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.