Is there a bijection between $\mathbb N$ and $\mathbb N^2$? 
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,

Yes. Imagine starting at $(1,1)$ and then zig-zagging diagonally across the quadrant. I'll leave you to formulate it.
Hint: for every natural number $>1$ there's a set of elements of $\mathbb{N}^2$ that add up to that number. For $2$, there's $(1,1)$. For $3$, there's $(2,1)$ and $(1,2)$...
