Question Number 113901 by frc2crc last updated on 16/Sep/20
$${when}\:{do}\:{I}\:{use}\:\mid{x}\mid \\ $$$${for}\:\sqrt{{x}^{\mathrm{2}} }? \\ $$
Commented by MJS_new last updated on 16/Sep/20
$$\mathrm{almost}\:\mathrm{always} \\ $$$${p}^{\mathrm{2}} \geqslant\mathrm{0}\forall{p}\in\mathbb{R}\:\Leftrightarrow\:{p}^{\mathrm{2}} \in\mathbb{R}^{+} \forall{p}\in\mathbb{R} \\ $$$$\sqrt{{q}}\geqslant\mathrm{0}\forall{q}\in\mathbb{R}^{+} \\ $$$$\Rightarrow\:\sqrt{{p}^{\mathrm{2}} }\geqslant\mathrm{0}\forall{p}\in\mathbb{R} \\ $$$$\Rightarrow\:\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid\:\forall{x}\in\mathbb{R} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{use}\:\mathrm{it}\:\mathrm{in}\:\mathrm{calculus}\:\left(\mathrm{integral}\:\mathrm{solving}\right. \\ $$$$\left.\mathrm{by}\:\mathrm{substitution}\right) \\ $$
Commented by frc2crc last updated on 16/Sep/20
$$\sqrt{\mathrm{4}}=\mathrm{2}\:{or}\:\mid\mathrm{2}\mid? \\ $$
Commented by bobhans last updated on 16/Sep/20
$$\sqrt{\mathrm{4}}\:=\:\mathrm{2}\:{and}\:\mid\mathrm{2}\mid\:=\:\mathrm{2}\: \\ $$
Commented by MJS_new last updated on 16/Sep/20
$$\sqrt{\mathrm{2}^{\mathrm{2}} }=\mid\mathrm{2}\mid=\mathrm{2} \\ $$$$\sqrt{\left(−\mathrm{2}\right)^{\mathrm{2}} }=\mid−\mathrm{2}\mid=\mathrm{2} \\ $$