Proving that the given function $F:\mathbb N\times \mathbb N\to\mathbb N$ is bijective Consider the function $F:\mathbb N\times\mathbb  N\to \mathbb N$ defined by 
$$F(a,b)=\frac{(a+b-2)(a+b-1)}{2} +a$$ How can I prove that it is a bijective function? I proved it using Partial derivatives by assuming the function to be continuous and defined on $\mathbb{R}$. But how can I prove it using definition of one-one and onto functions?
 A: Here assuming $\mathbb{N}=\{1,2,...\}$.
Notice that $G=2F(a,b)$ always falls between $(a+b-2)^2$ and $(a+b)^2$. In fact $G$ is always greater than the number halfway between $(a+b-2)^2$ and $(a+b-1)^2$ and less than the number halfway between $(a+b-1)^2$ and $(a+b)^2$. This determines $(a+b-1)^2$, hence $a+b$. Furthermore, $G$ is greater than $(a+b-1)^2$ if and only if $a>b$. The exact value of $G$ then determines $a$ and $b$ specifically. This shows it's injective.
To prove it's onto just notice that there are exactly $a+b-1$ many even numbers between those halfway points, i.e. the number of possibilities for $(a,b)$ for $a+b$ fixed.
A: Here is an argument for injectivity.
Consider the function $f:\mathbb{Z}^{\geq 2}\rightarrow \mathbb{Z}^{\geq 0}$ defined by $f(u) = (u-2)(u-1)/2$. Note that
$$\begin{align}
2(f(u+1) - f(u)) &= (u-1)u - (u-2)(u-1) \\
&= u^2 - u - u^2 + 3u - 2 \\
&= 2u - 2
\end{align}$$
so $f(u+1) - f(u) = u-1$. Therefore $f$ is an increasing function, and in general, if $v > u$ we have
$$f(v) - f(u) \geq u-1$$
Now suppose that $F(a,b) = F(c,d)$. Consider two cases.
Case 1 $a+b = c+d$
Then $0 = F(a,b) - F(c,d) = a-c$, so $a = c$, and therefore $b = d$.
Case 2 without loss of generality, assume $a+b > c+d$
Then
$$\begin{align}
0 &= F(a,b) - F(c,d) \\
&= f(a+b) + a - f(c+d) - c \\
&\geq (c+d-1) + a-c \\
&= a+d - 1
\end{align}$$
so $a + d \leq 1$. But this is impossible since both $a$ and $d$ are $\geq 1$.
