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Prove that |a| < b if and only if -b < a < b

User Petrnohejl
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Answer:

Since,


|x|=\left\{\begin{matrix}x &amp;\text{ if } x \geq 0 \\ -x &amp;\text{ if } x < 0\end{matrix}\right.

Here, the given equation is,

|a| < b

Case 1 : if a ≥ 0,

|a| < b ⇒ a < b

Case 2 : If a < 0,

|a| < b ⇒ -a < b ⇒ a > - b

( Since, when we multiply both sides of inequality by negative number then the sign of inequality is reversed. )

|a| < b ⇒ a < b or a > - b ⇒ -b < a < b

Conversely,

If -b < a < b

a < b or a > - b

⇒ a < b or -a < b

⇒ |a| < b

Hence, proved..

User Advantej
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