What is the Surjectivity of the Function f: Z^2 \rightarrow Z, f((n,m))=nm?

[DPMachine]

Homework Statement

I'm trying to show that the function below is surjective

f: Z^2 \rightarrow Z, f((n,m))=nm

Homework Equations

The Attempt at a Solution

1. Suppose z\in Z.

2. Then z=nm.

3. So n=\frac{z}{m} and m=\frac{z}{n}

4. So f((n,m))=\frac{z^2}{mn}=\frac{n^2m^2}{mn}=nm=z

I'm mostly wondering about step 2... Am I allowed to assume that? (z=nm?)

 

[aPhilosopher]

Why are you solving for z? Why not just use the fact that f(m, n) = mn = z?

 

[DPMachine]

Why are you solving for z? Why not just use the fact that f(m, n) = mn = z?

Right.. but what's the (n, m) in that case?

I'm trying to prove that for any arbitrary z in the codomain, there exists some (n, m) in the domain that satisfies the function. I'm trying to find (n, m) in terms of z.

Sorry if I didn't get your point.

 

[aPhilosopher]

I meant solving for n and m in terms of z in my previous post.
If you know that f(a, b) = ab and you know that z = mn then you have f(m, n) = z. You don't need it more complicated than that.

And you can assume that there's always a factorization z=mn, just choose z = 1*z.

 

[HallsofIvy]

Uhh, Given any integer z, aren't both (z,1) and (1,z) mapped into z? Isn't that enough?