Originally Posted by

**issacnewton** Hi

I am here after a long time. I want to prove that no natural number can be both even and odd. This is from basic analysis book. So I start by saying that let S be the set of natural numbers such that they are both even and odd.

$\displaystyle S=\{ n\in \mathbb{N}| \mbox{ n is even and n is odd} \} $

I assume the negation. So let S be non-empty. Since S is a subset of $\displaystyle \mathbb{N}$, by well ordering principle, S has a least element $\displaystyle k$. So

$\displaystyle k$ is both odd and even. $\displaystyle k = 2m$ and $\displaystyle k = 2n-1$ for some $\displaystyle m,n \in \mathbb{N}$. Which means $\displaystyle 2m = 2n-1$.

$\displaystyle \Rightarrow 2(m-n)=1$. Which means that $\displaystyle 1 $ is an even number. At this point can I say that I have reached a contradiction. Or do I first need to prove tthathat $\displaystyle 1 $ is an odd number to use that property