I know this seems like an easy proof, but I cant seem to directly prove it even though I know it's true. I need to prove that $2^a$ is even, and $3^b$ is odd, so I'm trying to prove that $2^a|2$, meaning that it is even, and $3^b|3$, meaning that it is odd.
I have that $2^a|2$ means that $2^a=m2$ for some integer $m$. I know that the definition of any even number $n$ is $n=2k$ for some integer $k$.
I have that $3^b|3$ means that $3^b=m3$ for some integer $m$. I know that the definition of any odd number $n$ is $n=2k+1$ for some integer $k$.
I am in desperate need of some direction, and I'm trying to be very precise and only use definitions as opposed to just stating that $2^a$ is even because OF COURSE it's even. Thank you!
Addition: So now i can see how I can prove $2^a$ is even. How can I prove $3^b$ is odd then? Although I can now prove that it is a multiple of $3$, this won't help to show that it is odd I see now. How could I prove that $3^b$ is odd?
$2^a = 2^{a-1} \color{red}{\times 2}$ which is manifestly even because it's explicitly $\color{red}{\text{a multiple of $2$}}$. (This holds unless $a=0$, of course, but then $2^a$ isn't even.)
To show that $3^b$ is odd, consider $$3^b = (2+1)^b = 2^b + \binom{b}{1} 2^{b-1} 1^1 + \binom{b}{2} 2^{b-2} 1^2 + \dots + \binom{b}{b-1} 2^1 1^{b-1} + \binom{b}{b} 1$$ which is $$2^b + \binom{b}{1} 2^{b-1} + \binom{b}{2} 2^{b-2} + \dots + \binom{b}{b-1} 2^1 + 1$$ which is $$2 \times \left[2^{b-1} + \binom{b}{1} 2^{b-2} + \binom{b}{2} 2^{b-3} + \dots + \binom{b}{b-1}\right] + 1$$ which is one more than a multiple of $2$.