NCERT Solutions Class 10th Maths Chapter – 1 Real Numbers Examples

Example 1. Consider the numbers 4ⁿ, where n is a natural number. Check whether there is any value of n for which 4ⁿ ends with the digit zero.

Solution: If the number 4ⁿ, for any n, were to end with the digit zero, then it would be divisible by 5.

That is, the prime factorisation of 4ⁿ would contain the prime 5.

This is not possible because 4ⁿ = (2)²ⁿ;

so the only prime in the factorisation of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4ⁿ.

So, there is no natural number n for which 4ⁿ ends with the digit zero.

You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it! This method is also called the prime factorisation method. Let us recall this method through an Example.

Example 2. Find the LCM and HCF of 6 and 20 by the prime factorisation method.

Solution: We have:
6 = 2¹ x 3¹ and 20 = 2 × 2 × 5 = 2² × 5¹
You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes.

Note that HCF(6, 20) = 2¹

Product of the smallest power of each common prime factor in the numbers.

LCM (6, 20) = 2² × 3¹ × 5¹ = Product of the greatest power of each prime factor, involved in the numbers.

From the example above, you might have noticed that HCF(6, 20) × LCM(6,20) = 6 × 20. In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.

We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.

Example 3. Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.

Solution: The prime factorisation of 96 and 404 gives:

96 = 2⁵ × 3, 404 = 2² × 101

Therefore, the HCF of these two integers is 2² = 4.
Also,

LCM (96, 404) = 96 × 404/HCF(96, 404)
= 96 × 404/4
= 9696

Example 4. Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.

Solution: We have:

6 = 2×3, 72 = 2³ x 3², 120 = 2³ × 3 × 5

Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2¹ x 3¹ = 2 × 3 = 6

2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.

So,
LCM (6, 72, 120) = 2³ x 3² x 5¹ = 360

Remark: Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM.

Example 5. Prove that √3 is irrational.

Solution: Let us assume, to the contrary, that √3 is rational.

That is, we can find integers a and b (≠ 0) such that √3 = a/b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b√3 = a

Squaring on both sides, and rearranging, we get 3b² = a²

Therefore, a² is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.

So, we can write a = 3c for some integer c.
Substituting for a, we get 3b² = 9c², that is, b² = 3c²

This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3)

Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √3 is rational.

So, we conclude that √3 is irrational.

Example 6. Show that 5 – √3 is irrational.

Solution: Let us assume, to the contrary, that at 5 – √3 is rational

That is, we can find coprime a and b (b ≠ 0) such that 5 – √3 = a/b

Therefore, 5 – a/b = √3.

Rearranging this equation, we get √3 = 5 – a/b = 5b – a/b

Since a and b are integers, we get 5 – a/b is rational, and so √3 is rational.

But this contradicts the fact that √3 is irrational.

This contradiction has arisen because of our incorrect assumption that 5 – √3 is rational.

So, we conclude that 5 – √3 is irrational.

Example 7. Show that 3√2 is irrational.

Solution: Let us assume, to the contrary, that 3√2 is rational.

That is, we can find coprime a and b (b ≠ 0) such that 3√2 = a/b

Rearranging, we get √2 = a/3b

Since 3, a and b are integers, a/3b is rational, and so √2 is rational.

But this contradicts the fact that √2 is irrational. So, we conclude that 3√2 is irrational.