How do I prove that the function $\displaystyle f:\mathbb{Z}^{+}\times \mathbb{Z}^{+}\rightarrow \mathbb{Z}^{+}$ defined as $\displaystyle f\left ( m,n \right )=\binom{m+n-1}{2}+m$ is an onto function by mathematical induction? The bracket meant "n choose r".

I started off this way:

Basis Step: $\displaystyle m=1, n=1$

$\displaystyle f(1,1) = \binom{1}{2}+1 = 0+1=1$

So f(1,1) is true since f(1,1)=1, the element "1" in the codomain has its preimages m=1 and n=1.

Then assume true for $\displaystyle \exists m,n \in \mathbb{Z}^{+} \; such \; that \; f(m,n) = k$.

Inductive Step:

Then there exists a $\displaystyle m,n \in \mathbb{Z}^{+}$ such that $\displaystyle f(m,n) = k+1$

So $\displaystyle \binom{m+n-1}{2}+m = k+1$

$\displaystyle \frac{(m+n-1)!}{2!(m+n-1-2)!}+m=k+1$

$\displaystyle \frac{(m+n-1)!}{2!(m+n-3)!}+m=k+1$

But how do I continue to show that this will be true for all k+1 and that this is an onto function?

Are there any "easier tricks/process" when doing induction? I am trying to keep doing induction until I get the hang of it but I am always stuck after the basis step.

Doesn't feel like a friendly technique.

Thanks for any help.