How do I show that a mapping is bijective? How do I show that $$ g: \mathbb{N} \times \mathbb{N} \to \mathbb{N}, (m,n) \to 2^{m-1}(2n-1)$$
is bijective?
Any tips or help for me? I'm kind of stuck.
EDIT: I know what bijective means but my problem is I don't know HOW to show that.
Thanks in advance :)
 A: For any $k \in \mathbb{N}$, $k$ can be written as $k= 2^pq$ where $p$ is the degree of $2$ in the prime factorization of $k$ and thus $q$ is odd. So the given mapping is a surjective.
It's is injective because of the uniqueness of prime factorization: if $k=2^{m_1}(2n_1 -1)= 2^{m_2}(2n_2 -1)$, then both $m_1$ and $m_2$ are the degree of $2$ in the prime factorization of $k$, so they are equal, so will be $n_1$ and $n_2$
A: If you can show that this function is injective and bijective, you won. So lets start with injective. We have to show: If $g(x)=g(y)$ then $x=y$.
So $2^{a-1}(2b-1) = 2^{m-1}(2n-1)$ implies that $2^{a-1}=2^{m-1}$ (and therefore $a =m$) because you cant divide the right term by on each side by $2$. This also implies $b=n$, so the function is injective.
Now for surjectivity we have to show that for every $y$ there is a $x$ such that $y=g(x)$. That means we have to show that we can write every number in $\mathbb N$ as $2^{m-1}(2n-1)$.
Since the prime factorization is unique, we can divide each number in to two factors: The first is a power of $2$ and the second one does not contain a power of $2$. And that factor does not contain $2$ it is odd an can therefore be written as $(2n-1)$ uniquely. So therefore the function is surjective.
Now we have surjectivity and injectivity, and therefore bijectivity.
