Spivak Question 14, Chapter 1, clarifying a statement.

Question

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.

Answer 1

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 $a$.

Answer 2

$-b \le a \le b \iff \left| a \right| \le b\tag1$

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 $\mathbb R.$)