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So I came upon this issue when trying to prove the following inequality:

|x| - |y| $\le$ |x - y|

one of the triangle inequalities.

So one way I thought of proving this is to square both sides. With a few simple steps, the inequality simplifies to |x||y| $\ge$ xy , which we know is true. But is this enough to really prove this inequality?

The other way to prove it would be with contradiction; ie, "Just suppose that |x| - |y| > |x - y| " and then show that this simplifies to a false inequality.

Are these two different approaches comparable in how deeply they prove the inequality?

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    Unfortunately, your first method of proof is completely wrong: if "squaring both sides gives a true result" were a valid method of proof, then $-1 = 1$. Not to mention that $|x| |y| \le xy$ is also wrong. –  May 08 '18 at 22:16
  • Also $|-1| \cdot |+1|$ is not less than $-1$ – AnalysisStudent0414 May 08 '18 at 22:17
  • The inequality was a typo, thanks for pointing that out. As for the proof in general, now that I look back I realize that I constructed the approach incorrectly. Instead, start with |x-y|. (|x-y|)^2 = (x-y)^2 $\le$ |x|^2 - 2|x||y| +|y|^2 That's the way I meant to do this –  May 08 '18 at 22:21
  • $x \le y \not \implies x^2 \le y^2$ and $x^2 \le y^2 \not \implies x \le y$ as signs may not be consistent. – fleablood May 08 '18 at 22:23
  • So you're saying that the method I offered in the comments was incorrect? I found a textbook that does it this way; however, it is a book specifically for an introductory course, so it may be assuming less than you are. –  May 08 '18 at 22:28

2 Answers2

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Are these two different approaches comparable in how deeply they prove the inequality?

You basically ask if between two different kind of proofs one is "better" or "more valid" than the other. The answer is No. Proof is proof: As long as the reasoning is logically correct, a proof proves the statement it is a proof of. A proof by contradiction is as good as any other.

On the other hand, different proofs for the same statement (regardless if one of them is by contradiction or not) can provide different insight into a problem or statement, but that is subjective and not measurable.

EDIT: I should mention that there is a philosophical branch of mathematics called "Constructivism", where proof by contradiction and axiom of choice are not allowed. If there are merits in that philosophy, I let for you to decide. However, the majority of the mathematical doesn't feel obliged to follow the constructionistic ways and will accept proofs by contradiction without problem.

SK19
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  • With their comments on my sloppiness I was able to edit the original example. Thank you for addressing the question, despite that! So I guess my intuition was that an algebraic construction of something "felt" more rigorous than contradiction. But based on what I'm reading, a proof by contradiction is still using algebra - just from the other point of view. In general though, are statements considered proven if we only have a proof by contradiction? Or does existence of contradiction imply existence of other proofs? –  May 08 '18 at 22:26
  • @kasa I added a paragraph regarding the constructive aspect. Also, if there is one proof, there are infinite proofs, because you can add unnecessary arguments to your proof. To ponder how many fundamentally different proofs of a statement there are would mean another question (that I will not be able to answer). – SK19 May 08 '18 at 22:36
  • @kasa A proof by contradiction (nothing to do with whether you use algebra or not) is just a much a proof as one that works directly from hypothesis to conclusion. But as this answer points out, a direct proof may (and usually does) offer insight into why the theorem is true. Silly example: I know you can get from my daughter's house to mine by car, since if you couldn't she wouldn't have been able to visit me yesterday. But a proof showing the route to take would be much more informative. – Ethan Bolker May 08 '18 at 22:38
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Proof by contradiction is best used when you believe a statement is false, and are unable to find an algebraic proof, or when you see an obvious counterexample that can immediately tell you that the statement is false. The most commonplace use of it I've seen is in proof of irrationality.

Is it stronger/weaker than a standard proof? Absolutely not. To prove a statement false is to prove a statement false, whether you find one counterexample or infinitely many, the statement is still false.

Rhys Hughes
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  • What exactly is an "algebraic proof"? – SK19 May 08 '18 at 22:38
  • Your answer suggests that proof by contradiction is what you do to prove something false. I don't think that's a good way to look at it. For example, to prove that a polynomial of odd degree has a root, you can assume it doesn't and get a contradiction by invoking the intermediate value theorem. Proof by contradiction is a good strategy here because actually showing how to calculate a root would be harder (though possible). – Ethan Bolker May 08 '18 at 22:45