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

### (i) 1/âˆš2

Solution:

Let assume that 1/âˆš2 is a rational number
Let us assume 1/âˆš2 = r where r is a rational number
1/r = âˆš2
We have assume that r is a rational number, 1/r = âˆš2 is also a rational number
But as we know that âˆš2 is an irrational number
So what we have assume is wrong.
So we can say that, 1/âˆš2 is an irrational number.

### (ii) 7âˆš5

Solution:

Letâ€™s assume that 7âˆš5 is a rational number.
Again assume that two positive integers a and b.
7âˆš5 = a/b here a and b are co-primes
â‡’ âˆš5 = a/7b
â‡’ âˆš5 is rational [ a and b are integers â‡’ a/7b is a rational number]
This shows that âˆš5 is irrational. So, our assumption is wrong.
So we can say that 7âˆš5 is an irrational number.

### (iii) 6 + âˆš2

Solution:

Letâ€™s assume on that 6+âˆš2 is a rational number.
Then, there are co prime positive integers a and b
6 + âˆš2 = a/b
â‡’ âˆš2 = a/b â€“ 6
â‡’ âˆš2 = (a â€“ 6b)/b
â‡’ âˆš2 is rational [(a-6b)/b is a rational number]
This contradicts that âˆš2 is irrational. So, our assumption is incorrect.
So we can say that 6 + âˆš2 is an irrational number.

### (iv) 3 âˆ’ âˆš5

Solution:

Letâ€™s assume on that 3-âˆš5 is a rational number.
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 [(a+3b)/b is a rational number]
This contradicts that âˆš5 is irrational. our assumption is incorrect.
So we can say that 3-âˆš5 is an irrational number.

### (i) 2/âˆš7

Solution:

Letâ€™s assume that 2/âˆš7 is a rational number.
There exist co-prime positive integers a and b
2/âˆš7 = a/b
â‡’ âˆš7 = 2b/a
â‡’ âˆš7 is rational [2b/a is a rational number]
This contradicts that âˆš7 is irrational. So, we can say that our assumption is incorrect.
So we can say that, 2/âˆš7 is an irrational number.

### (ii) 3/(2âˆš5)

Solution:

Letâ€™s assume that 3/(2âˆš5) is a rational number.
There exist co â€“ prime positive integers a and b
3/(2âˆš5) = a/b
â‡’ âˆš5 = 3b/2a
â‡’ âˆš5 is rational [3b/2a is a rational number]
This contradicts that âˆš5 is irrational. So, our assumption is incorrect.
So we can say that, 3/(2âˆš5) is an irrational number.

### (iii) 4 + âˆš2

Solution:

Letâ€™s assume on the contrary that 4 + âˆš2 is a rational number.
There exist co prime positive integers a and b
4 + âˆš2 = a/b
â‡’ âˆš2 = a/b â€“ 4
â‡’ âˆš2 = (a â€“ 4b)/b
â‡’ âˆš2 is rational [(a â€“ 4b)/b is a rational number]
This contradict that âˆš2 is irrational. So, our assumption is incorrect.
So we can say that 4 + âˆš2 is an irrational number.

### (iv) 5âˆš2

Solution:

Letâ€™s assume on that 5âˆš2 is a rational number.
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 [a/5b is a rational number]
This contradicts that âˆš2 is irrational. So, our assumption is incorrect.
So we can say that, 5âˆš2 is an irrational number.

### Question 3. Show that 2 âˆ’ âˆš3 is an irrational number.

Solution:

Letâ€™s assume that 2 â€“ âˆš3 is a rational number.
There exist co prime positive integers a and b
2 â€“ âˆš3= a/b
â‡’ âˆš3 = 2 â€“ a/b
â‡’ âˆš3 = (2b â€“ a)/b
â‡’ âˆš3 is rational [(2b â€“ a)/b is a rational number]
This contradicts that âˆš3 is irrational. So, our assumption is incorrect.
So we can say that, 2 â€“ âˆš3 is an irrational number.

### Question 4. Show that 3 + âˆš2 is an irrational number.

Solution:

Letâ€™s assume on that 3 + âˆš2 is a rational number.
There exist co prime positive integers a and b
3 + âˆš2= a/b
â‡’ âˆš2 = a/b â€“ 3
â‡’ âˆš2 = (a â€“ 3b)/b
â‡’ âˆš2 is rational [(a â€“ 3b)/b is a rational number]
This contradicts that âˆš2 is irrational. So, our assumption is incorrect.
So we can say that, 3 + âˆš2 is an irrational number.

### Question 5. Prove that 4 âˆ’ 5âˆš2 is an irrational number.

Solution:

Letâ€™s assume that 4 â€“ 5âˆš2 is a rational number.
There exist co prime positive integers a and b
4 â€“ 5âˆš2 = a/b
â‡’ 5âˆš2 = 4 â€“ a/b
â‡’ âˆš2 = (4b â€“ a)/(5b)
â‡’ âˆš2 is rational [(4b â€“ a)/5b is a rational number]
This contradicts that âˆš2 is irrational. So, our assumption is wrong.
So we can say that, 4 â€“ 5âˆš2 is an irrational number.

