How to prove that this function is a surjection?

[Anezka]

If f were a function of 1 variable only, then this would be straight forward as I can try to find its inverse by reversing the operations defined in f. I know I need to show that for any given positive integer,p, there exists two positive integers, m and n such that 1/2(m+n−2)(m+n−1)+n=p. However, I am not so sure what to do in the case of two variables. Do I have to show that if I assume that if m or n are some fixed positive integer, then there is a solution for p?

 

[mfb]

Did you draw a sketch? That should make clear how the function "works". With that you can also find explicit n,m for every p if you want.

 

[PeroK]

If f were a function of 1 variable only, then this would be straight forward as I can try to find its inverse by reversing the operations defined in f. I know I need to show that for any given positive integer,p, there exists two positive integers, m and n such that 1/2(m+n−2)(m+n−1)+n=p. However, I am not so sure what to do in the case of two variables. Do I have to show that if I assume that if m or n are some fixed positive integer, then there is a solution for p?

View attachment 224576

Did anyone ever mention that your $n$ looks very similar to your $m$?

Anyway, the definition of a bijection for a function, $f$, of two variables is:

1) $f$ is a surjection: $\forall \ p, \ \exists m, n \$ such that $f(m, n) = p$
2) $f$ is an injection: $f(n_1, m_1) = f(n_2, m_2) \ \Rightarrow \ n_1 = n_2 \$ and $m_1 = m_2$