Could someone point me in the right direction for this proof? I need to prove that Q (rational numbers) is countable by applying the function $f(m/n) = 2^m3^n$ with $m,n$ being relatively prime numbers.
I honestly have no idea where to start.  Any pointers would be great!

Injectivity
$f\left(\dfrac{m_1}{n_1}\right)=f\left(\dfrac{m_2}{n_2}\right) \implies 2^{m_1}3^{n_1}=2^{m_2}3^{n_2}$
Case 1 $m_1\geq m_2, n_1\leq n_2$
$\therefore 2^{m_1}3^{n_1}=2^{m_2}3^{n_2} \implies 2^{m_1-m_2}=3^{n_2-n_1} \implies m_1=m_2, \ n_1=n_2$
Case 2 $m_1\geq m_2, n_1\geq n_2$
$\therefore 2^{m_1}3^{n_1}=2^{m_2}3^{n_2} \implies \left(2^{m_1-m_2}\right)\left(3^{n_1-n_2}\right)=1 \implies m_1=m_2, \ n_1=n_2$
Case 3 $m_1\leq m_2, n_1\geq n_2$
Similar analysis as Case 1.
Case 4 $m_1\leq m_2, n_1\leq n_2$
Similar analysis as Case 2.
$\\$
Surjectivity
Obvious from the function.
$\therefore f$ is bijective.
