Show that a multivariate function is bijective

[Mr Davis 97]

Homework Statement

Show that $f: \mathbb{Z}^{+} \times \mathbb{Z}^{+} \longrightarrow \mathbb{Z}^{+}$where $\displaystyle f(m,n) = \frac{(m+n-2)(m+n-1)}{2}+m$ is bijective

Homework Equations

The Attempt at a Solution

First, we show that $f(a,b) = f(c,d) \implies a=c \land b=d$.

$f(a,b) = f(c,d)$

$\displaystyle \frac{(a+b-2)(a+b-1)}{2}+a = \frac{(c+d-2)(c+d-1)}{2}+c$

After simplification, we find that

$(a+b)^2 + a - b = (c+d)^2 + c - d$

Comparing sides, then $a + b = c+d, ~a-b = c-d$
Then $a=c,~b=d$

Second, we show that for all positive integers a, there exist an m and an n such that f(m,n) = a.
We see that $f(m,2-m) = m$, so the function maps to every element in the codomain

Does this show that the function is bijective?

 

[Delta2]

For the second part 2-m might not necessarily be a positive integer which seems to be a problem since the function is defined for pairs of positive integers.

For first part this conclusion

Comparing sides, then a+b=c+d, a−b=c−da+b=c+d, a−b=c−da + b = c+d, ~a-b = c-d

doesn't necessarily follow from the previous line. It might be correct but needs some additional justification me thinks.