Question Number 159079 by LEKOUMA last updated on 12/Nov/21
$${Prove}\:{that}\:\:\mathrm{2}+\sqrt{\mathrm{3}}\:,\:\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{3}}\:\:{and}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{5}}\: \\ $$$${are}\:{the}\:{number}\:{irrational} \\ $$
Answered by mindispower last updated on 13/Nov/21
$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\in{Q}\Leftrightarrow\sqrt{\mathrm{3}}\in{Q} \\ $$$$\sqrt{\mathrm{3}}=\frac{{a}}{{b}},{a}\wedge{b}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{b}^{\mathrm{2}} ={a}^{\mathrm{2}} ….\mathrm{1}\Rightarrow\mathrm{3}\mid{a}\Rightarrow{a}=\mathrm{3}{s} \\ $$$$\mathrm{1}\Leftrightarrow{b}^{\mathrm{2}} =\mathrm{3}{s}^{\mathrm{2}} \Rightarrow\mathrm{3}\mid{b}\:…\mathrm{2} \\ $$$$\left(\mathrm{1}\right)\&\left(\mathrm{2}\right)\Rightarrow\mathrm{3}\mid{a}\wedge{b}=\mathrm{1}\:{absurd}\Rightarrow\sqrt{\mathrm{3}}\notin{Q} \\ $$$$\left(\mathrm{2}\right)\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{3}}=\frac{{a}}{{b}}\Rightarrow\sqrt{\mathrm{3}}=\frac{\mathrm{3}{a}+\mathrm{2}{b}}{{b}}\in{Q}\:{absurd}\:{previous}\:{proof} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{5}}=\frac{\sqrt{\mathrm{3}}+\mathrm{5}}{−\mathrm{21}}\in{Q}\Rightarrow\sqrt{\mathrm{3}}\in{Q}\:{absurd} \\ $$