Prove that 1 2 is irrational.

Let us assume that 1 2 is a rational number. Then, there exists positive co-primes a and b such that, 1 2 = a b ⇒ b 2 = a ⇒ b = a 2 ⇒ 2 = b a Here, b a is rational as a and b are integers. This implies that 2 is rational which contradicts to the fact that 2 is irrational. Hence, our assumption was wrong and 1 2 is irrational. Hence, proved.

Prove that 3 + 5 is irrational.

Let us assume that 3 + 5 is rational. Then, there exists positive co-primes a and b such that, 3 + 5 = a b ⇒ 5 = a b - 3 Squaring both sides, ⇒ ( 5 ) 2 = a b - 3 2 ⇒ 5 = a 2 b 2 - 2 a 3 b + 3 ⇒ 2 = a 2 b 2 - 2 a 3 b ⇒ a 2 - 2 b 2 b 2 = 2 3 a b ⇒ 3 = a 2 - 2 b 2 2 a b Since a and b are integers, a 2 - 2 b 2 2 a b is rational. So, 3 is also rational. Now, this contradicts the fact that 3 is irrational. Thus, our assumption was wrong. Hence 3 + 5 is irrational.

Prove that 6 + 2 is irrational.

Given, 6 + 2 Let us assume that 6 + 2 is rational. Then, there exists positive co-primes a , b such that, 6 + 2 = a b ⇒ 2 = a b - 6 ⇒ 2 = a - 6 b b Since a and b are integers, a - 6 b b is also rational and hence, 2 should be rational. This contradicts the fact that 2 is irrational. Therefore, our assumption is false and 6 + 2 is irrational. Hence, proved.

Prove that 5 is irrational.

Given, 5 Let us assume that 5 is a rational number. Then, there exists positive co-primes a , b such that, 5 = a b ⇒ 5 b = a Squaring both sides, ⇒ 5 b 2 = a 2 . . . . ( i ) Here, a 2 is a multiple of 5 . So, a should also be a multiple of 5 . ⇒ a = 5 c Putting this value of a in ( i ) , ⇒ 5 b 2 = ( 5 c ) 2 ⇒ 5 b 2 = 25 c 2 ⇒ b 2 = 5 c 2 Here, b 2 is a multiple of 5 . So, b should also be a multiple of 5 . Here, a and b are the multiples of 5 , which contradicts our assumption that a and b are co-prime. Hence, 5 is irrational.

Prove that 3 + 2 5 is irrational.

Given, 3 + 2 5 Let us assume that 3 + 2 5 is a rational number. Then, there exists positive co-primes a , b such that, 3 + 2 5 = a b ⇒ 2 5 = a b - 3 ⇒ 5 = a - 3 b 2 b Since, a and b are integers, a - 3 b 2 b is also rational and hence, 5 should be rational. This contradicts the fact that 5 is irrational. Therefore, our assumption is false and 3 + 2 5 is an irrational number. Hence, proved.

Prove that p + q is an irrational, where p , q are primes.

Let us assume that p + q is a rational number. Then, there exist co-prime positive integers a and b such that, p + q = a b ; b ≠ 0 ⇒ a b - p = q Squaring both sides, ⇒ a b - p 2 = ( q ) 2 ⇒ a 2 b 2 - 2 a b p + p = q ⇒ a 2 b 2 + p - q = 2 a b p ⇒ a 2 + p - q b 2 b 2 = 2 a b p ⇒ a 2 - p - q b 2 2 a b = p Here, L.H.S. is a rational number while p is irrational because p is prime. Hence, our assumption is wrong. Thus, p + q is an irrational number.

E M B I B E

Mathematics Class 10>Chapter 1 - Real Numbers>Exercise

Telangana Board Solutions for Chapter: Real Numbers, Exercise 13: Exercise

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: Real Numbers, Exercise 13: Exercise

Attempt the practice questions on Chapter 1: Real Numbers, Exercise 13: Exercise with hints and solutions to strengthen your understanding. Mathematics Class 10 solutions are prepared by Experienced Embibe Experts.