Q. 174.8( 4 Votes )

Prove that <span

Answer :

Let us assume that √3 is a rational number.

That 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.


So, √3b = a


3b2 = a2 (Squaring on both sides)


Therefore, a2 is divisible by 3.


‘a’ is also divisible by 3.


So, we can write a = 3c for some integer c.


Equation (1) becomes,


3b2 = (3c)2


3b2 = 9c2


b2 = 3c2


This 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.


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