Prove that 2 - 3 is an irrational number.

Let us assume that 2 - 3 is a rational number. Now, let 2 - 3 = a b , where a , b are co-prime numbers and b ≠ 0 ⇒ 3 = 2 - a b = 2 b - a b Since a , b are integers, then 2 b - a b is also a rational number ⇒ 3 is a rational number But, we know that, 3 is an irrational number. This shows our initial assumption was incorrect. Therefore, 2 - 3 is an irrational number.

Prove that 3 + 2 is an irrational number.

Let us assume that 3 + 2 is a rational number. Now, let 3 + 2 = a b , where a , b are co-prime numbers and b ≠ 0 ⇒ 2 = a b - 3 = a - 3 b b Since a , b are integers, then a - 3 b b is also a rational number ⇒ 2 is a rational number But, we know that, 2 is an irrational number. This shows our initial assumption was incorrect. Therefore, 3 + 2 is an irrational number.

Prove that 4 - 5 2 is an irrational number.

Let us assume that 4 - 5 2 is a rational number. Now, let 4 - 5 2 = a b , where a , b are co-prime numbers and b ≠ 0 ⇒ 5 2 = 4 - a b = 4 b - a b ⇒ 2 = 4 b - a 5 b Since a , b are integers, then 4 b - a 5 b is also a rational number ⇒ 2 is a rational number But, we know that, 2 is an irrational number. This shows our initial assumption was incorrect. Therefore, 4 - 5 2 is an irrational number.

Prove that 2 3 - 1 is an irrational number.

Let us assume that 2 3 - 1 is a rational number. Now, let 2 3 - 1 = a b , where a , b are co-prime numbers and b ≠ 0 ⇒ 2 3 = 1 + a b = a + b b ⇒ 3 = a + b 2 b Since a , b are integers, then a + b 2 b is also a rational number ⇒ 3 is a rational number But, we know that, 3 is an irrational number. This shows our initial assumption was incorrect. Therefore, 2 3 - 1 is an irrational number.

Prove that 2 - 3 5 is an irrational number.

Let us assume that 2 - 3 5 is a rational number. Now, let 2 - 3 5 = a b , where a , b are co-prime numbers and b ≠ 0 ⇒ 3 5 = 2 - a b = 2 b - a b ⇒ 5 = 2 b - a 3 b Since a , b are integers, then 2 b - a 3 b is also a rational number ⇒ 5 is a rational number But, we know that, 5 is an irrational number. This shows our initial assumption was incorrect. Therefore, 2 - 3 5 is an irrational number.

Prove that 5 + 3 is an irrational number.

Assume that 5 + 3 is a rational number ' a ' ⇒ a = 5 + 3 Squaring both sides, ⇒ a 2 = 5 + 3 2 ⇒ a 2 = 5 + 3 + 2 5 3 ⇒ a 2 = 8 + 2 15 ⇒ 2 15 = a 2 - 8 ⇒ 15 = a 2 - 8 2 Now, a 2 - 8 2 is a rational number, since a is a rational number. ⇒ 15 is a rational number But, we know that, 15 is an irrational number. Hence, our assumption is wrong. Therefore, 5 + 3 is an irrational number.

Prove that for any prime positive integer p , p is an irrational number.

Let p is a rational number. As per the definition of rational numbers, we can express p = a b , {where a , b are co-prime numbers and b ≠ 0 }. On squaring both sides of equation, we get, p 2 = a 2 b 2 ⇒ p = a 2 b 2 ⇒ a 2 = p b 2 . . . . . . i Which means that, a 2 is a multiple of p , ⇒ p divides a Now, let a = p m ⇒ a 2 = p 2 m 2 Now putting this value of a 2 in equation i , p 2 m 2 = p b 2 ⇒ p m 2 = q 2 ⇒ p divides b 2 ⇒ p divides b This means that, p is a common divisor of a and b , which contradicts the assumption done in the first step. { a , b are co-primes numbers}. Therefore, p is an irrational number.

Prove that there is no positive integer n for which n - 1 + n + 1 is rational.

Suppose there exists a positive integer n for which n - 1 + n + 1 is a rational number. Let p q = n - 1 + n + 1 , where p and q are positive integers and q ≠ 0 ⇒ q p = 1 n - 1 + n + 1 ⇒ q p = 1 n - 1 + n + 1 × n - 1 - n + 1 n - 1 - n + 1 = n - 1 - n + 1 ( n - 1 ) - ( n + 1 ) = n - 1 - n + 1 - 2 ⇒ 2 q p = n + 1 - n - 1 Thus, we have, ( n - 1 + n + 1 ) + ( n + 1 - n - 1 ) = p q + 2 q p ⇒ 2 n + 1 = p 2 + 2 q 2 p q ⇒ n + 1 = p 2 + 2 q 2 2 p q - - - - i Also, we have, ( n - 1 + n + 1 ) - ( n + 1 - n - 1 ) = p q - 2 q p ⇒ 2 n - 1 = p 2 - 2 q 2 p q ⇒ n - 1 = p 2 - 2 q 2 2 p q - - - - i i From i and i i , we get, n + 1 and n - 1 are rational. [Since p , q are integers, p 2 + 2 q 2 2 p q and p 2 - 2 q 2 2 p q are rationals]. From i and i i , we also get, n + 1 and n - 1 are perfect square of positive integers. Now, ( n + 1 ) - ( n - 1 ) = 2 , which is not possible since any two perfect squares differ by at least 3 . Hence, there is no positive integer n for which n - 1 + n + 1 is a rational number. [proved]

E M B I B E

Modern's abc+ of Mathematics for Class 10>Chapter 1 - Real Numbers>EXERCISE

J P Mohindru and Bharat Mohindru Solutions for Chapter: Real Numbers, Exercise 4: EXERCISE

Author:J P Mohindru & Bharat Mohindru

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Real Numbers, Exercise 4: EXERCISE

Attempt the free practice questions on Chapter 1: Real Numbers, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. Modern's abc+ of Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.