Prove that for all real numbers $a$ and $b$, $|a| ≤ b$ iff $-b ≤ a ≤ b$ The problem is from exercise 12 of chapter 3.5 of Velleman's How To Prove It.
For the (----&gt;) direction, I realise we are supposed to break the proof into two cases.
So, suppose $$|a|≤ b $$ Case 1:   $a=|a| $ which implies $a≤b.\;$ Now suppose $-b&gt;a,\;$which implies $-a&gt;b,\;$ but we know that $a≤b,\;$ so $-b&gt;a\;$ is a contradiction. Therefore, $-b≤a.\;$Since we now know that $a≤b\;$ and $-b≤a,\;$we can conclude that $-b ≤ a ≤ b.$
Case 2: $-a=|a|,\;$which implies  $-a≤b,\;$which is equivalent to $-b≤a.$
This is where I'm stuck at. From the assumption of case 2, we have deduced that $-b≤a. \;$ But I don't know how to prove that $a≤b\;$ to reach the stated goal of $-b ≤ a ≤ b \:$ for this case. Assuming $a ≤ b \;$ is false (that is, assuming $a&gt;b$) and trying to reach a contradiction doesn't seem to work.
I'm sure the solution is simple but I've spent 2 hours on this with no luck. ;
Thank you for your time.
EDIT: I should have clarified that since the text is about proof strategies, the author intended the reader to use proof by cases in this section of the book to get a better grasp of that particular strategy. Even so, thank you all so much for all the different ways of approaching the proof that you suggested! I've learned new things today!
For case 2, I ended up using the fact that $a&lt;0\,$ to prove that $a≤b. \,$ That is, since $a&lt;0\,$ iff $-a&gt;0,\,$ and since $b≥-a,\,$ then $b&gt;0.\,$ Thus we have $a&lt;0&lt;b,\,$ which implies $a≤b.\,$
Therefore, we can conclude that $-b ≤ a ≤ b.$
You can also use this maximum properties $\quad\begin{cases}\max(x,y)+z=\max(x+z,y+z)\\\max(u,v)\le w\iff (u\le w)\text{ and }(v\le w)\end{cases}$
$|a|=\max(a,-a)\le b\iff \max(a,-a)-b\le 0\iff \max(a-b,-a-b)\le 0 \iff\begin{cases}a-b\le 0\\-a-b\le 0\end{cases}\iff \begin{cases}a\le b\\-b\le a\end{cases}\iff -b\le a\le b$
This way of proceeding somehow 'hides' the multiple cases discussion.
I think the following reasoning is shorter.
For $b\geq0$ it's obvious by the definition of the absolute value:
$|a|$ it's a distance between the point, which is corresponded to $a$ on the number axis, and the origin, which gives immediately: $$|a|\leq b\Leftrightarrow -b\leq a\leq b$$
But for $b&lt;0$ the statement $$|a|\leq b\Leftrightarrow -b\leq a\leq b$$ is also obvious and we are done!