squareroot 3SolutionLet us assume that 3 is rational Then 3

Solution

Let us assume that 3 is rational. Then 3 can be represented as a/b , where a and b are integers having no common factors. Now, on squaing both the sides, we get 3 = a²/b² or, 3b²=a². Hence a² is divisible by 3, so that a is also divisible by 3 (fundamental theorem of arithmetic). Let a = 3k, where k is an integer. Then we have 3b²= (3k)² = 9k² or b² = 3k² so that b = ± k 3 . This means that a and b have k as a common factor, which is a contradiction.Hence 3 is irrational.