Then there is a 1:1 mapping between the Natural numbers and the Even numbers.

Proof. Consider the map f(n) := 2n. Then for every integer n, there exists an even integer f(n). Consider two distinct integers j and k, then f(j) = 2j and f(k) = 2k, and since j does not equal k, we know that 2j does not equal 2k. Thus, f is a 1:1 map.

Suppose that there exists some $N \in \mathbb{Z}$ which has the property $N \geq n \forall n \in \mathbb{Z}$. This means that $N$ is at least as large as any integer.

Now consider that for any $n \in \mathbb{Z}$, we have $(n + 1) \in \mathbb{Z}$ as well. Also, since $0 \lt 1$, we have $(0 + N) \lt (1 + N)$ so that $N \lt (N + 1)$. Thus, we have constructed an integer, $(N + 1)$ which is greater than $N$. Since $N$ was assumed to be the greatest integer, we have shown by contradiction that such a property cannot hold. Thus, there is no largest integer.