Question Number 72900 by aliesam last updated on 04/Nov/19
$${prove}\:{that}\: \\ $$$$ \\ $$$$−\mid{a}\mid\leqslant{a}\leqslant\mid{a}\mid \\ $$$$ \\ $$$${a}\:{is}\:{a}\:{real}\:{number} \\ $$
Answered by MJS last updated on 04/Nov/19
$$\forall{r}\in\mathbb{R}:\:\mid{r}\mid:=\begin{cases}{{r};\:{r}\geqslant\mathrm{0}}\\{−{r};\:{r}<\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:\begin{cases}{{a}\geqslant\mathrm{0}:\:\pm\mid{a}\mid=\pm{a}\:\Rightarrow\:−\mid{a}\mid\leqslant{a}=\mid{a}\mid}\\{{a}<\mathrm{0}:\:\pm\mid{a}\mid=\mp{a}\:\Rightarrow\:−\mid{a}\mid={a}<\mid{a}\mid}\end{cases} \\ $$$$\Rightarrow\:\forall{a}\in\mathbb{R}:\:−\mid{a}\mid\leqslant{a}\leqslant\mid{a}\mid \\ $$