T10:16:34

See below.

$y = log_2 3 hArr 2^y = 3$ If $y$ is rational then $y = n/m$ with ${n,m} in ZZ$ and $m ne 0$ or

$2^(n/m) = 3 rArr 2^n = 3^m$

This last equality is an absurd because $3^m$ is odd and $2^n$ is even.

Concluding $y = log_2 3$ is not rational, being then irrational.

How do you prove log_2 3log_2 3 is irrational number ?