Prove that the given number is irrational. 1 2

To prove 1 2 is an irrational number. Take, 1 2 = p q where p and q are coprime ⇒ q = 2 p On squaring both sides, we get ⇒ q 2 = 2 p 2 ⇒ q 2 = 2 p 2 .......... i Here we can say that when q 2 is divisible by 2 , then q is also divisible by 2 . So, q = 2 r (where r is an integer) Now putting the value of q in eq. i , we get ⇒ 2 r 2 = 2 p 2 ⇒ 4 r 2 = 2 p 2 ⇒ r 2 = 2 p 2 4 ⇒ r 2 = p 2 2 So when p 2 is divided by 2 it gives r 2 , means p is also divisible by 2 . But we know that p and q are coprime. This is a contradiction and has arisen due to our wrong assumption. Therefore, 1 2 is an irrational number.

Prove that the given number is irrational. 3 + 5

To prove 3 + 5 is an irrational number. Let 3 + 5 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers So, 3 + 5 = p q ⇒ 3 = p q - 5 On squaring both the sides, ⇒ 3 2 = p q - 5 2 ⇒ 3 = p 2 q 2 - 2 5 p q + 5 ⇒ 2 5 p q = p 2 q 2 + 5 - 3 ⇒ 2 5 p q = p 2 q 2 + 2 ⇒ 5 p q = p 2 2 q 2 + 2 2 ⇒ 5 p q = p 2 2 q 2 + 1 ⇒ 5 p q = p 2 + 2 q 2 2 q 2 ⇒ 5 = p 2 + 2 q 2 2 q 2 × q p ⇒ 5 = p 2 + 2 q 2 2 p q Here when p & q are integer then p 2 + 2 q 2 2 p q is a rational number. It means 5 is also a rational number. This is a contradiction as 5 is an irrational number. This is a contradiction and has arisen due to our wrong assumption. Hence our assumption is false. Therefore, 3 + 5 is an irrational number.

Prove that the given number is irrational. 3 + 2 5

To prove 3 + 2 5 is an irrational number. Let 3 + 2 5 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers. So, 3 + 2 5 = p q ⇒ 2 5 = p q - 3 ⇒ 2 5 = p - 3 q q ⇒ 5 = p - 3 q 2 q Here when p & q are integer then p - 3 q 2 q is a rational number. It means 5 is also a rational number. But this contradicts the fact that 5 is an irrational number. This is a contradiction and has arisen due to our wrong assumption. Hence, our assumption is false. Therefore, 3 + 2 5 is an irrational number.

To divide 5 11 by 3 33 Now, 5 11 3 33 = 5 11 3 3 × 11 Using the property a b = a × b = 5 11 3 × 3 × 11 = 5 3 3 = 5 3 3 × 3 3 = 5 3 3 3 × 3 = 5 3 3 × 3 = 5 3 9 Therefore, 5 11 3 33 = 5 3 9 .

Solve 2 2 + 3 3 3 + 1 - 2 2 - 3 3 3 - 1

Given, 2 2 + 3 3 3 + 1 - 2 2 - 3 3 3 - 1 ⇒ 2 2 + 3 3 3 + 1 × 3 3 - 1 3 3 - 1 - 2 2 - 3 3 3 - 1 × 3 3 + 1 3 3 + 1 ⇒ 6 2 + 3 3 - 1 3 3 - 1 - 6 2 - 3 3 + 1 3 3 - 1 ⇒ 6 2 3 - 2 + 3 - 3 6 - 6 2 3 + 2 - 3 - 3 6 ⇒ 6 2 3 + 1 - 3 6 - 6 2 3 - 1 - 3 6 ⇒ 2 18 + 6 - 18 6 - 2 18 - 6 - 18 6 ⇒ 2 18 + 6 - 18 - 2 18 + 6 + 18 6 ⇒ 6 + 6 6 ⇒ 2 6 6 ⇒ 6 3 Hence, the required answer is 6 3 .

Find a rational number between 2 and 3 .

We know the decimal form of 2 and 3 as: 2 = 1 . 4142135 . . . . . 3 = 1 . 73205080 . . . . . . As we know that rational numbers are those decimal numbers which are either terminating or repeating. So, you can given any terminating or repeating decimal number between 1 . 4142135 . . . and 1 . 73205080 . . . as your answer. So, numbers between 2 and 3 are just greater than 1 . 41 and less than 1 . 73 . Thus answer can be any of the following rationals: 1 . 415 , 1 . 416 , 1 . 417 , 1 . 532 , 1 . 582 , 1 . 60 , 1 . 62 , 1 . 70 , 1 . 72 etc.

Prove that the given number is irrational. 5

To prove 5 is an irrational number. Let 5 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers. 5 = p q ⇒ 5 q = p On squaring both sides, ⇒ 5 q 2 = p 2 ⇒ 5 q 2 = p 2 .......... i ⇒ q 2 = p 2 5 Here we can say that when p 2 is divisible by 5 , then p is also divisible by 5 . So p = 5 r (where r is an integer) Now putting the value of p in eq i , 5 r 2 = 5 q 2 ⇒ 25 r 2 = 5 q 2 ⇒ r 2 = 5 q 2 25 = q 2 5 So, when q 2 is divided by 5 it gives r 2 , means q is also divisible by 5 . But p & q are coprime. This is a contradiction and has arisen due to our wrong assumption. Hence, 5 is irrational number.

Prove that the given number is irrational. 6 + 2

To prove 6 + 2 is an irrational number. Let 6 + 2 be a rational number. We know that a rational number can be written in the form of p q where p and q are coprime integers. So, 6 + 2 = p q ⇒ 2 = p q - 6 ⇒ 2 = p - 6 q q Here when p & q are integer then p - 6 q q is a rational number. It means 2 is also a rational number. But this contradicts the fact that 2 is an irrational number. This is a contradiction and has arisen due to our wrong assumption. Hence, our assumption is false. Therefore, 6 + 2 is an irrational number.

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Real Numbers>Exercise

Embibe Experts Solutions for Chapter: Real Numbers, Exercise 1: Exercise

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

Attempt the practice questions on Chapter 1: Real Numbers, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.