One-one and onto map from $\mathbb{N}\times \mathbb{N} \rightarrow \mathbb{N} $. Can you tell any one-one & onto map from $\mathbb{N}\times \mathbb{N}$ to $\mathbb{N}$?
I can prove that these have same cardinality but am unable to think of a mapping.
 A: How about: $$h:\mathbb N\times\mathbb N\rightarrow \mathbb N,~(m,n)\mapsto 2^{m-1}(2n-1)$$
A: A mapping is not obvious! Define $\phi(k)=1+2+ \cdots k=\frac{k(k+1)}{2}$. $h(m,n)=\phi(m+n-2) +m$ is a bijection. It's on page $358$ of Bartle and Sherbert if you have it.
A: Here's my answer. It is 
not original
 , of course (thanks to Noah Schweber for providing the link) -- I'll just be walking through one angle on the reasoning behind it. It's similar to the one that mich95 posted, but I was already partway through this when (s)he posted so I thought I might as well finish what I'd started!
There is one element $(m,n) \in \mathbb{N} \times \mathbb{N}$ such that $m+n = 0$ (namely, $(0,0)$). Map this to $0$.
There are two elements $(m,n) \in \mathbb{N} \times \mathbb{N}$ such that $m+n = 1$ (namely, $(0,1)$ and $(1,0)$). Map these to $1$ and $2$.
(I'm sure you get the idea by now but let's just do one more.)
There are three elements $(m,n) \in \mathbb{N} \times \mathbb{N}$ such that $m+n = 2$ (namely, $(0,2)$, $(1,1)$, and $(2,0)$). Map these to $3,4,5$.
In general, the elements $(m,n) \in \mathbb{N} \times \mathbb{N}$ such that $m+n = k$ are $(0,k), (1,k-1), \cdots, (k,0)$. These will map to $\frac{k(k+1)}{2}, \frac{k(k+1)}{2} +1, \cdots, \frac{k(k+1)}{2} + k = \frac{(k+1)(k+2)}{2}-1$ under this scheme (easily verified by induction).
Let $\phi: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ be the mapping described above. Then for $(m,n) \in \mathbb{N}$, we have
$$\phi(m,n) = \frac{(m+n)(m+n+1)}{2} + m$$ 
