Prove that for all real numbers $a$ and $b$, $|a| \leq b$ iff $-b \leq a \leq b$

Question

This is an exercise from Velleman's "How To Prove It". I do not have a specific question, but I'm just unsure about the structure for breaking proofs into cases. Since I'm self-studying, other comments would be appreciated as well!

Prove that for all real numbers $a$ and $b$, $|a| \leq b$ iff $-b \leq a \leq b$

Proof: Let $a$ and $b$ be arbitrary real numbers. Suppose $|a| \leq b$. We then consider cases.

Case 1. $a < 0$. It follows that $|a|=-a\leq b$. Multiplying this inequality by $-1$ gives $a \geq -b$. Since $a < 0$, $-a > 0$, so we can combine the inequalities to get $-b \leq a < 0 < -a \leq b$. Thus, $-b \leq a \leq b$.

Case 2. $a \geq 0$. Then $|a|=a \leq b$, and $-a \leq 0 \leq a \leq b$. It then follows that $0 \leq b$, so $-b \leq 0$. Then $-b \leq 0 \leq a \leq b$. Thus, $-b \leq a \leq b$.

Now suppose $-b \leq a \leq b$. Again we will consider cases.

Case 1. $a < 0$. Then $-b \leq a < 0$. Multiplying the inequality by -1 gives $b \geq -a > 0$. Since $a < 0$, $|a| = -a \leq b$.

Case 2. $a \geq 0$. Then it follows immediately from the givens that $|a| = a \leq b$. In every case we have shown that $|a| \leq b$. Since $a$ and $b$ were arbitrary, $\forall a \in \mathbb{R} \forall b \in \mathbb{R} (|a| \leq b \leftrightarrow -b \leq a \leq b)$. $\square$

Answer 1

It looks fine, although you can simplify the proof a bit by making the following argument before you break into cases: knowing that $0 \le |a|$ (which should be regarded as a known consequence of the definition of absolute value), and applying the hypothesis $|a| \le b$, it follows that $0 \le b$.

Answer 2

Your proof is fine, a comment though: it wouldn't make any sense if $b<0$, so simplify the problem by assuming $b\geq0$.

Another way to think about this: the absolute value is nothing but the distance of the number from $0$. If $|a|\leq b$, then $a$ is closer to $0$ than what $b$ and $-b$ are (since they have absolute value equal to $b$. So $a$ is contained in the interval $[-b,b]$ which is what you want to prove. This works the other way around too, to prove the converse.