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

Chapter 1 of Class 10 Mathematics, titled "Real Numbers" is foundational for the understanding the various concepts in algebra and number theory. This chapter explores the properties of the real numbers including their operations the Euclidean algorithm and fundamental theorem of arithmetic. Exercise 1.5 in this chapter focuses on the applying these concepts through the various problems enhancing students' problem-solving skills.

Real Numbers

The Real numbers include all rational and irrational numbers encompassing integers, fractions and decimals. They form the basis for the many mathematical operations and are used to the represent continuous quantities. The Real numbers can be visualized on a number line where every point corresponds to a unique real number.

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.

Read more :

• Real-Life Applications of Real Numbers
• Real Numbers

Summary

Exercise 1.5 of RD Sharma's Class 10 Mathematics textbook focuses on the concept of finding square roots and cube roots of real numbers. This section covers techniques for calculating square roots and cube roots of perfect squares and cubes, as well as estimating roots for non-perfect squares and cubes. Students learn to simplify expressions involving square roots and cube roots, solve equations containing roots, and apply these concepts to real-world problems. The exercise also includes questions on the properties of square roots and cube roots, such as their behavior with negative numbers and fractional exponents.