Class 10 RD Sharma Solutions- Chapter 1 Real Numbers β Exercise 1.5
Question 1. Show that the following numbers are irrational.
(i) 1/β2
Solution:
Let assume that 1/β2 is a rational number
Let us assume 1/β2 = r where r is a rational number
1/r = β2
We have assume that r is a rational number, 1/r = β2 is also a rational number
But as we know that β2 is an irrational number
So what we have assume is wrong.
So we can say that, 1/β2 is an irrational number.
(ii) 7β5
Solution:
Letβs assume that 7β5 is a rational number.
Again assume that two positive integers a and b.
7β5 = a/b here a and b are co-primes
β β5 = a/7b
β β5 is rational [ a and b are integers β a/7b is a rational number]
This shows that β5 is irrational. So, our assumption is wrong.
So we can say that 7β5 is an irrational number.
(iii) 6 + β2
Solution:
Letβs assume on that 6+β2 is a rational number.
Then, there are co prime positive integers a and b
6 + β2 = a/b
β β2 = a/b β 6
β β2 = (a β 6b)/b
β β2 is rational [(a-6b)/b is a rational number]
This contradicts that β2 is irrational. So, our assumption is incorrect.
So we can say that 6 + β2 is an irrational number.
(iv) 3 β β5
Solution:
Letβs assume on that 3-β5 is a rational number.
There exist co prime positive integers a and b such that
3-β5 = a/b
β β5 = a/b + 3
β β5 = (a + 3b)/b
β β5 is rational [(a+3b)/b is a rational number]
This contradicts that β5 is irrational. our assumption is incorrect.
So we can say that 3-β5 is an irrational number.
Question 2. Prove that the following numbers are irrationals.
(i) 2/β7
Solution:
Letβs assume that 2/β7 is a rational number.
There exist co-prime positive integers a and b
2/β7 = a/b
β β7 = 2b/a
β β7 is rational [2b/a is a rational number]
This contradicts that β7 is irrational. So, we can say that our assumption is incorrect.
So we can say that, 2/β7 is an irrational number.
(ii) 3/(2β5)
Solution:
Letβs assume that 3/(2β5) is a rational number.
There exist co β prime positive integers a and b
3/(2β5) = a/b
β β5 = 3b/2a
β β5 is rational [3b/2a is a rational number]
This contradicts that β5 is irrational. So, our assumption is incorrect.
So we can say that, 3/(2β5) is an irrational number.
(iii) 4 + β2
Solution:
Letβs assume on the contrary that 4 + β2 is a rational number.
There exist co prime positive integers a and b
4 + β2 = a/b
β β2 = a/b β 4
β β2 = (a β 4b)/b
β β2 is rational [(a β 4b)/b is a rational number]
This contradict that β2 is irrational. So, our assumption is incorrect.
So we can say that 4 + β2 is an irrational number.
(iv) 5β2
Solution:
Letβs assume on that 5β2 is a rational number.
There exist positive integers a and b such that
5β2 = a/b where, a and b, are co-primes
β β2 = a/5b
β β2 is rational [a/5b is a rational number]
This contradicts that β2 is irrational. So, our assumption is incorrect.
So we can say that, 5β2 is an irrational number.
Question 3. Show that 2 β β3 is an irrational number.
Solution:
Letβs assume that 2 β β3 is a rational number.
There exist co prime positive integers a and b
2 β β3= a/b
β β3 = 2 β a/b
β β3 = (2b β a)/b
β β3 is rational [(2b β a)/b is a rational number]
This contradicts that β3 is irrational. So, our assumption is incorrect.
So we can say that, 2 β β3 is an irrational number.
Question 4. Show that 3 + β2 is an irrational number.
Solution:
Letβs assume on that 3 + β2 is a rational number.
There exist co prime positive integers a and b
3 + β2= a/b
β β2 = a/b β 3
β β2 = (a β 3b)/b
β β2 is rational [(a β 3b)/b is a rational number]
This contradicts that β2 is irrational. So, our assumption is incorrect.
So we can say that, 3 + β2 is an irrational number.
Question 5. Prove that 4 β 5β2 is an irrational number.
