The theorem given is:
If $n$ is a natural number then $n$ can be written in the form $2a + 3b$ for some integers $a$ and $b$.
How would I prove this by induction? I've had a go at proving this but I don't know if my technique is sound.
The base case would be when n = 1 = 2(-1) + 3(1) (if we take the natural numbers as excluding 0). Then if I assume n = 2a + 3b is true, n+1=2a+3b+1. Therefore n+1=2a+3b+2(-1)+3(1) which can be written as n+1=2(a-1)+3(b+1) which should conclude the proof.
Is this a proper proof or is there some other way of doing it? How would I prove the theorem if I took the natural numbers to include 0 (i.e. could I still use 1=2(-1)+3(1) when it would no longer be the base case)?
Thanks for your help.