- #1
cbarker1
Gold Member
MHB
- 349
- 23
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.
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.