RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.5

Irrational numbers are the numbers which cannot be expressed in the standard form of p/q. Proving the irrationality of numbers is clearly explained in this exercise. The subject experts at BYJU’S have created the RD Sharma Solutions Class 10 to make students understand the correct procedure to solve the exercise problems. So, students can access the RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.5 PDF, which is available below, and prepare well for the upcoming Class 10 board exam.

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Access Answers to RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.5

1. Show that the following numbers are irrational.

(i) 1/√2

Solution:

Consider 1/√2 is a rational number

Let us assume 1/√2 = r where r is a rational number

On further calculation, we get

1/r = √2

Since r is a rational number, 1/r = √2 is also a rational number

But we know that √2 is an irrational number

So, our supposition is wrong.

Hence, 1/√2 is an irrational number.

(ii) 7√5

Solution:

Let’s assume, on the contrary, that 7√5 is a rational number. Then, there exist positive integers a and b such that

7√5 = a/b, where a and b are co-primes

⇒ √5 = a/7b

⇒ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number]

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, 7√5 is an irrational number.

(iii) 6 + √2

Solution:

Let’s assume, on the contrary, that 6+√2 is a rational number. Then, there exist coprime positive integers a and b such that

6 + √2 = a/b

⇒ √2 = a/b – 6

⇒ √2 = (a – 6b)/b

⇒ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 6 + √2 is an irrational number.

(iv) 3 − √5

Solution:

Let’s assume, on the contrary, that 3-√5 is a rational number. Then, there exist coprime positive integers a and b such that

3-√5 = a/b

⇒ √5 = a/b + 3

⇒ √5 = (a + 3b)/b

⇒ √5 is rational [∵ a and b are integers ∴ (a+3b)/b is a rational number]

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, 3-√5 is an irrational number.

2. Prove that the following numbers are irrationals.

(i) 2/√7

Solution:

Let’s assume, on the contrary, that 2/√7 is a rational number. Then, there exist coprime positive integers a and b such that

2/√7 = a/b

⇒ √7 = 2b/a

⇒ √7 is rational [∵ 2, a and b are integers ∴ 2b/a is a rational number]

This contradicts the fact that √7 is irrational. So, our assumption is incorrect.

Hence, 2/√7 is an irrational number.

(ii) 3/(2√5)

Solution:

Let’s assume, on the contrary, that 3/(2√5) is a rational number. Then, there exist coprime positive integers a and b such that

3/(2√5) = a/b

⇒ √5 = 3b/2a

⇒ √5 is rational [∵ 3, 2, a and b are integers ∴ 3b/2a is a rational number]

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, 3/(2√5) is an irrational number.

(iii) 4 + √2

Solution:

Let’s assume, on the contrary, that 4 + √2 is a rational number. Then, there exist coprime positive integers a and b such that

4 + √2 = a/b

⇒ √2 = a/b – 4

⇒ √2 = (a – 4b)/b

⇒ √2 is rational [∵ a and b are integers ∴ (a – 4b)/b is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 4 + √2 is an irrational number.

(iv) 5√2

Solution:

Let’s assume, on the contrary, that 5√2 is a rational number. Then, there exist positive integers a and b such that

5√2 = a/b, where a and b are co-primes

⇒ √2 = a/5b

⇒ √2 is rational [∵ a and b are integers ∴ a/5b is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 5√2 is an irrational number.

3. Show that 2 − √3 is an irrational number.

Solution:

Let’s assume, on the contrary, that 2 – √3 is a rational number. Then, there exist coprime positive integers a and b such that

2 – √3= a/b

⇒ √3 = 2 – a/b

⇒ √3 = (2b – a)/b

⇒ √3 is rational [∵ a and b are integers ∴ (2b – a)/b is a rational number]

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, 2 – √3 is an irrational number.

4. Show that 3 + √2 is an irrational number.

Solution:

Let’s assume, on the contrary, that 3 + √2 is a rational number. Then, there exist coprime positive integers a and b such that

3 + √2= a/b

⇒ √2 = a/b – 3

⇒ √2 = (a – 3b)/b

⇒ √2 is rational [∵ a and b are integers ∴ (a – 3b)/b is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 3 + √2 is an irrational number.

