Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that if x and y are real numbers, then |xy|=|x||y|. Hint check all the cases.

I tried assuming the left hand side equals the right hand side if we remove absolute values. Also, tried using the fact that |x| can equal x and -x and |y| can equal y and -y.

share|improve this question
What you have tried? –  Sami Ben Romdhane Feb 25 at 16:46
Let me start of by saying im struggling with this course a lot and my attempts may seem rather unintelligent. I tried subtracting one side and somehow using triangle in equality, also tried assuming the left side equals the right when removing absolute values which I realize does nothing. –  Achilles Feb 25 at 16:48
It's still in your best interest to mention what you tried, or even simply to expand on what you don't understand, and/or what you've covered thus far in your course, and/or any definitions you think might help (example: how to we define the absolute value of a variable?). Any information that will help us help you, especially information that helps us see that you are trying to help yourself, too. –  amWhy Feb 25 at 16:51
The left-hand side won't equal the right hand side in general if you remove the absolute values. But the idea of removing the absolute values is a good one, and the hint tells you how to do this - $|x|$ is always equal to $x$ or $-x$ depending on whether $x\geq0$ or $x<0$. So making a choice of sign for both $x$ and $y$ lets you take the absolute values away, and there are four possible pairs of sign choices... –  Matt Pressland Feb 25 at 16:52
Thanks, for the added info, Achilles. I didn't see your comment when I was writing my comment. That's precisely the kind of information I was referring to. There's no penalty for trying, even if you think you're off. Most of us run into problems where we don't "get it right" until after getting it wrong. It's called learning. ;-) –  amWhy Feb 25 at 16:57
show 6 more comments

closed as off-topic by Thomas, Sami Ben Romdhane, Yiorgos S. Smyrlis, Hurkyl, Davide Giraudo Feb 25 at 17:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thomas, Sami Ben Romdhane, Yiorgos S. Smyrlis, Hurkyl, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

2 Answers

up vote 4 down vote accepted

There is no need to check the cases. Since $|a|=\sqrt{a^2}$, you have $|xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y^2}=|x|\,|y|$.

share|improve this answer
Is this just an alternate method, or the proper way to answer? This makes perfect sense to me after seeing it I just am a weak thinker I suppose. –  Achilles Feb 25 at 17:16
Alternate. Both ways are correct. –  LeoTheKub Feb 25 at 17:17
Don't put yourself down, Achilles. Experience comes with practice. –  LeoTheKub Feb 25 at 17:26
True , thanks for the advice Wlodek. –  Achilles Feb 25 at 17:29
I would add that if the question gives a hint, you should probably use it. Otherwise it may look like plagiarism. –  Cruncher Feb 25 at 21:14
add comment


Check all cases!

  • $x \geq 0,\; y\geq 0$
  • $x \geq 0, \;y < 0$
  • $x < 0, \;y\geq 0$
  • $x < 0,\; y< 0$.
share|improve this answer
I think I would rather use strictly > or < 0, and show a case for x or y = 0. +1 though, I think this is the best answer given the wording of the question –  Cruncher Feb 25 at 21:16
Nice hint! It deserves the nice answer badge and more! –  Sami Ben Romdhane Feb 26 at 14:37
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.