Prove that if $x \in \mathbb{R}$ and $y \in \mathbb{R}$, then $xy \leq |xy|$.

Question

Proof (Attempt):

Case $1$ : $x \geq 0$ and $y \geq 0$. Then $xy \geq 0$ so $xy =|xy|$.

Case $2$ : $x \geq 0$ and $y < 0$. Then $xy \leq 0$ so $xy \leq |xy|$.

Case $3$ : $x < 0$ and $y \geq 0$. Then $xy \leq 0$ so $xy \leq |xy|$.

Case $4$ : $x < 0$ and $y < 0$. Then $xy > 0$ so $xy = |xy|$.

This is what I came up with, though I wonder if I should concern the case where either one of them is zero separately as another case. My question is, does this proof seem okay or should I include sub cases to account for when $x$ or $y$ is zero?

You can instead prove that for all $z \in \mathbb{R}$, $z \leq |z|$, which only requires two cases. Then since $xy \in \mathbb{R}$, it follows that $xy \leq |xy|$. Your proof is correct, though, and already encompasses the cases when $x=0$ or $y=0$. — kccu, Dec 08 '16 at 21:34
It doesn't hurt to mention that if either $x$ or $y$ are $0$, then the product is also $0$, so that $xy = 0 = |0| = |xy|$. But otherwise, good proof! :) — layman, Dec 08 '16 at 21:36
I don't see any need to include more cases-- in fact, you could make your proof shorter. Your use of "$\leq 0$" and "$\geq 0$" already account for when your variables are $0$. — NoName, Dec 08 '16 at 21:47
Okay so if one proves that $x \leq |x|$ then we are done because $x$ here is a real number so when you do $xy$, the product is just another real number which then you can apply the previous argument of $x \leq |x|$. — PiFarmer86, Dec 08 '16 at 22:10

score 2 · Answer 1 · answered Dec 08 '16 at 21:36

2

The simplest is to use one of the possible definitions of the absolute value on $\mathbf R$: $$\lvert a\rvert=\max(a, -a).$$

answered Dec 08 '16 at 21:36

Bernard

175,478

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OP was asking for someone to review their proof. Your answer suggests an alternate path to the proof, rather than a review of OP's proof. – layman Dec 08 '16 at 21:36
You're right. I think it's better, in absolute value questions, to avoid considering cases when it's possible. – Bernard Dec 08 '16 at 21:50
This doesn't avoid case consideration, you still have to separate between positive and negative to be able to tell what the maximum is. – Git Gud Dec 08 '16 at 22:01
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Yes, but for this precise question, it's included in the definition, since $\lvert a\rvert=\max(a,-a)$ implies $a\le \lvert a\rvert$ for any $a$. – Bernard Dec 08 '16 at 22:08
You're correct in this respect. – Git Gud Dec 08 '16 at 22:51
I take back what I said. You're using cases in the definition of $\max$. – Git Gud Dec 09 '16 at 00:17
Why, yes, but not in proving the O.P.'s inequality. This definition uses cases once and for all. For instance, proving the triangle inequality becomes much simpler. – Bernard Dec 09 '16 at 00:29

Prove that if $x \in \mathbb{R}$ and $y \in \mathbb{R}$, then $xy \leq |xy|$.

1 Answers1