Prove that, 6 + 2 is irrational.

Let us assume that 6 + 2 is a rational number. Thus, 6 + 2 = p q ,where p and q are coprime numbers and q ≠ 0 We can write, 2 = p q - 6 2 = p - 6 q q ⇒ p - 6 q q is a rational number. But we know that 2 is an irrational number, it is contradicts to our assumption. Hence, 6 + 2 is irrational number.

Prove that, 6 + 2 is irrational.

Let us assume that 6 + 2 is a rational number. Thus, 6 + 2 = p q ,where p and q are coprime numbers and q ≠ 0 We can write, 2 = p q - 6 2 = p - 6 q q ⇒ p - 6 q q is a rational number. But we know that 2 is an irrational number, it is contradicts to our assumption. Hence, 6 + 2 is irrational number.

Prove that 7 5 is irrational.

Given, 7 5 Let us assume that 7 5 is a rational number. Then, there exists positive co-primes a , b such that, 7 5 = a b ⇒ 7 5 b = a Squaring both sides, ⇒ 245 b 2 = a 2 . . . . ( i ) Here, a 2 is a multiple of 245 . So, a should also be a multiple of 245 . ⇒ a = 245 c , where c is the other factor of a . Putting this value of a in ( i ) , ⇒ 245 b 2 = ( 245 c ) 2 ⇒ 245 b 2 = 245 2 c 2 ⇒ b 2 = 245 c 2 Here, b 2 is a multiple of 245 . So, b should also be a multiple of 245 . Here, a and b are the multiples of 245 , which contradicts our assumption that a and b are co-prime. Hence, 7 5 is irrational.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Real Numbers>Jharkhand Board-2018

Embibe Experts Solutions for Chapter: Real Numbers, Exercise 1: Jharkhand Board-2020

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Real Numbers, Exercise 1: Jharkhand Board-2020

Attempt the free practice questions on Chapter 1: Real Numbers, Exercise 1: Jharkhand Board-2020 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.