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Answer» Let us assume that √7\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}\\sqrt7=\\frac ab\\\\b\\sqrt7=a\\\\on\\;squaring\\;both\\;sides\\;we\\;get\\\\7b^2=a^2\\end{array}}{/tex}Therefore, a2 is divisible by 7,\xa0it follows that a is also divisible by 7.So, we can write a = 7c for some integer c.Substituting for a, we get 7b2 = 49c2, that is, b\u200b\u200b\u200b\u200b\u200b\u200b2\xa0= 7c2This means that b2 is divisible by 7, and so b is also divisible by 7\xa0Therefore, a and b have at least 7\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 √7\xa0is rational.So, we conclude that √7\xa0is irrational.
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