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Prove that root 3 is irrational number |
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Answer» Let us assume that √3 is a rational number.then, as we know a rational number should be in the form of p/q\xa0where p and q are co- prime number.So,√3 = p/q { where p and q are co- prime}√3q = pNow, by squaring both the side\xa0we get,(√3q)² = p²3q² = p² ........ ( i )So,if 3 is the factor of p²then, 3 is also a factor of p ..... ( ii )=> Let p = 3m { where m is any integer }squaring both sidesp² = (3m)²p² = 9m²\xa0putting the value of p² in equation ( i )3q² = p²\xa03q² = 9m²q² = 3m²So,if 3 is factor of q²then, 3 is also factor of qSince3 is factor of p & q bothSo, our assumption that p & q are co- prime is wrong Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are coprime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. |
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