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. Our subjects experts at BYJUâ€™S have created the RD Sharma Solutions Class 10 to make students understand the correct procedure to solve the exercise problems. In case you need any reference to any question of this exercise, you can access the RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.5 for which PDF is available below.

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

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

**1. Show that the following numbers are irrational.**

**(i) 1/âˆš2**

**Solution: **

Consider 1/âˆš2 is a rational number

Let us assume 1/âˆš2 = r where r is a rational number

On further calculation we get

1/r = âˆš2

Since r is a rational number, 1/r = âˆš2 is also a rational number

But we know that âˆš2 is an irrational number

So our supposition is wrong.

Hence, 1/âˆš2 is an irrational number.

**(ii) 7âˆš5**

**Solution: **

Letâ€™s assume on the contrary that 7âˆš5 is a rational number. Then, there exist positive integers a and b such that

7âˆš5 = a/b where, a and b, are co-primes

â‡’ âˆš5 = a/7b

â‡’ âˆš5 is rational [âˆµ 7, a and b are integers âˆ´ a/7b is a rational number]

This contradicts the fact that âˆš5 is irrational. So, our assumption is incorrect.

Hence, 7âˆš5 is an irrational number.

**(iii) 6 + âˆš2**

**Solution:**

Letâ€™s assume on the contrary that 6+âˆš2 is a rational number. Then, there exist co prime positive integers a and b such that

6 + âˆš2 = a/b

â‡’ âˆš2 = a/b â€“ 6

â‡’ âˆš2 = (a â€“ 6b)/b

â‡’ âˆš2 is rational [âˆµ a and b are integers âˆ´ (a-6b)/b is a rational number]

This contradicts the fact that âˆš2 is irrational. So, our assumption is incorrect.

Hence, 6 + âˆš2 is an irrational number.

**(iv)**^{ }**3 âˆ’ âˆš5**

**Solution:**

Letâ€™s assume on the contrary that 3-âˆš5 is a rational number. Then, there exist co prime positive integers a and b such that

3-âˆš5 = a/b

â‡’ âˆš5 = a/b + 3

â‡’ âˆš5 = (a + 3b)/b

â‡’ âˆš5 is rational [âˆµ a and b are integers âˆ´ (a+3b)/b is a rational number]

This contradicts the fact that âˆš5 is irrational. So, our assumption is incorrect.

Hence, 3-âˆš5 is an irrational number.

**2. Prove that the following numbers are irrationals.**

**(i) 2/âˆš7**

**Solution: **

Letâ€™s assume on the contrary that 2/âˆš7 is a rational number. Then, there exist co-prime positive integers a and b such that

2/âˆš7 = a/b

â‡’ âˆš7 = 2b/a

â‡’ âˆš7 is rational [âˆµ 2, a and b are integers âˆ´ 2b/a is a rational number]

This contradicts the fact that âˆš7 is irrational. So, our assumption is incorrect.

Hence, 2/âˆš7 is an irrational number.

**(ii) 3/(2âˆš5)**

**Solution:**

Letâ€™s assume on the contrary that 3/(2âˆš5) is a rational number. Then, there exist co â€“ prime positive integers a and b such that

3/(2âˆš5) = a/b

â‡’ âˆš5 = 3b/2a

â‡’ âˆš5 is rational [âˆµ 3, 2, a and b are integers âˆ´ 3b/2a is a rational number]

This contradicts the fact that âˆš5 is irrational. So, our assumption is incorrect.

Hence, 3/(2âˆš5) is an irrational number.

**(iii) 4 + âˆš2**

**Solution:**

Letâ€™s assume on the contrary that 4 + âˆš2 is a rational number. Then, there exist co prime positive integers a and b such that

4 + âˆš2 = a/b

â‡’ âˆš2 = a/b – 4

â‡’ âˆš2 = (a – 4b)/b

â‡’ âˆš2 is rational [âˆµ a and b are integers âˆ´ (a – 4b)/b is a rational number]

This contradicts the fact that âˆš2 is irrational. So, our assumption is incorrect.

Hence, 4 + âˆš2 is an irrational number.

