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Prove that√8 is a irrational?

Answer» Let us assume that √8\xa0is rational.That is, we can find integers a and b (≠0) such that\xa0a and b are co-prime{tex}\\style{font-family:Arial}{\\begin{array}{l}\\sqrt8=\\frac ab\\\\b\\sqrt8=a\\\\on\\;squaring\\;both\\;sides\\;we\\;get\\\\8b^2=a^2\\end{array}}{/tex}Therefore, a2 is divisible by 8,\xa0it follows that a is also divisible by 8.So, we can write a = 8c for some integer c.Substituting for a, we get 8b2 = 64c2, that is, b\u200b\u200b\u200b\u200b\u200b\u200b2\xa0= 8c2This means that b2 is divisible by 8, and so b is also divisible by 8\xa0Therefore, a and b have at least 8\xa0as a common factor.But this contradicts the fact that a and b are co-prime.This contradiction has arisen because of our incorrect assumption that √8\xa0is rational.So, we conclude that √8\xa0is irrational.
Firsty let root 8 is rational Then root8/1=p/q(p and q r coprime anf q is not equal to 0)Then cross multiplication then we obtainP=root8qSquare on both sideP2=(root8q)2Then we obtain P2=8q2 because root se square cancel ho jaygaThen we can say that p2 divide 8exactly and p will also divide 8 exaclty Then by euclid division lemmaP=8q+0Then square on both sideP2=(8q)2 Then put the value of p2 8q2=64q28q2=8(8q2)Then 8 se 8 cancel ho jaygaThenq2=8q2Then we can say that q2 will divide 8 exactly and q will divide exactly Then the common factor of p and p is 8 then it is contraduction to our supposition. So, our supposition is wrong . Hence root 8 is irrational no.


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