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Question Number 166910 by rexford last updated on 02/Mar/22

given that is prime,proof that (√p) is   irrational

$${given}\:{that}\:{is}\:{prime},{proof}\:{that}\:\sqrt{{p}}\:{is}\: \\ $$$${irrational} \\ $$

Commented by rexford last updated on 02/Mar/22

thank you

$${thank}\:{you} \\ $$

Answered by mr W last updated on 02/Mar/22

assume (√p)=rational=(a/b) with gcd(a,b)=1  ⇒p=(a^2 /b^2 ) ⇒p is a perfect square  ⇒contradiction    actually (√p) is always irrational, if  p is not a perfect square. therefore  (√2), (√3), (√5), (√6), (√7), (√8), (√(10)), (√(11)), (√(12)),  (√(13)), (√(14)), (√(15)), (√(17)),(√(18)), (√(19)), ... are all  irrational numbers.

$${assume}\:\sqrt{{p}}={rational}=\frac{{a}}{{b}}\:{with}\:{gcd}\left({a},{b}\right)=\mathrm{1} \\ $$$$\Rightarrow{p}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:\Rightarrow{p}\:{is}\:{a}\:{perfect}\:{square} \\ $$$$\Rightarrow{contradiction} \\ $$$$ \\ $$$${actually}\:\sqrt{{p}}\:{is}\:{always}\:{irrational},\:{if} \\ $$$${p}\:{is}\:{not}\:{a}\:{perfect}\:{square}.\:{therefore} \\ $$$$\sqrt{\mathrm{2}},\:\sqrt{\mathrm{3}},\:\sqrt{\mathrm{5}},\:\sqrt{\mathrm{6}},\:\sqrt{\mathrm{7}},\:\sqrt{\mathrm{8}},\:\sqrt{\mathrm{10}},\:\sqrt{\mathrm{11}},\:\sqrt{\mathrm{12}}, \\ $$$$\sqrt{\mathrm{13}},\:\sqrt{\mathrm{14}},\:\sqrt{\mathrm{15}},\:\sqrt{\mathrm{17}},\sqrt{\mathrm{18}},\:\sqrt{\mathrm{19}},\:...\:{are}\:{all} \\ $$$${irrational}\:{numbers}. \\ $$

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