**(iv) 5âˆš2**

**Solution:**

Letâ€™s assume on the contrary that 5âˆš2 is a rational number. Then, there exist positive integers a and b such that

5âˆš2 = a/b where, a and b, are co-primes

â‡’ âˆš2 = a/5b

â‡’ âˆš2 is rational [âˆµ a and b are integers âˆ´ a/5b is a rational number]

This contradicts the fact that âˆš2 is irrational. So, our assumption is incorrect.

Hence, 5âˆš2 is an irrational number.

**3. Show thatÂ 2 âˆ’ âˆš3Â is an irrational number.**

**Solution: **

Letâ€™s assume on the contrary that 2 – âˆš3 is a rational number. Then, there exist co prime positive integers a and b such that

2 – âˆš3= a/b

â‡’ âˆš3 = 2 – a/b

â‡’ âˆš3 = (2b – a)/b

â‡’ âˆš3 is rational [âˆµ a and b are integers âˆ´ (2b – a)/b is a rational number]

This contradicts the fact that âˆš3 is irrational. So, our assumption is incorrect.

Hence, 2 – âˆš3 is an irrational number.

**4. Show thatÂ 3 + âˆš2Â is an irrational number**.

**Solution: **

Letâ€™s assume on the contrary that 3 + âˆš2 is a rational number. Then, there exist co prime positive integers a and b such that

3 + âˆš2= a/b

â‡’ âˆš2 = a/b – 3

â‡’ âˆš2 = (a – 3b)/b

â‡’ âˆš2 is rational [âˆµ a and b are integers âˆ´ (a – 3b)/b is a rational number]

This contradicts the fact that âˆš2 is irrational. So, our assumption is incorrect.

Hence, 3 + âˆš2 is an irrational number.

**5. Prove thatÂ 4 âˆ’ 5âˆš2Â is an irrational number.**

**Solution: **

Letâ€™s assume on the contrary that 4 – 5âˆš2 is a rational number. Then, there exist co prime positive integers a and b such that

4 – 5âˆš2 = a/b

â‡’ 5âˆš2 = 4 – a/b

â‡’ âˆš2 = (4b â€“ a)/(5b)

â‡’ âˆš2 is rational [âˆµ 5, a and b are integers âˆ´ (4b – a)/5b is a rational number]

This contradicts the fact that âˆš2 is irrational. So, our assumption is incorrect.

Hence, 4 – 5âˆš2 is an irrational number.

**6. Show thatÂ 5 âˆ’ 2âˆš3Â is an irrational number.**

**Solution: **

Letâ€™s assume on the contrary that 5 – 2âˆš3 is a rational number. Then, there exist co prime positive integers a and b such that

5 – 2âˆš3 = a/b

â‡’ 2âˆš3 = 5 – a/b

â‡’ âˆš3 = (5b â€“ a)/(2b)

â‡’ âˆš3 is rational [âˆµ 2, a and b are integers âˆ´ (5b – a)/2b is a rational number]

This contradicts the fact that âˆš3 is irrational. So, our assumption is incorrect.

Hence, 5 – 2âˆš3 is an irrational number.

**7. Prove thatÂ 2âˆš3 âˆ’ 1Â is an irrational number.**

**Solution: **

Letâ€™s assume on the contrary that 2âˆš3 â€“ 1 is a rational number. Then, there exist co prime positive integers a and b such that

2âˆš3 – 1 = a/b

â‡’ 2âˆš3 = a/b + 1

â‡’ âˆš3 = (a + b)/(2b)

â‡’ âˆš3 is rational [âˆµ 2, a and b are integers âˆ´ (a + b)/2b is a rational number]

This contradicts the fact that âˆš3 is irrational. So, our assumption is incorrect.

Hence, 2âˆš3 â€“ 1 is an irrational number.

