TITLE: Is there a bijection between $\mathbb N$ and $\mathbb N^2$?
QUESTION [11 upvotes]: Is there a bijection between $\mathbb N$ and $\mathbb N^2$?

If I can show $\mathbb N^2$ is equipotent to $\mathbb N$, I can show that $\mathbb Q$ is countable. Please help. Thanks,

REPLY [13 votes]: Here is a non constructive proof.
Of course, there is an injection of $\mathbb{N}$ into $\mathbb{N}\times\mathbb{N}$.
Now consider the map
$$
(x,y)\longmapsto 2^x\cdot 3^y
$$
By the fundamental theorem of arithmetic, this is an injection of $\mathbb{N}\times\mathbb{N}$ into $\mathbb{N}$.
By Cantor-Bernstein-Schroeder, there exists a bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$.
Edit: I added this answer because the only constructive proof I knew was the zigzag along the diagonals, which I find slightly tedious to write down. But thanks to Andre Nicolas, now I know a straightforward and easy-to-write-down bijection. Which is actually a little step away from the injection I use above.

REPLY [3 votes]: Yes there exists a quadratic bijection. 
See this