2nd way is by long division method.It shows that root3 has non terminating non repeating decimal expension.
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 co-prime.√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 co-prime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational.
1 hi way h prove krne ka
It has only one way not 3