Why Is My Proof Of Irrationality Of Sqrt 2 Wrong?
This proof uses the odd/even dichotomy and proof by contradiction but in a different way than the usual fraction irreducibility argument.
Assume sqrt(2) is rational
Integer b must be either even or odd.
Case where b is even i.e. b=2p
Case where b is odd i.e. b=2q+1
There are no 2 integers which differ by half. This is the required contradiction to prove the initial assumption (root 2 is rational) is invalid.
This proof doesn't seem to depend on sqrt(2). So it can be used to prove any number (e.g. 5) is irrational.
Actually I think I get it now. My assumption that 'a' is the same across both the odd and even cases is a mistake.