Question Number 166910 by rexford last updated on 02/Mar/22
$${given}\:{that}\:{is}\:{prime},{proof}\:{that}\:\sqrt{{p}}\:{is}\: \\ $$$${irrational} \\ $$
Commented by rexford last updated on 02/Mar/22
$${thank}\:{you} \\ $$
Answered by mr W last updated on 02/Mar/22
$${assume}\:\sqrt{{p}}={rational}=\frac{{a}}{{b}}\:{with}\:{gcd}\left({a},{b}\right)=\mathrm{1} \\ $$$$\Rightarrow{p}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\Rightarrow{p}\:{is}\:{a}\:{perfect}\:{square} \\ $$$$\Rightarrow{contradiction} \\ $$$$ \\ $$$${actually}\:\sqrt{{p}}\:{is}\:{always}\:{irrational},\:{if} \\ $$$${p}\:{is}\:{not}\:{a}\:{perfect}\:{square}.\:{therefore} \\ $$$$\sqrt{\mathrm{2}},\:\sqrt{\mathrm{3}},\:\sqrt{\mathrm{5}},\:\sqrt{\mathrm{6}},\:\sqrt{\mathrm{7}},\:\sqrt{\mathrm{8}},\:\sqrt{\mathrm{10}},\:\sqrt{\mathrm{11}},\:\sqrt{\mathrm{12}}, \\ $$$$\sqrt{\mathrm{13}},\:\sqrt{\mathrm{14}},\:\sqrt{\mathrm{15}},\:\sqrt{\mathrm{17}},\sqrt{\mathrm{18}},\:\sqrt{\mathrm{19}},\:…\:{are}\:{all} \\ $$$${irrational}\:{numbers}. \\ $$