# Thread: Proof using Well Ordering

1. ## Proof using Well Ordering

This is a real baby proof, but I'm not sure I'm doing the well ordering proof right, as its my first attempt.

Is this the basic structure of a well ordering proof?

2. Originally Posted by davismj

This is a real baby proof, but I'm not sure I'm doing the well ordering proof right, as its my first attempt.

Is this the basic structure of a well ordering proof?

I feel as though you are way overdoing it. Let $S=\left\{n\in\mathbb{N}:a^{n+1}\ne a\cdot a^n\right\}$. Assume that $S\ne\varnothing$, then by the well-ordering principle $S$ has a least element $k$. Thus, $a^{k+1}\ne a^k\cdot a$. Dividing both sides we see that $a^{k}\ne a^{k-1}\cdot a\implies k-1\in S$. This contradicts the minimality of $k$ and thus it follows that $S=\varnothing$.