Question Number 158567 by MathsFan last updated on 06/Nov/21
$$\:\mathrm{show}\:\mathrm{that}\:\sqrt{\mathrm{3}}\:\mathrm{is}\:\mathrm{an}\: \\ $$$$\:\mathrm{irrarional}\:\mathrm{number} \\ $$
Commented by mr W last updated on 06/Nov/21
$${assume}\:\sqrt{\mathrm{3}}=\frac{{p}}{{q}}\:{with}\:{gcd}\left({p},{q}\right)=\mathrm{1} \\ $$$${p}^{\mathrm{2}} =\mathrm{3}{q}^{\mathrm{2}} \\ $$$$\Rightarrow{p}=\mathrm{3}{k} \\ $$$${q}^{\mathrm{2}} =\mathrm{3}{k}^{\mathrm{2}} \\ $$$$\Rightarrow{q}=\mathrm{3}{h} \\ $$$$\Rightarrow{gcd}\left({p},{q}\right)=\mathrm{3}\: \\ $$$$\Rightarrow{contradition}\:{with}\:{gcd}\left({p},{q}\right)=\mathrm{1} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\neq\frac{{p}}{{q}} \\ $$
Commented by MathsFan last updated on 06/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$