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Prove $$|xy| = |x|\cdot |y| $$

Proofs abound, but the most elegant I have seen include steps like this. (I understand this does not complete the proof, but the shown part includes the issue that I am not completely understanding).

$${(|xy|)}^{2} = {(xy)}^{2} = {x}^{2}{y}^{2} = (|x| \cdot |y|)^{2} $$

What's been unclear to me is the assertion made in the last step ie allowing us to assert $(|x| \cdot |y|)^{2}$ as opposed to asserting only $(x \cdot y)^{2}$ as was done in the second step. Is this nothing more than strictly applying the definition of Absolute Value to a number known to be $\geq 0$? For example, can one simply state that if $a \geq 0$ then $a = |a|$

user1115542
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What you said would only give us that $x^2 = |x^2|$, we would still need to prove that we can take out the square.

Here we are using two facts. First is that $x^2 = (-x)^2$, and hence both are equal to $|x|^2$. On the other hand, we using the fact that $(ab)^k = a^k b^k$. So what we should have done was

$$x^2y^2 = |x|^2|y|^2 = (|x||y|)^2.$$

Rishi Sonthalia
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  • So,just to be absolutely clear in my mind, the moment you declare, for example, $|x|$, where $x$ is any number, then $x$ by definition represents $2$ ranges of numbers. One when $x \leq 0$ and one when $x \geq 0$. – user1115542 Oct 03 '21 at 12:10
  • I edited my question to clarify that this does not show the complete proof. – user1115542 Oct 03 '21 at 13:05
  • @user1115542 No; if $x$ is indeed an arbitrary number ("any number"), then writing $|x|$ does not change that (what $x$ is); however, the relationship between $|x|$ and $x$ does of course depend on whether $x<0$ or $x\geq0.$ Check out this recent Answer of mine, and compare statements (A) to (E) there. – ryang Oct 03 '21 at 13:42
  • @RyanG Very nice. So, to bring your answer back to my question then. Spivak ( and many others) repeatedly refer to the fact that, for example, $|a| = \sqrt a^{2}$, the implication being, my understanding of course, that squaring a number yields a positive value. But even though I think I understand why this is done, there is clearly something I am missing. – user1115542 Oct 03 '21 at 14:13
  • @RyanG Sorry about that. So, specifically, in the proof stated above, which is directly from Spivak, it is the last step that is the "fuzzy" part ie ${x}^{2}{y}^{2} = (|x| \cdot |y|)^{2}$, and again why is this "valid" ie adding the absolute value signs. – user1115542 Oct 03 '21 at 14:44
  • @RyanG The correct answers simply mean I am missing something. But, I believe I understand the basics. It will gel at some point. I did like both of your answers, and have upvoted both of them. Again... thank you for your input. – user1115542 Oct 03 '21 at 15:18
  • I meant both your comment and Rishi's answer. – user1115542 Oct 03 '21 at 15:27
  • @RyanG I think I might have inadvertently deleted your comment... not sure how I managed to do that. If so, just emphasizes there is a steep learning curve not to offend your helpers :-) !!!! – user1115542 Oct 03 '21 at 15:32
  • @RyanG. Well, it was perfect timing. I spent a while trying to figure out how to explain it to you!!!! And, I am beginning to understand what it is that caused this whole issue to be “fuzzy”. I was not separating the notion that the Absolute value for any value, say $a$ is a single positive value $a$ from the method to obtain that value, in essence the definition of Absolute value. If that does not make sense, then rest assured I am getting closer to clarity!!! – user1115542 Oct 03 '21 at 16:00
  • @user1115542 I do understand: it is easy to get confused because when we write these things informally we may not pay enough attention to the difference between identity/equation, proving/solving, etc., so when trying to rigorously clarify concepts it becomes harder to judge between overthinking it and underthinking it. Anyway, here are two recent Answers I wrote that may be orthogonally relevant: this and this. – ryang Oct 03 '21 at 16:14
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    @RyanG Those are great answers. (Upvoted both). And of course, from my perspective, the more one starts using Absolute Values, the more sense it makes. But, as you alluded to, the reason for doing all these proofs is to really probe one’s understanding. And, probe it does!!! – user1115542 Oct 03 '21 at 16:21
  • Thanks Rishi… who would think such a short proof could devour so much ink on my part. – user1115542 Oct 03 '21 at 19:19