Question Number 157098 by mathocean1 last updated on 19/Oct/21
$${n}\:\in\:\mathbb{N}^{\ast} \:;\:{n}\:{is}\:{not}\:{a}\:{square}\:{of}\:{any} \\ $$$${integer}.\:{Show}\:{that}\:\sqrt{{n}}\:\notin\:{Q}\:. \\ $$
Answered by mindispower last updated on 19/Oct/21
$$\sqrt{{n}}=\frac{{p}}{{q}},\:{p} {q}=\mathrm{1}\Rightarrow \\ $$$$\Rightarrow{q}^{\mathrm{2}} {n}={p}^{\mathrm{2}} \\ $$$$\Rightarrow{q}^{\mathrm{2}} \mid{p}^{\mathrm{2}} ,{p} {q}=\mathrm{1}\Rightarrow{p}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{1} \\ $$$${q}^{\mathrm{2}} \mid{p}^{\mathrm{2}} ,{q}^{\mathrm{2}} \mid{p}^{\mathrm{2}} \Rightarrow{q}^{\mathrm{2}} \mid{p}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{1}\Rightarrow{q}=\mathrm{1} \\ $$$$\sqrt{{n}}={p}\in\mathbb{N}\:\Rightarrow{n}={p}^{\mathrm{2}} \:{but}\:{n}\:{is}\:{not}\:{square}\Rightarrow\sqrt{{n}}\notin{Q} \\ $$
Commented by mathocean1 last updated on 22/Oct/21
$${thanks}\:{mindispower} \\ $$