Question

How to prove/disprove   is surjective?

F:2*Z >Zf (m, п) т

Answer 1

A function f with domain X and co-domain Y is surjective if for every  y in Ythere exists at least one x in X such that f(x) = y

we have where   and

Let's say we have an element y in the co-domain where y Z . We need to find two integers p and q such that p Z and q Z  and |p| - |q| = y

We have two cases :-

(i) If y >= 0. In this case, we take p = y+1 and q = 1 . |p| - |q| = y , thus we have found p and q such that f(p,q) = y .      

(ii) If y < 0 . In this case, we take p = 0 and q = y. |p| - |q| = 0 - |y| = y , thus we have found p and q such that f(p,q) = y

In both the above scenarios, we have found p and q in the domain X for any y in the co-domain Y . Thus, f(m,n) = |m| - |n|   is a surjective function.