proving a function is a bijective Is $f(x)=2^x$ a bijection from $f:\mathbb{Q} \rightarrow \mathbb{Q}^+$? If my understanding of how codomains work, this would only include all $x$ values that have a positive $y$. So...--prophetes.ai

proving a function is a bijective Is $f(x)=2^x$ a bijection from $f:\mathbb{Q} \rightarrow \mathbb{Q}^+$? If my understanding of how codomains work, this would only include all $x$ values that have a positive $y$. So would this then be bijective since the area that would make this not surjective not be accounted for?

I think in terms of being surjective, the kicker here is the choice of $\mathbb{Q}$: Is every positive fraction the result of raising 2 to some rational power? Or, is $x=\log_2(y)$ always rational?