# Proof using Well Ordering

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• Feb 27th 2010, 06:23 PM
davismj
Proof using Well Ordering
http://i47.tinypic.com/6h5cmw.jpg

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?

Thanks in advance.
• Feb 27th 2010, 10:17 PM
Drexel28
Quote:

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?

Thanks in advance.

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

It needs some preening, but you get the idea.