Prove that $|\mathbb{N} \times \mathbb{N}| \leq |\mathbb{N}|$ Claim: $|\mathbb{N} \times \mathbb{N}| \leq |\mathbb{N}|$
Proof:
For all  $ x \in \mathbb{N}, y \in \mathbb{N}$, we can define a function $f(x,y) = x + \frac{(x+y-1)(x+y-2)}{2}$ 
We now prove that $f$ is an injection. We will show that if $f(x,y) = f(u,v)$ then $x+y$ = $u+v$ 
Suppose $f(x,y) = f(u,v)$ but $x+y \neq u+v$. Then either $x+y > u+v$ or $u+v > x+y$. 
Case 1: $x+y > u+v$
Then $x+y = u+v + \delta$ for some $\delta > 0$.
$$x + \frac{(x+y -1)(x+y-2)}{2} = u + \frac{(u+v -1)(u+v-2)}{2} + \delta$$
Now, I don't know what to do. 

Here's a way of proving that result that's easier to understand:
$$
\begin{array}{cccccccccc}
1,1 & & 1,2 & \rightarrow & 1,3 & & 1,4 & \rightarrow & 1,5 & & 1,6 & \rightarrow \\
\downarrow & \nearrow & & \swarrow & & \nearrow & & \swarrow & & \nearrow \\
2,1 & & 1,1, & & 2,3 & & 2,4 & & 2,5 & & \cdots \\
& \swarrow & & \nearrow & & \swarrow & & \nearrow \\
3,1 & & 3,2 & & 3,3 & & 3,4 & & 3,5 & & \cdots \\
\downarrow & \nearrow & & \swarrow & & \nearrow \\
4,1 & & 4,2 & & 4,3 & & 4,4, & & 4,5 & & \cdots \\
& \swarrow & & \nearrow \\
5,1 & & 5,2 & & 5,3 & & 5,4 & & 5,5 & & \cdots \\
\downarrow & \nearrow \\
6,1 & & \vdots & & \vdots & & \vdots & & \vdots
\end{array}
$$
There's a first member of $\mathbb N\times\mathbb N$, then a second, then a third, and so on.  No matter which member of $\mathbb N\times\mathbb N$ you pick, I know, without knowing which one you picked, that after some finite number of steps, I will reach it by following the arrows.  
