RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5: Numbers that cannot be stated in the usual form of p/q are referred to as irrational numbers. This exercise explains how to prove the irrationality of numbers. Kopykitab’s subject experts have prepared the RD Sharma Solutions Class 10 to help students understand how to do workout problems correctly. If you need help with any of the questions in this exercise, you can use the RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.5, which is available in PDF format below.
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RD Sharma Class 10 Solutions Chapter 1 Exercise 1.5 PDF
Access answers to RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.5- Important Question with Answers
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.
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