Prove that√3 is irrational

1.	Prove that√3 is irrational
Answer» Let us assume that √3 is a rational number.That is, we can find integers\xa0a\xa0and\xa0b\xa0(≠ 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\xa0(Squaring on both sides) → (1)Therefore, a2\xa0is divisible by 3Hence ‘a’ is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2\xa0=(3c)2⇒\xa03b2\xa0= 9c2∴ b2\xa0= 3c2This means that b2\xa0is 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.Now√2 = 1.4142...√3 = 1.7321...1.45, 1.5, 1.55, 1.6, 1.65, 1.7 lies between √2 and √3.Hence the rational numbers between √2 and √3 are:145/100, 15/10, 155/100, 16/10, 165/100 and 17/10 Let root3 is a rational no and let its simplest form a/b ,but b is not equal to 0Root 3 =a/b => 3=a^2/b^2 (sq.both sides) => 3b=a^2 (3 divides 3 b^2) 3 divides aLet a=3c for c is some integer 3b^2 =9c^2 => b^2 = 3c^2 => 3 divides b^2So,root 3 is an irrational. PROVED

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