If $−1https://mathhelpboards.com/pre-algebra-algebra-2/range-values-inequalities-19110.html
I came across this thread while looking for some inequality practice problems for the GRE. (I paraphrase the thread so please read the link before answering my question)
However, there is one step in the provided explanation that I don't understand. 
Question Asked in Other Thread:


*

*If $−1<a−b<10$,and$−3≤b≤1$ then what inequality represents the range
of values of $a^2$?
Plugging in $-1$ and $1$ for $b$ yields $-4<a<11$. 
Since the boundary is for $a^2$, the range would be $16<a^2<121$.
But why would this be incorrect?
Solution Offered in Other Thread:


*

*Given that:
$−4<a<11$
this implies:
$0≤|a|<11$
This is the step that I personally am having trouble with. Why does this imply $0≤|a|<11$? I didn't see that right away. If any one could possibly explain this or provide some visual intuition it would be greatly appreciated. 
 A: If $a < 0$ then $|a| = -a$.  So as $-4 < a < 11$ if $-4 < a < 0$ then $0\le |a| < 4$.
If $a \ge 0$ then $|a| = a$.  So if $0 \le a < 11$ then $0 \le |a| < 11$.
So either way, $0 \le |a| < 11$.
.....
or note $0\le a < k\iff -k < a < k$.
So note:   $-4 < a < 11$ means $-\max(4,11) < -4 < a < 11 = \max (4,11)$.
So $-\max(4,11) < a < \max(4,11)$ 
so $0 \le |a| <\max(4,11) = 11$.
Which should be obvious if you draw it.
.....
Take a number line where $a$ is between $-4$ to the left and $+11$ to the right.  Then to get the absolute value of $a$, we fold it on the $0$ and superimpose the negatives on top of the positives.  Whereever the $a$ was to begin with, it's no between the $0$ and the $11$.  
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$-1 < a-b < 10$ means $a < 10 + b$.  So the largest possible  $a$ value will occur when $b$ is the largest is can be.  $-3\le b \le 1$.  So $a < 10 + b \le 10 + 1=11$.  So if $a$ is non-negative $a^2 < 11^2$.
Also $-1< a- b < 10$ means $b-1 < a$. so the lowest possible value of $a$ is when $b$ is as small as possible.  $-3\le b \le 1$ so $-4<a $.  So if $a$ is negative $a^2 < 4^2$.
SO either way $a^2 < 11^2$.  (and if $a = 0$ we can have $a^2 = 0$.  So the range of $a^2$ is $[0,121]$.
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"Since the boundary is for $a^2$, the range would be $16<a^2<121$."
No  $-4 < a$ does not mean $(-4)^2 < a^2$.  Negative numbers "flip" the signs.
$-4 < a$ instead means that $(-4)^2 > a^2$.
So $-4 < a < 11$ means $-4 < a \le 0$ or $0 \le a < 11$.
$-4 < a \le 0$ means $0 \le a^2 < (-4)^2$ and $0 \le a < 11$ means $0 \le a^2 < 11^2$.
So either $0 \le a^2 < 16$ or $0 \le a^2 < 121$.  But that's redundant.
$0 \le a^2 < 121$.
A: What you get wrong:

The conclusion $-4<a\Rightarrow (-4)^2<a^2$ does not true for $-4<a\le4$.

$|a|\ge 0$ for all $a\in\mathbb{R}$.
If you are given $a<b<c$ such that $a<0$, $c>0$ and $|a|<|c|$ or more precisely, $-a<c$, then what can be implied here is $0\le |b| < c$.
However, If you are given $a<b<c$ such that $a<0$, $c>0$ and $|a|>|c|$ or more precisely, $-a>c$, then what can be implied here is $0\le |b| < -a$ or $0\le |b|<|a|$.
I can give you two examples:


*

*$-4<|a|<11\Rightarrow 0\le |a|<11$, this is because we split into two cases:


*

*If $-4<a<0$ then $|a|=-a$, so $-4<-a\Rightarrow |a|<4$.

*If $0\le a<11$ then $|a|=a$, so |a|<11$.


*$-12<|a|<9\Rightarrow 0\le |a|<12$, this is because we split into two cases:


*

*If $-12<a<0$ then $|a|=-a$, so $-12<-a\Rightarrow |a|<12$.

*If $0\le a<9$ then $|a|=a$, so $|a|<9$.
Just combine the range of two cases for each example and you will get your answer.
