# Why does this step work in this proof?

I'm trying to learn discrete math and am brushing up on proofs by reading Richard Hammack's Book of Proof. I'm tripped up on this proof... I understand that it's contrapositive, and why contrapositive is the best approach, but I'm not sure why you should multiply both sides of $y-x > 0$ by the positive value $x^2+y^2$. Can anyone explain this to me?

Thanks!

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Because it works. Rabbit out of the hat. I was told Gauss loved pulling rabbits out of his hat. From his wiki: "Gauss usually declined to present the intuition behind his often very elegant proofs—he preferred them to appear "out of thin air" and erased all traces of how he discovered them." – Git Gud Jan 21 '13 at 22:11

This is an example of reverse engineering.

We first realize that in the original problem, we have $y^3+yx^2$, which could be factored as $y(x^2+y^2)$ and $x^3 + xy^2$ could be factored as $x(x^2+y^2)$.

Hence, to prove the contrapositive of the statement, we take that $y>x$ and multiply throughout by $x^2+y^2$ to get our desired conclusion.

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Thanks, makes sense. I was hoping there was some rule I was forgetting that I could use to approach problems like this in the future. Though, this works. – user56763 Jan 21 '13 at 22:28
Most proofs require a bit (or a lot, or an awful lot) of scratch paper. Your random scribblings (or even foul language) doesn't need to appear to the world. It is so much better to present the polished result, and when somebody (like you here) asks, you look at them with supreme disdain (they don't have to know it took you weeks of hard work to come up with that "simple" idea ;-). If you can, grab a copy of Pólya's "How to solve it". – vonbrand Jan 23 '13 at 1:51
To continue on that, there's a lot of proofs that are solved by multiplying by 1 or subtracting 0. The part that takes mounds of scratch paper can be realizing that the "1" you should multiply by is, for example, (x-i)/(x-i). – Patrick M Jul 11 '13 at 1:31

Contrapositive is not necessarily the best approach. We have $y(x^2+y^2)\le x(x^2+y^2)$. If $x=y=0$, then clearly $y\le x$.

If $x$ and $y$ are not both $0$, then $x^2+y^2\gt 0$, and so by division by the positive number $x^2+y^2$, or if you prefer by multiplication by $\frac{1}{x^2+y^2}$, we get $y\le x$.

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That's how I would do it. – user7530 Jan 21 '13 at 22:28
Ah, cool. Thanks. – user56763 Jan 21 '13 at 22:28
Though it is "easier" to prove that $$ab > ac \text{ and } a>0 \text{ implies }b>c$$ in an ordered field by proving the contrapositive. – user17762 Jan 21 '13 at 22:30
Use whatever comes more natural. But remember that the proof is for your fellow humans to understand. – vonbrand Jan 23 '13 at 1:52