Is the given function injective, surjective?

Question

The following function is obviously injective, surjective and bijective, but I'm not really sure how to prove this. N = {0,1,2,...}

$f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$, $f(x,y)=x+y$

Answer 1

In this case, the function is clearly surjective, yes. Note that for any $n \in \mathbb{N}$ you have $(0,n) \in \mathbb{N} \times \mathbb{N}$ for which $f(0,n)=0+n=n$, as just one potential example.

However, injectivity fails for various pairings. Just as one specific example, $f(1,2)=f(0,3)=3$. However, $(1,2)\neq(0,3)$, thus not injective (and in turn, not bijective).