Background
Hello, I'm teaching myself proofs, and am unsure whether or not my proof of $\forall a,b \in \mathbb{R}(|a| \leq b \iff -b \leq a \leq b)$ is correct. Your feedback is greatly appreciated.
Proof
Proof. $(\rightarrow)$ Suppose $a,b \in \mathbb{R}$ and $ |a| \leq b$. We consider both cases.
Case 1. $a \geq 0$. Then $|a| = a \leq b$ by definition of absolute value. Since $a \geq 0$, then $-a \leq 0$ by multiplying the inequality by $-1$. Similarly, multiplying $a \leq b$ by $-1$ gives us $-a \geq -b$, or equivalently $-b \leq -a$. Since $-a \leq 0$ and $a \geq 0$, or equivalently $0 \leq a$, then $-a \leq a$ by transitivity. We now have the following inequality,
$$
-b \leq -a \leq a \leq b
$$
Which means $-b \leq a \leq b$ by transitivity.
Case 2. $a < 0$. Then, $|a| = -a \leq b$ by definition of absolute value. Multiplying both sides of the inequality by $-1$, we get $a \geq -b$, or equivalently, $-b \leq a$. Since $-b \leq a$ and $a < 0$, then $-b < 0$. Also, since $-b < 0$, multiplying by $-1$ means $b > 0$, or equivalently $0 < b$. But since $a < 0$, then $-a > 0$. By transitivity, $a < 0 < -a$, means $a < -a$. Note that $-a \leq b$. So we have,
$$
-b \leq a < -a \leq b
$$
$\color{blue}{\text{I am unsure about this part. How do I introduce equality?}}$ Therefore, $-b \leq a < -a \leq b$. Since $a < 0$ and $-a > 0$, the only time $a = -a$ is when $a = -a = 0$. So,
$$
-b \leq 0 \leq b
$$
Or, $-b \leq a \leq b$ by substitution.
Since we've exhausted all cases, if $|a| \leq b$ then $-b \leq a \leq b$ for all $a,b \in \mathbb{R}$.
$(\leftarrow)$ Suppose $a,b \in \mathbb{R}$ and $-b \leq a \leq b$. Then, $a \geq -b$ and $a \leq b$. We must show $|a| \leq b$. We consider two cases.
Case 1. $a < 0$. Multiplying $a \geq -b$ by $-1$, we get $-a \leq b$. By definition of absolute value, $-a = |a| \leq b$.
Case 2. $a \geq 0$. Since $a \leq b$, then $a = |a| \leq b$ by definition of absolute value.
Therefore, $|a| \leq b$ when $-b \leq a \leq b$ for all $a,b \in \mathbb{R}$. $\qquad \Box$
Question
I'm really self-concious about my ability to do this sort of stuff, so I hope I haven't butchered this. The blue highlighted part is where I'm most uncertain. I've seen some books (e.g. Rosen's Discrete Math book), use the following definition for absolute value:
$$
|a| = a \text{ when } a \geq 0 \text{, and } -a \text{ when } a \leq 0.
$$
In other words, he uses the $\geq$ and $\leq$ relations in both cases. Is this okay? I've always seen the definition as $a < 0$ for one case, and $a \geq 0$ for the other. Does this matter? If it does, how do you introduce the equality like I had to for Case 2 in the $(\rightarrow)$ proof? Also, when using transitivity, and you have something like
$$
-b \leq a < -a \leq b
$$
I'm guessing you can't conclude $-b \leq a \leq b$. Since we haven't established that $a = -a$, so we can't say more than what we're given, which is that $a < -a$, so we could only say $-b \leq a < b$. Is my reasoning correct here?
Best Answer
The two definitions of absolute value are equivalent. Some would argue that the one that you’re using is better style, because the cases are disjoint, but the one that you found in Rosen is also acceptable, since the two parts of the definition agree when both apply, i.e., when $a=0$.
There is no need to introduce equality after you arrive at
$$-b\le a<-a\le b\;:\tag{1}$$
that immediately implies that $a<b$, and if $a<b$, then certainly $a\le b$, so from $(1)$ you can immediately conclude that $-b\le a\le b$.
I would probably have organized the proof of $(\leftarrow)$ a little differently, doing Case 2 first, since it’s truly trivial: if $a\ge 0$, then $|a|=a$, and we’re assuming that $-b\le a\le b$, so $-b\le|a|\le b$. That, however, is a matter of taste.