prove the √2 is a irrational number by method of contradiction
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How can we prove that root 2 is an irrational number by the method of contradiction? Let us assume the contrary, that √2 is rational. So we can find integers r and s (not = 0) such that √2 = r/s suppose r and s have common factor other than 1. then we divide by common factor to get √2 = a/b where a and b are co-prime. b√2 = a squaring both sides, we get 2b^2 = a^2 . Therefore 2 divides a^2. Therefore 2 will also divide a. We can write a= 2c for some integer c. Substituting for a, we get 2b^2 = 4c^2 that is b^2 = 2c^2. this means that 2 divides b^2 so 2 will also divide b. Therefore and b have at least 2 as a common factor. This contradicts the fact that a and b have no common factors other than 1. This contradiction has come because our incorrect assumption of √2 as rational. Therefore √2 is irrational Hope it helps u definitely
Let root 2 be rational and let its simplest form be a/b. Then, a and b are the integers having no common factors other than 1, and b not equal to 0. Root2 = a/b --» 2 = a^2/ b^2 [ on squaring both sides] --» 2b^2 = a^2 ---» 2 divides a^2 [ 2 divides 2b^2] ---» 2 divides a [ 2 is a prime and divides a^2] Let a =2c Putting a = 2c in (i) 2b^2 = 4c^2 ---» b^2 = 2c^2 ---» 2 divides b^2 2 divides b [ 2 is prime and 2 divides b^2] Thus, 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This contradiction arises by assuming that root2 is rational. Hence, Root 2 is irrational. Hope it helps Mark me as brainliest Thanks...