**8. Prove thatÂ 2 âˆ’ 3âˆš5Â is an irrational number. **

**Solution: **

Letâ€™s assume on the contrary that 2 – 3âˆš5 is a rational number. Then, there exist co prime positive integers a and b such that

2 – 3âˆš5 = a/b

â‡’ 3âˆš5 = 2 – a/b

â‡’ âˆš5 = (2b â€“ a)/(3b)

â‡’ âˆš5 is rational [âˆµ 3, a and b are integers âˆ´ (2b – a)/3b is a rational number]

This contradicts the fact that âˆš5 is irrational. So, our assumption is incorrect.

Hence, 2 – 3âˆš5 is an irrational number.

**9. Prove thatÂ **âˆš5 + âˆš3**Â is irrational.**

**Solution: **

Letâ€™s assume on the contrary that âˆš5 + âˆš3 is a rational number. Then, there exist co prime positive integers a and b such that

âˆš5 + âˆš3 = a/b

â‡’ âˆš5 = (a/b) – âˆš3

â‡’ (âˆš5)^{2} = ((a/b) – âˆš3)^{2} [Squaring on both sides]

â‡’ 5 = (a^{2}/b^{2}) + 3 â€“ (2âˆš3a/b)

â‡’ (a^{2}/b^{2}) – 2 = (2âˆš3a/b)

â‡’ (a/b) – (2b/a) = 2âˆš3

â‡’ (a^{2} – 2b^{2})/2ab = âˆš3

â‡’ âˆš3 is rational [âˆµ a and b are integers âˆ´ (a^{2} – 2b^{2})/2ab is rational]

This contradicts the fact that âˆš3 is irrational. So, our assumption is incorrect.

Hence, âˆš5 + âˆš3 is an irrational number.

**10. Prove thatÂ âˆš2 + âˆš3Â is irrational.**

**Solution:**

Letâ€™s assume on the contrary that âˆš2 + âˆš3 is a rational number. Then, there exist co prime positive integers a and b such that

âˆš2 + âˆš3 = a/b

â‡’ âˆš2 = (a/b) – âˆš3

â‡’ (âˆš2)^{2} = ((a/b) – âˆš3)^{2} [Squaring on both sides]

â‡’ 2 = (a^{2}/b^{2}) + 3 â€“ (2âˆš3a/b)

â‡’ (a^{2}/b^{2}) + 1 = (2âˆš3a/b)

â‡’ (a/b) + (b/a) = 2âˆš3

â‡’ (a^{2} + b^{2})/2ab = âˆš3

â‡’ âˆš3 is rational [âˆµ a and b are integers âˆ´ (a^{2} + 2b^{2})/2ab is rational]

This contradicts the fact that âˆš3 is irrational. So, our assumption is incorrect.

Hence, âˆš2 + âˆš3 is an irrational number.

**11.Â Prove that for any prime positive integer p,Â **âˆšp**Â is an irrational number.**

**Solution: **

Consider âˆšp as a rational number

Assume âˆšp = a/b where a and b are integers and b â‰ 0

By squaring on both sides

p = a^{2}/b^{2}

pb = a^{2}/b

p and b are integers pb= a^{2}/b will also be an integer

But we know that a^{2}/b is a rational number so our supposition is wrong

Therefore, âˆšp is an irrational number.

**12. If p, q are prime positive integers, prove thatÂ **âˆšp + âˆšq**Â is an irrational number.**

**Solution:**

Letâ€™s assume on the contrary that âˆšp + âˆšq is a rational number. Then, there exist co prime positive integers a and b such that

âˆšp + âˆšq = a/b

â‡’ âˆšp = (a/b) – âˆšq

â‡’ (âˆšp)^{2} = ((a/b) – âˆšq)^{2} [Squaring on both sides]

â‡’ p = (a^{2}/b^{2}) + q â€“ (2âˆšq a/b)

â‡’ (a^{2}/b^{2}) – (p+q) = (2âˆšq a/b)

â‡’ (a/b) – ((p+q)b/a) = 2âˆšq

â‡’ (a^{2} – b^{2}(p+q))/2ab = âˆšq

â‡’ âˆšq is rational [âˆµ a and b are integers âˆ´ (a^{2} – b^{2}(p+q))/2ab is rational]

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

Hence, âˆšp + âˆšq is an irrational number.