# Proof by contradiction and the well ordering principle

I have a question regarding proof by contradiction and the well ordering principle on integers.

Okay lets says for example an arbitrary value $$p$$ is a positive integer and which I assume is the smallest positive integer for some property.

Then lets say I find an arbitrary value $$r$$, also a positive integer, which I calculate to be less than $$p$$ for the same property. Then I have a contradiction.

So my question is how is this possible. Lets substitute $$1$$ for $$p$$ and it satisfies the property, then how can I find $$r$$ since there is no smaller positive integer? I'm not sure if $$0$$ is considered a positive integer but if it is then substitute $$p$$ for $$0$$.

• Another term for this is proof by infinite descent. I don’t have time to write an answer but you may find something helpful here en.m.wikipedia.org/wiki/Proof_by_infinite_descent – spaceisdarkgreen Mar 1 at 20:18
• It will also help to get more of a feel for the hypothetical nature of proofs by contradiction by studying other proofs by contradiction that might be easier to understand. – spaceisdarkgreen Mar 1 at 20:21

The point of the proof by contradiction is that you can find no such $$r$$. That's the contradiction. You assume something you want to be false, then prove something that can't be possible. It may be helpful to give a specific example of a proof that's confusing you.
• Yeah, so in that proof you assume that $\sqrt{2}$ is irrational. Then using well ordering you can find a least $q$ such that $q\sqrt{2}$ is an integer. Then without assuming anything else, you can find $r < q$ a positive integer such that $r \sqrt{2}$ is a positive integer, which is a contradiction by minimality of $q$. – bitesizebo Mar 1 at 20:17