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Question Number 127400 by mohammad17 last updated on 29/Dec/20

$${prove}\:{that}\::\mid\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\mid\leqslant\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid \\$$

Commented by mohammad17 last updated on 29/Dec/20

$$? \\$$

Commented by liberty last updated on 29/Dec/20

$${observasi}\:{the}\:{inequality}\:\mid\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid\:\leqslant\:\mid\:{z}_{\mathrm{1}} \mid\:+\:\mid{z}_{\mathrm{2}} \:\mid \\$$$${is}\:{known}\:{as}\:{triangle}\:{inequality}\:\:.{Now}\:{from} \\$$$${the}\:{identity}\:{z}_{\mathrm{1}} \:=\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +\left(−{z}_{\mathrm{2}} \right) \\$$$${gives}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +\left(−{z}_{\mathrm{2}} \right)\mid\:\leqslant\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid+\mid−{z}_{\mathrm{2}} \mid \\$$$${since}\:\mid{z}_{\mathrm{2}} \mid=\mid−{z}_{\mathrm{2}} \mid\:{then}\:\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\:\leqslant\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid...\left({i}\right) \\$$$${but}\:{because}\:{z}_{\mathrm{1}} +{z}_{\mathrm{2}} ={z}_{\mathrm{2}} +{z}_{\mathrm{1}} \:,\:\left({i}\right)\:{can}\:{be}\:{written}\: \\$$$${in}\:{the}\:{alternative}\:{form}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{2}} +{z}_{\mathrm{1}} \mid\:\geqslant\:\mid{z}_{\mathrm{2}} \mid−\mid{z}_{\mathrm{1}} \mid\:=−\left(\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\right) \\$$$${or}\:\mid{z}_{\mathrm{2}} \mid−\mid{z}_{\mathrm{1}} \mid\:=\:\mid\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\mid\:...\left({ii}\right) \\$$$${replacing}\:{z}_{\mathrm{2}} \:{by}\:−{z}_{\mathrm{2}} \:{we}\:{get}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:\geqslant\:\mid\:\mid{z}_{\mathrm{1}} \mid−\mid−{z}_{\mathrm{2}} \mid\:\mid \\$$$${or}\:\mid{z}_{\mathrm{1}} −{z}_{\mathrm{2}} \mid\:\geqslant\:\mid\:\mid{z}_{\mathrm{1}} \mid−\mid{z}_{\mathrm{2}} \mid\:\mid \\$$