5. Prove that 4 − 5√2 is an irrational number.

Solution:

Let’s assume, on the contrary, that 4 – 5√2 is a rational number. Then, there exist coprime positive integers a and b such that

4 – 5√2 = a/b

⇒ 5√2 = 4 – a/b

⇒ √2 = (4b – a)/(5b)

⇒ √2 is rational [∵ 5, a and b are integers ∴ (4b – a)/5b is a rational number]

This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

Hence, 4 – 5√2 is an irrational number.

6. Show that 5 − 2√3 is an irrational number.

Solution:

Let’s assume, on the contrary, that 5 – 2√3 is a rational number. Then, there exist coprime positive integers a and b such that

5 – 2√3 = a/b

⇒ 2√3 = 5 – a/b

⇒ √3 = (5b – a)/(2b)

⇒ √3 is rational [∵ 2, a and b are integers ∴ (5b – a)/2b is a rational number]

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, 5 – 2√3 is an irrational number.

7. Prove that 2√3 − 1 is an irrational number.

Solution:

Let’s assume, on the contrary, that 2√3 – 1 is a rational number. Then, there exist coprime positive integers a and b such that

2√3 – 1 = a/b

⇒ 2√3 = a/b + 1

⇒ √3 = (a + b)/(2b)

⇒ √3 is rational [∵ 2, a and b are integers ∴ (a + b)/2b is a rational number]

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, 2√3 – 1 is an irrational number.

8. Prove that 2 − 3√5 is an irrational number.

Solution:

Let’s assume, on the contrary, that 2 – 3√5 is a rational number. Then, there exist co-prime positive integers a and b such that

2 – 3√5 = a/b

⇒ 3√5 = 2 – a/b

⇒ √5 = (2b – a)/(3b)

⇒ √5 is rational [∵ 3, a and b are integers ∴ (2b – a)/3b is a rational number]

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, 2 – 3√5 is an irrational number.

9. Prove that √5 + √3 is irrational.

Solution:

Let’s assume, on the contrary, that √5 + √3 is a rational number. Then, there exist coprime positive integers a and b such that

√5 + √3 = a/b

⇒ √5 = (a/b) – √3

⇒ (√5)² = ((a/b) – √3)² [Squaring on both sides]

⇒ 5 = (a²/b²) + 3 – (2√3a/b)

⇒ (a²/b²) – 2 = (2√3a/b)

⇒ (a/b) – (2b/a) = 2√3

⇒ (a² – 2b²)/2ab = √3

⇒ √3 is rational [∵ a and b are integers ∴ (a² – 2b²)/2ab is rational]

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, √5 + √3 is an irrational number.

10. Prove that √2 + √3 is irrational.

Solution:

Let’s assume, on the contrary, that √2 + √3 is a rational number. Then, there exist coprime positive integers a and b such that

√2 + √3 = a/b

⇒ √2 = (a/b) – √3

⇒ (√2)² = ((a/b) – √3)² [Squaring on both sides]

⇒ 2 = (a²/b²) + 3 – (2√3a/b)

⇒ (a²/b²) + 1 = (2√3a/b)

⇒ (a/b) + (b/a) = 2√3

⇒ (a² + b²)/2ab = √3

⇒ √3 is rational [∵ a and b are integers ∴ (a² + 2b²)/2ab is rational]

This contradicts the fact that √3 is irrational. So, our assumption is incorrect.

Hence, √2 + √3 is an irrational number.

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

Solution:

Consider √p as a rational number

Assume √p = a/b where a and b are integers and b ≠ 0

By squaring on both sides

p = a²/b²

pb = a²/b

p and b are integers pb= a²/b will also be an integer

But we know that a²/b is a rational number, so our supposition is wrong

Therefore, √p is an irrational number.

12. If p and q are prime positive integers, prove that √p + √q is an irrational number.

Solution:

Let’s assume, on the contrary, that √p + √q is a rational number. Then, there exist coprime positive integers a and b such that

√p + √q = a/b

⇒ √p = (a/b) – √q

⇒ (√p)² = ((a/b) – √q)² [Squaring on both sides]

⇒ p = (a²/b²) + q – (2√q a/b)

⇒ (a²/b²) – (p+q) = (2√q a/b)

⇒ (a/b) – ((p+q)b/a) = 2√q

⇒ (a² – b²(p+q))/2ab = √q

⇒ √q is rational [∵ a and b are integers ∴ (a² – b²(p+q))/2ab is rational]

This contradicts the fact that √q is irrational. So, our assumption is incorrect.

Hence, √p + √q is an irrational number.