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Prove that ( root 3 ) is an irrational number.

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$eginarrayl extLet us assume on the contrary that sqrt3 ext is a rational number.\ extThen, there exist positive integers p and q such that\ sqrt3=fracpq extwe have taken, p and q, are co-prime i.e. their HCF is 1\ extNow, sqrt3=fracpq => 3q^2=p^2 , extit means that 3 extdivied p^2 ext, so it dived p also.\ extLet p=3c for some integer c.\ p^2=9c^2, =>3q^2=9c^2 => q^2=3c^2\ extit means that 3 extdivied q^2 ext, so it dived q also.\ extwe observe that p and q have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.\ extHence, sqrt3 extis an irrational number.endarray$

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Deependra Verma

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