Question Number 1865 by Denbang last updated on 18/Oct/15
$${proof}\:{that}\:\sqrt{\mathrm{2}}\:{is}\:{an}\:{irrational}\:{number} \\ $$$$ \\ $$
Answered by 123456 last updated on 18/Oct/15
$$\mathrm{suppuse}\:\mathrm{by}\:\mathrm{absurf}\:\mathrm{that}\:\sqrt{\mathrm{2}}\in\mathbb{Q},\:\mathrm{then} \\ $$$$\exists\left({p},{q}\right)\in\mathbb{Z},{q}\neq\mathrm{0}\:\mathrm{such}\:\mathrm{that}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}},\left({p},{q}\right)=\mathrm{1} \\ $$$$\mathrm{then} \\ $$$$\mathrm{2}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\Leftrightarrow{p}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \\ $$$$\mathrm{then}\:{p}^{\mathrm{2}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right)\Leftrightarrow{p}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{wich}\:\mathrm{imply}\:\mathrm{that} \\ $$$${p}=\mathrm{2}{k},{k}\in\mathbb{Z} \\ $$$$\mathrm{then} \\ $$$${p}^{\mathrm{2}} =\left(\mathrm{2}{k}\right)^{\mathrm{2}} =\mathrm{4}{k}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \Leftrightarrow{q}^{\mathrm{2}} =\mathrm{2}{k}^{\mathrm{2}} \\ $$$${q}^{\mathrm{2}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right)\Leftrightarrow{q}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{so}\:{p}\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2}\right)\:\mathrm{and}\:{q}\equiv\mathrm{0}\left(\mathrm{mod}\right) \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{imply}\:\left({p},{q}\right)\neq\mathrm{1},\:\mathrm{absurd}\:\blacksquare \\ $$