Question Number 159015 by Armindo last updated on 11/Nov/21
Answered by mr W last updated on 12/Nov/21
$${say}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\:{with}\:{gcd}\left({p},{q}\right)=\mathrm{1} \\ $$$$\mathrm{2}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} } \\ $$$${p}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \:\Rightarrow{p}={even}=\mathrm{2}{k} \\ $$$$\mathrm{4}{k}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \\ $$$${q}^{\mathrm{2}} =\mathrm{2}{k}^{\mathrm{2}} \:\Rightarrow{q}={even}=\mathrm{2}{h} \\ $$$$\Rightarrow{both}\:{p}\:{and}\:{q}\:{are}\:{even},\:{gcd}\left({p},{q}\right)=\mathrm{2} \\ $$$$\Rightarrow{contradiction}\:{with}\:{gcd}\left({p},{q}\right)=\mathrm{1} \\ $$