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

• Last Updated : 13 Jan, 2021

### (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.

### (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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