Proof using Well Ordering

**davismj** · February 27th 2010, 06:23 PM

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.

**Drexel28** · February 27th 2010, 10:17 PM

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 $S=\left\{n\in\mathbb{N}:a^{n+1}\ne a\cdot a^n\right\}$ . 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 $S=\varnothing$ .

It needs some preening, but you get the idea.

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