Solution:
Letβs assume that 4 β 5β2 is a rational number.
There exist co prime positive integers a and b
4 β 5β2 = a/b
β 5β2 = 4 β a/b
β β2 = (4b β a)/(5b)
β β2 is rational [(4b β a)/5b is a rational number]
This contradicts that β2 is irrational. So, our assumption is wrong.
So we can say that, 4 β 5β2 is an irrational number.
Question 6. Show that 5 β 2β3 is an irrational number.
Solution:
Letβs assume on that 5 β 2β3 is a rational number.
There exist co prime positive integers a and b
5 β 2β3 = a/b
β 2β3 = 5 β a/b
β β3 = (5b β a)/(2b)
β β3 is rational [(5b β a)/2b is a rational number]
This contradicts that β3 is irrational. So, our assumption is wrong.
So we can say that, 5 β 2β3 is an irrational number.
Question 7. Prove that 2β3 β 1 is an irrational number.
Solution:
Letβs assume that 2β3 β 1 is a rational number.
There exist co prime positive integers a and b
2β3 β 1 = a/b
β 2β3 = a/b + 1
β β3 = (a + b)/(2b)
β β3 is rational [(a + b)/2b is a rational number]
This contradicts that β3 is irrational. So, our assumption is wrong.
So we can say that, 2β3 β 1 is an irrational number.
Question 8. Prove that 2 β 3β5 is an irrational number.
Solution:
Letβs assume on that 2 β 3β5 is a rational number.
There exist co prime positive integers a and b such that
2 β 3β5 = a/b
β 3β5 = 2 β a/b
β β5 = (2b β a)/(3b)
β β5 is rational [(2b β a)/3b is a rational number]
This contradicts that β5 is irrational. So, our assumption is wrong.
So we can say that, 2 β 3β5 is an irrational number.
Question 9. Prove that β5 + β3 is irrational.
Solution:
Letβs assume on that β5 + β3 is a rational number.
There exist co prime positive integers a and b
β5 + β3 = a/b
β β5 = (a/b) β β3
β (β5)2 = ((a/b) β β3)2 [Squaring on both sides]
β 5 = (a2/b2) + 3 β (2β3a/b)
β (a2/b2) β 2 = (2β3a/b)
β (a/b) β (2b/a) = 2β3
β (a2 β 2b2)/2ab = β3
β β3 is rational [(a2 β 2b2)/2ab is rational]
This contradicts that β3 is irrational. So, our assumption is wrong.
so we can say that, β5 + β3 is an irrational number.
Question 10. Prove that β2 + β3 is irrational.
Solution:
Letβs assume on that β2 + β3 is a rational number.
There exist co prime positive integers a and b.
β2 + β3 = a/b
β β2 = (a/b) β β3
β (β2)2 = ((a/b) β β3)2 [Squaring on both sides]
β 2 = (a2/b2) + 3 β (2β3a/b)
β (a2/b2) + 1 = (2β3a/b)
β (a/b) + (b/a) = 2β3
β (a2 + b2)/2ab = β3
β β3 is rational [(a2 + 2b2)/2ab is rational]
This contradicts that β3 is irrational. So, our assumption is wrong.
So we can say that, β2 + β3 is an irrational number.
Question 11. Prove that for any prime positive integer p, βp is an irrational number.
Solution:
Assume that βp as a rational number
Again Assume that βp = a/b where a and b are integers and b β 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, βp is an irrational number.
Question 12. If p, q are prime positive integers, prove that βp + βq is an irrational number.
Solution:
Letβs assume on the contrary that βp + βq is a rational number.
Then, there exist co prime positive integers a and b such that
βp + βq = a/b
β βp = (a/b) β βq
β (βp)2 = ((a/b) β βq)2 [Squaring on both sides]
β p = (a2/b2) + q β (2βq a/b)
β (a2/b2) β (p+q) = (2βq a/b)
β (a/b) β ((p+q)b/a) = 2βq
β (a2 β b2(p+q))/2ab = βq
β βq is rational [(a2 β b2(p+q))/2ab is rational]
This contradicts that βq is irrational. So, our assumption is wrong.
so we can say that, βp + βq is an irrational number.