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Question Number 45712 by malwaan last updated on 15/Oct/18

$$\mathrm{prove}\mathrm{that}:$$$$\sqrt{2}\mathrm{is}\mathrm{irrationl}\mathrm{number}$$

Commented by maxmathsup by imad last updated on 15/Oct/18

$$letsuppose\sqrt{2}=\frac{p}{q}withpandqintegrsnnaturaland\mathrm{\Delta }(p,q)=1\Rightarrow$$$$2=\frac{p^2}{q^2}\Rightarrow p^2=2q^2\Rightarrow 2/p^2\Rightarrow 2/p\Rightarrow \mathrm{\exists }m\in N/p=2m\Rightarrow 4m^2=2q^2\Rightarrow q^2=2m^2\Rightarrow$$$$2/q^2\Rightarrow 2/q\Rightarrow 2\in D_p\cap D_{q}butthisisimpossiblebecause\mathrm{\Delta }(p,q)=1$$$$finally\sqrt{2}\notin Q.$$

Commented by malwaan last updated on 15/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 16/Oct/18

$$youarewelcomesir.$$

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