Show that ( 5 − 2 3 ) is irrational.

Assuming x = 5 − 2 3 to be a rational number. x = − 2 3 ⇒ x 2 = ( 5 − 2 3 ) 2 x 2 = 5 2 + ( 2 3 ) 2 − 2 ( 5 ) ( 2 3 ) x 2 = 25 + 12 − 20 3 x 2 − 37 = − 20 3 ⇒ 37 − x 2 20 = 3 Since x is a rational number, x 2 is also a rational number. 37 − x 2 is a rational number 37 − x 2 20 is a rational number and 3 is a rational number But we know 3 is an irrational number. Which is a contradiction. Hence, our assumption is wrong. Thus, ( 5 − 2 3 ) is an irrational number.

Prove that 5 2 is irrational.

Assuming 5 2 to be a rational number. ∴ 5 2 = p q , where p and q are some integers and HCF ( p , q ) = 1 . . . ( 1 ) 5 2 q = p ( 5 2 q ) 2 = p 2 2 25 q 2 = p 2 p 2 is divisible by 2 p is divisible by 2 . . . ( 2 ) Let p = 2 m , where m is some integer. ∴ 5 2 = 2 m ( 5 2 q ) 2 = ( 2 m ) 2 2 25 q 2 = 4 m 2 25 q 2 = 2 m 2 q 2 is divisible by 2 q is divisible by 2 . . . ( 3 ) From 2 and 3 , 2 is a common factor of both p and q , which contradicts 1 . Hence, our assumption is wrong. Thus, 5 2 is irrational.

E M B I B E

Secondary School Mathematics for Class 10>Number Systems>Chapter 1 - Real Numbers>Exercise 1D

R S Aggarwal and Veena Aggarwal Solutions for Exercise 4: Exercise 1D

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 4: Exercise 1D

Attempt the free practice questions from Exercise 4: Exercise 1D with hints and solutions to strengthen your understanding. Secondary School Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.