TITLE: prove there is no rational r satisfying $2^r=3$
QUESTION [7 upvotes]: I first assumed that there exists a rational $r=\frac{a}{b}$ such that $2^r=3$.
..and I can't make a progress after this.
can anyone help me out?

REPLY [4 votes]: More generally,
suppose that
$a^n = b^m$
where all variables
are positive integers
and there is a prime $p$
such that
$p$ divides $a$
and
$p$ dies not divide $b$.
Then
$p$ divides $a^n$
and $p$ does not divide $b^m$,
which is a contradiction.
Therefore,
if $a^n = b^m$,
then every prime that divides $a$
must divide $b$
and vice versa.
More general results can be stated
(by looking at the complete 
prime factorizations),
but this is a reasonable generalization
of your problem.