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Question Number 166910 by rexford last updated on 02/Mar/22
given that is prime,proof that (√p) is   irrational
giventhatisprime,proofthatpisirrational
Commented by rexford last updated on 02/Mar/22
thank you
thankyou
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.
assumep=rational=abwithgcd(a,b)=1p=a2b2pisaperfectsquarecontradictionactuallypisalwaysirrational,ifpisnotaperfectsquare.therefore2,3,5,6,7,8,10,11,12,13,14,15,17,18,19,areallirrationalnumbers.

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