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Answer» Let us assume that √3 is 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}\\sqrt3=\\frac ab\\\\b\\sqrt3=a\\\\on\\;squaring\\;both\\;sides\\;we\\;get\\\\3b^2=a^2\\end{array}}{/tex}Therefore, a2 is divisible by 3,\xa0it follows that a is also divisible by 3.So, we can write a = 3c for some integer c.Substituting for a, we get 3b2 = 9c2, that is, b\u200b\u200b\u200b\u200b\u200b\u200b2\xa0= 3c2This means that b2 is divisible by 3, and so b is also divisible by 3\xa0Therefore, a and b have at least 3 as 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 √3 is rational.So, we conclude that √3 is irrational.
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