Real Numbers Exercise 1.2 Class 10
Real Numbers Exercise 1.2 Class 10 1. Prove that √5 is irrational Solution: Assume that √5 is rational. Then it can be written in the form 5 = a b \sqrt{5} = \frac{a}{b} where a and b are integers , b ≠ 0 b \neq 0 b = 0 and HCF(a, b) = 1 . Squaring both sides, 5 = a 2 b 2 5 = \frac{a^2}{b^2} a 2 = 5 b 2 a^2 = 5b^2 This means a 2 a^2 a 2 is divisible by 5 , so a must also be divisible by 5 . Let a = 5 k a = 5k Substitute in the equation: ( 5 k ) 2 = 5 b 2 (5k)^2 = 5b^2 25 k 2 = 5 b 2 25k^2 = 5b^2 b 2 = 5 k 2 b^2 = 5k^2 So b is also divisible by 5 . Thus a and b are both divisible by 5 , which contradicts the fact that HCF(a, b) = 1 . Therefore our assumption is wrong. Hence, √5 is irrational. 2. Prove that 3 + 2 5 is irrational Solution: Assume that 3 + 2 5 3 + 2\sqrt5 is rational. Then 2 5 = ( 3 + 2 5 ) − 3 2\sqrt5 = (3 + 2\sqrt5) - 3 The right side is rational, so 2 5 2\sqrt5 becomes rational . Dividing by 2, 5 = 2 5 2 This means √5 is rati...