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Class 10 RD Sharma Solutions- Chapter 1 Real Numbers – Exercise 1.5

  • Last Updated : 13 Jan, 2021

Question 1. Show that the following numbers are irrational.

(i) 1/√2

Solution:

Let assume that 1/√2 is a rational number 
Let us assume 1/√2 = r where r is a rational number 
1/r = √2 
We have assume that r is a rational number, 1/r = √2 is also a rational number 
But as we know that √2 is an irrational number 
So what we have assume is wrong. 
So we can say that, 1/√2 is an irrational number.

(ii) 7√5

Solution:

Let’s assume that 7√5 is a rational number. 
Again assume that two positive integers a and b. 
7√5 = a/b here a and b are co-primes 
⇒ √5 = a/7b 
⇒ √5 is rational [ a and b are integers ⇒ a/7b is a rational number] 
This shows that √5 is irrational. So, our assumption is wrong. 
So we can say that 7√5 is an irrational number.

(iii) 6 + √2

Solution:

Let’s assume on that 6+√2 is a rational number. 
Then, there are co prime positive integers a and b 
6 + √2 = a/b 
⇒ √2 = a/b – 6 
⇒ √2 = (a – 6b)/b 
⇒ √2 is rational [(a-6b)/b is a rational number] 
This contradicts that √2 is irrational. So, our assumption is incorrect. 
So we can say that 6 + √2 is an irrational number.

(iv) 3 − √5

Solution:

Let’s assume on that 3-√5 is a rational number. 
There exist co prime positive integers a and b such that 
3-√5 = a/b 
⇒ √5 = a/b + 3 
⇒ √5 = (a + 3b)/b 
⇒ √5 is rational [(a+3b)/b is a rational number] 
This contradicts that √5 is irrational. our assumption is incorrect. 
So we can say that 3-√5 is an irrational number.

Question 2. Prove that the following numbers are irrationals.

(i) 2/√7

Solution:



Let’s assume that 2/√7 is a rational number. 
There exist co-prime positive integers a and b 
2/√7 = a/b 
⇒ √7 = 2b/a 
⇒ √7 is rational [2b/a is a rational number] 
This contradicts that √7 is irrational. So, we can say that our assumption is incorrect. 
So we can say that, 2/√7 is an irrational number.

(ii) 3/(2√5)

Solution:

Let’s assume that 3/(2√5) is a rational number. 
There exist co – prime positive integers a and b 
3/(2√5) = a/b 
⇒ √5 = 3b/2a 
⇒ √5 is rational [3b/2a is a rational number] 
This contradicts that √5 is irrational. So, our assumption is incorrect. 
So we can say that, 3/(2√5) is an irrational number.

(iii) 4 + √2

Solution:

Let’s assume on the contrary that 4 + √2 is a rational number. 
There exist co prime positive integers a and b 
4 + √2 = a/b 
⇒ √2 = a/b – 4 
⇒ √2 = (a – 4b)/b 
⇒ √2 is rational [(a – 4b)/b is a rational number] 
This contradict that √2 is irrational. So, our assumption is incorrect. 
So we can say that 4 + √2 is an irrational number.

(iv) 5√2

Solution:

Let’s assume on that 5√2 is a rational number. 
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/5b is a rational number] 
This contradicts that √2 is irrational. So, our assumption is incorrect. 
So we can say that, 5√2 is an irrational number.

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

Solution:

Let’s assume that 2 – √3 is a rational number. 
There exist co prime positive integers a and b 
2 – √3= a/b 
⇒ √3 = 2 – a/b 
⇒ √3 = (2b – a)/b 
⇒ √3 is rational [(2b – a)/b is a rational number] 
This contradicts that √3 is irrational. So, our assumption is incorrect. 
So we can say that, 2 – √3 is an irrational number.

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

Solution:

Let’s assume on that 3 + √2 is a rational number. 
There exist co prime positive integers a and b 
3 + √2= a/b 
⇒ √2 = a/b – 3 
⇒ √2 = (a – 3b)/b 
⇒ √2 is rational [(a – 3b)/b is a rational number] 
This contradicts that √2 is irrational. So, our assumption is incorrect. 
So we can say that, 3 + √2 is an irrational number.

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

Solution:

Let’s assume that 4 – 5√2 is a rational number. 
There exist co prime positive integers a and b 
4 – 5√2 = a/b 
⇒ 5√2 = 4 – a/b 
⇒ √2 = (4b – a)/(5b) 
⇒ √2 is rational [(4b – a)/5b is a rational number] 
This contradicts that √2 is irrational. So, our assumption is wrong. 
So we can say that, 4 – 5√2 is an irrational number.

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

Solution:

Let’s assume on that 5 – 2√3 is a rational number. 
There exist co prime positive integers a and b 
5 – 2√3 = a/b 
⇒ 2√3 = 5 – a/b 
⇒ √3 = (5b – a)/(2b) 
⇒ √3 is rational [(5b – a)/2b is a rational number] 
This contradicts that √3 is irrational. So, our assumption is wrong. 
So we can say that, 5 – 2√3 is an irrational number.

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

Solution:

Let’s assume that 2√3 – 1 is a rational number. 
There exist co prime positive integers a and b 
2√3 – 1 = a/b 
⇒ 2√3 = a/b + 1 
⇒ √3 = (a + b)/(2b) 
⇒ √3 is rational [(a + b)/2b is a rational number] 
This contradicts that √3 is irrational. So, our assumption is wrong. 
So we can say that, 2√3 – 1 is an irrational number.

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

Solution:

Let’s assume on that 2 – 3√5 is a rational number. 
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 [(2b – a)/3b is a rational number] 
This contradicts that √5 is irrational. So, our assumption is wrong. 
So we can say that, 2 – 3√5 is an irrational number.

Question 9. Prove that √5 + √3 is irrational.

Solution:

Let’s assume on that √5 + √3 is a rational number. 
There exist co prime positive integers a and b 
√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 [(a2 – 2b2)/2ab is rational] 
This contradicts that √3 is irrational. So, our assumption is wrong. 
so we can say that, √5 + √3 is an irrational number.

Question 10. Prove that √2 + √3 is irrational.

Solution:

Let’s assume on that √2 + √3 is a rational number. 
There exist co prime positive integers a and b. 
√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 [(a2 + 2b2)/2ab is rational] 
This contradicts that √3 is irrational. So, our assumption is wrong. 
So we can say that, √2 + √3 is an irrational number.

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

Solution:

Assume that √p as a rational number 
Again Assume that √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 assumption is wrong 
So, √p is an irrational number.

Question 12. If p, 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 co prime 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 [(a2 – b2(p+q))/2ab is rational] 
This contradicts that √q is irrational. So, our assumption is wrong. 
so we can say that, √p + √q is an irrational number.

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