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Part (a) proves $\left| a \right| = \left| -a \right|$ The actual proof in Spivak's workbook is pretty straight forward. Let's accept this as done.
Part (b) proves $-b \le a \le b \iff \left| a \right| \le b$. There are many questions that prove this many ways in Stack exchange, including of course questions about Spivak's approach. Again, for the sake of argument, let's accept this proved.

Now, Spivak says

it follows that $-\left| a \right| \le a \le \left| a \right|$

A questioner asked for an explanation, and the most upvoted answer was

let $b = |a|$ and you are done.

Could someone explain the obvious here? Yes, indeed, if you substitute $|a|$ for $b$ it provides the asked for proof, but the notion that one can simply replace a variable with an absolute value is, I believe what the previous questioner and I are asking about. And, almost inevitably, it means that there is something we, or I in this case, am missing in my understanding of absolute values.
If this question is a little confusing, it probably means I am missing something obvious, hence the question.

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The confusion is not in what absolute values are, but in what variables are.

The statement $$ -b \le a \le b \iff |a| \le b $$ (with an implied "for all $a,b \in \mathbb R$ at the beginning) means that if we have any values of $a$ and $b$ for which one of the two sides holds, we can deduce that the other side also holds for them. For example, we could take $a =-3$ and $b=5$, and check the right-hand side: $|{-3}| \le 5$. Then we could immediately conclude the left-hand side: that $-5 \le -3 \le 5$.

What we can also do is, for any value of $a$, pick the value $b = |a|$. There is nothing special about absolute value here: we could have picked $b = a^2$, or $b = e^{\sqrt{\log a}}$, or any other expression in $a$. But picking $b = |a|$ is convenient, because then the right-hand side $|a| \le b$ simplifies to $|a| \le |a|$, which we know holds for any $a$.

Therefore the left-hand side, which simplifies to $-|a| \le a \le |a|$, must also hold for all $a$.

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  • $\begingroup$ Very nice. Thank you. Your explanation is what I needed, especially the part of the convenience of choosing $|a|$ and how that fits in. Thank you again. $\endgroup$ Oct 17 at 20:53
  • $\begingroup$ Misha...please indulge me in seeing if indeed I have gotten this right..Once proven ( your "statement") we get to use it. Your point adds that which I missed, ie if it holds for one side it holds for the other. Key point, for me at any rate. So, now picking $b = |a|$, and I assume we can use the $=$ here as the original proof uses $\le$ sets down that which we know, as you point out, hence we can immediately apply that to the left side. Misha, you have added those "hooks" to Spivak's statement, giving it much greater depth. Thank you. $\endgroup$ Oct 17 at 22:36
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    $\begingroup$ Yeah, I think you've got it. It might be easier to think about the whole thing if we use a different variable like $x$. "Setting $a=x$ and $b=|x|$, we know that $|a| \le b$ because $|x| \le |x|$, so we conclude $-b\le a\le b$ or $-|x|\le x\le|x|$" is no different from "Setting $a=-3$ and $b=5$, we know that $|a|\le b$ because $|{-3}| \le 5$, so we conclude $-b\le a\le b$ or $-5\le-3\le5$". $\endgroup$ Oct 17 at 23:16
  • $\begingroup$ Got it. Thanks for taking the time. :-) $\endgroup$ Oct 17 at 23:22
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$-b \le a \le b \iff \left| a \right| \le b\tag1$

it follows that $-\left| a \right| \le a \le \left| a \right|\tag2$

Could someone explain the obvious here? Yes, indeed, if you substitute $|a|$ for $b$ it provides the asked for proof, but the notion that one can simply replace a variable with an absolute value is, I believe what the previous questioner and I are asking about.

Statement $(1)$ reads:$$\text{for each }(a,b)\in\mathbb R^2,\quad -b \le a \le b \iff \left| a \right| \le b.$$ Therefore, $$\text{for each }(a,b)\in\mathbb R^2,\quad \left| a \right| \le b\implies-b \le a \le b.$$ In particular: $$\text{for each }(a,|a|)\in\mathbb R^2,\quad \left| a \right| \le |a|\implies-|a| \le a \le |a|.$$ In other words: $$\text{for each }a\in\mathbb R,\quad \left| a \right| \le |a|\implies-|a| \le a \le |a|.$$ Thus, $$\text{for each }a\in\mathbb R,$$ since $\left| a \right|\le |a|$ is indeed true, by Modus Ponens, $$-|a| \le a \le |a|,$$ as required.

(Notice that $|a|$ equals $a$ when the latter is positive, and equals $-a$ when the latter is negative; in other words, $|a|$—like $a,b,-a$ and $-b$—is just a variable on the universe of discourse $\mathbb R.$)

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  • $\begingroup$ Thank you for your answer. It helps having a nice overview as provided by you. $\endgroup$ Oct 17 at 20:55
  • $\begingroup$ Ooops...your comment disappeared but this is still relevant!! Ryan, yes most certainly. And, looking at these proofs in great detail really does help, me at any rate. Moreover, glossing over them does not make them clearer later on!! Esp with Spivak who routinely digs out what was previously done. The real joy of reading answers from the real mathematicians, is getting to understand the way they think. That's a great gift, thanks. $\endgroup$ Oct 17 at 22:40
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    $\begingroup$ @user1115542 All I said was something along the lines of “really slowing down, going back to the basics, asking and stating the "obvious" sometimes does help with known blind spots.” (And I've just expanded the above Answer.) Yes, MSE is really a gem for mathematics learners! $\endgroup$
    – ryang
    Oct 18 at 5:27
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    $\begingroup$ You're welcome, @user1115542 . 1. "For each $(a,b){\in}\mathbb R^2$" is just a fancy way of saying "$\forall a{\in}\mathbb R\;\forall b{\in}\mathbb R$". 2. I extracted the Right-to-Left portion(the reverse direction) of the bi-implication because to obtain the required final result (using Modus Ponens), we need only that portion. $\endgroup$
    – ryang
    Oct 18 at 17:59
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    $\begingroup$ Thanks Ryan. I am getting a crash course in mathematical notation. :-) (Including, $\forall$, which needs no explanation :-) ). OK...got it. Between yourself and Misha, you have certainly helped me, and no doubt others who have stumbled at the same point, also trying to be methodical with all Spivak's problems. $\endgroup$ Oct 18 at 18:22

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