### Question 6. Show that 5 âˆ’ 2âˆš3 is an irrational number.

Solution:

Letâ€™s assume on that 5 â€“ 2âˆš3 is a rational number.
There exist co prime positive integers a and b
5 â€“ 2âˆš3 = a/b
â‡’ 2âˆš3 = 5 â€“ a/b
â‡’ âˆš3 = (5b â€“ a)/(2b)
â‡’ âˆš3 is rational [(5b â€“ a)/2b is a rational number]
This contradicts that âˆš3 is irrational. So, our assumption is wrong.
So we can say that, 5 â€“ 2âˆš3 is an irrational number.

### Question 7. Prove that 2âˆš3 âˆ’ 1 is an irrational number.

Solution:

Letâ€™s assume that 2âˆš3 â€“ 1 is a rational number.
There exist co prime positive integers a and b
2âˆš3 â€“ 1 = a/b
â‡’ 2âˆš3 = a/b + 1
â‡’ âˆš3 = (a + b)/(2b)
â‡’ âˆš3 is rational [(a + b)/2b is a rational number]
This contradicts that âˆš3 is irrational. So, our assumption is wrong.
So we can say that, 2âˆš3 â€“ 1 is an irrational number.

### Question 8. Prove that 2 âˆ’ 3âˆš5 is an irrational number.

Solution:

Letâ€™s assume on that 2 â€“ 3âˆš5 is a rational number.
There exist co prime positive integers a and b such that
2 â€“ 3âˆš5 = a/b
â‡’ 3âˆš5 = 2 â€“ a/b
â‡’ âˆš5 = (2b â€“ a)/(3b)
â‡’ âˆš5 is rational [(2b â€“ a)/3b is a rational number]
This contradicts that âˆš5 is irrational. So, our assumption is wrong.
So we can say that, 2 â€“ 3âˆš5 is an irrational number.

### Question 9. Prove that âˆš5 + âˆš3 is irrational.

Solution:

Letâ€™s assume on that âˆš5 + âˆš3 is a rational number.
There exist co prime positive integers a and b
âˆš5 + âˆš3 = a/b
â‡’ âˆš5 = (a/b) â€“ âˆš3
â‡’ (âˆš5)2 = ((a/b) â€“ âˆš3)2 [Squaring on both sides]
â‡’ 5 = (a2/b2) + 3 â€“ (2âˆš3a/b)
â‡’ (a2/b2) â€“ 2 = (2âˆš3a/b)
â‡’ (a/b) â€“ (2b/a) = 2âˆš3
â‡’ (a2 â€“ 2b2)/2ab = âˆš3
â‡’ âˆš3 is rational [(a2 â€“ 2b2)/2ab is rational]
This contradicts that âˆš3 is irrational. So, our assumption is wrong.
so we can say that, âˆš5 + âˆš3 is an irrational number.

### Question 10. Prove that âˆš2 + âˆš3 is irrational.

Solution:

Letâ€™s assume on that âˆš2 + âˆš3 is a rational number.
There exist co prime positive integers a and b.
âˆš2 + âˆš3 = a/b
â‡’ âˆš2 = (a/b) â€“ âˆš3
â‡’ (âˆš2)2 = ((a/b) â€“ âˆš3)2 [Squaring on both sides]
â‡’ 2 = (a2/b2) + 3 â€“ (2âˆš3a/b)
â‡’ (a2/b2) + 1 = (2âˆš3a/b)
â‡’ (a/b) + (b/a) = 2âˆš3
â‡’ (a2 + b2)/2ab = âˆš3
â‡’ âˆš3 is rational [(a2 + 2b2)/2ab is rational]
This contradicts that âˆš3 is irrational. So, our assumption is wrong.
So we can say that, âˆš2 + âˆš3 is an irrational number.

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

Solution:

Assume that âˆšp as a rational number
Again Assume that âˆšp = a/b where a and b are integers and b â‰  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, âˆšp is an irrational number.

### Question 12. If p, q are prime positive integers, prove that âˆšp + âˆšq 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) â€“ âˆšq
â‡’ (âˆšp)2 = ((a/b) â€“ âˆšq)2 [Squaring on both sides]
â‡’ p = (a2/b2) + q â€“ (2âˆšq a/b)
â‡’ (a2/b2) â€“ (p+q) = (2âˆšq a/b)
â‡’ (a/b) â€“ ((p+q)b/a) = 2âˆšq
â‡’ (a2 â€“ b2(p+q))/2ab = âˆšq
â‡’ âˆšq is rational [(a2 â€“ b2(p+q))/2ab is rational]
This contradicts that âˆšq is irrational. So, our assumption is wrong.
so we can say that, âˆšp + âˆšq is an irrational number.

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