MHB Proving an absolute value inequality

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The discussion focuses on proving the inequality if |a| ≤ b, then -b ≤ a ≤ b, for real numbers a and b. The definition of absolute value is clarified, distinguishing cases where a is non-negative and negative. In Case I, it is noted that if a ≥ 0, then |a| = a, leading to a ≤ b, while in Case II, if a < 0, then |a| = -a, resulting in -b < 0 ≤ a. Participants express confusion regarding the logical structure of the proof, particularly the incorrect assertion that a > b in Case I. The need for a clearer, more coherent proof is emphasized.
cbarker1
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If $\left| a \right| \le b$, then $-b\le a\le b$.
Let $a,b \in\Bbb{R}$ The definition of the absolute value is $ \left| x \right|= x, x\ge 0$ and $\left| x \right|=-x, x< 0$, where x is some real number.

Case I:$a\ge 0$, $\left| a \right|=a>b$

Case II: a<0, $\left| a \right|=-a<b$the solution is $-b<0\le a\le b$

I work on a number line. yet I still have trouble with the proof.
 
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At which point of the proof are you facing difficulties?
 
Well, I, for one, have trouble understanding the proof since I don't follow its logical structure, and not just because $a>b$ should be $a\le b$ in case I. Maybe someone can write a more coherent proof.
 
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