Real Numbers Class 10: Notes, Proofs and Important Questions

This guide covers everything in Real Numbers Class 10, the first chapter of NCERT Class 10 Mathematics. You will learn how prime factorisation gives us HCF and LCM, why numbers like 4ⁿ can never end in 0, and how to prove that √2, √3 and √5 are irrational. Solved questions and common mistakes are at the end.

What the chapter is about

Real numbers include both rational and irrational numbers.

• A rational number can be written as a/b, where a and b are integers and b ≠ 0.
• An irrational number cannot be written in this form.

The whole chapter rests on one idea: prime numbers are the building blocks of positive integers. For example, 18 = 2 × 3 × 3. Changing the order of these blocks does not change the number, and no other collection of prime blocks makes 18.

We use this idea in two ways. First, prime factors explain HCF and LCM. Second, we use a proof by contradiction: we assume a square root is a fraction and show that this assumption cannot work.

The Fundamental Theorem of Arithmetic

A composite number is a positive integer greater than 1 that has at least one divisor other than 1 and itself.

Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

So 2 × 3 × 5 × 7 and 7 × 5 × 3 × 2 are the same factorisation. By convention, we write the prime factors in ascending order.

How to find a prime factorisation:

1. Start with the composite number, for example 32760.
2. Make a factor tree, breaking the number into its smallest prime factors step by step.
3. Write it as a product of primes: 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13.
4. Group equal primes using powers: 32760 = 2³ × 3² × 5 × 7 × 13.

Another example: 253 = 11 × 23.

HCF and LCM by prime factorisation

• HCF = product of the smallest power of each common prime factor. It is the largest number that divides all the given numbers exactly.
• LCM = product of the greatest power of each prime factor involved. It is the smallest number that is a multiple of all the given numbers.

For any two positive integers a and b:

HCF(a, b) × LCM(a, b) = a × b

Example: 6 and 20. 6 = 2¹ × 3¹ and 20 = 2² × 5¹. The only common prime is 2, and its smallest power is 2¹, so HCF = 2. The primes involved are 2, 3 and 5, with greatest powers 2², 3¹ and 5¹. So LCM = 4 × 3 × 5 = 60. Check: 2 × 60 = 120 and 6 × 20 = 120.

The rule is only for two numbers. For 12, 18 and 30: 12 = 2² × 3, 18 = 2 × 3², 30 = 2 × 3 × 5. HCF = 2 × 3 = 6 and LCM = 2² × 3² × 5 = 180. But 6 × 180 = 1080, while 12 × 18 × 30 = 6480. They are not equal.

Using prime factorisation to answer "can it end in 0?"

A number ends with the digit 0 only if its prime factorisation contains both 2 and 5, because 2 × 5 = 10.

• 4ⁿ = (2²)ⁿ = 2²ⁿ. The only prime factor is 2. There is no 5, so 4ⁿ can never end with 0 for any natural number n.
• 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. The prime factors are only 2 and 3. Since the factorisation is unique, 5 can never appear, so 6ⁿ cannot end with 0.

Divisibility of squares by primes

Theorem: Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Why it works: write a = p₁p₂...pₙ as primes. Then a² = p₁²p₂²...pₙ². If a prime p divides a², then by uniqueness of factorisation p must be one of p₁, ..., pₙ. So p divides a.

Example: 3 divides 81 = 9², and 3 also divides 9.

The condition that p is prime matters. 4 divides 36 = 6², but 4 does not divide 6.

Proving √2, √3 and √5 are irrational

Two numbers are coprime if their only common factor is 1, that is, their HCF is 1.

Proof that √2 is irrational (by contradiction):

1. Assume √2 is rational, so √2 = a/b, where a and b are coprime integers and b ≠ 0.
2. Square both sides: 2 = a²/b², so a² = 2b².
3. So 2 divides a², and therefore 2 divides a. Write a = 2c.
4. Substitute: (2c)² = 2b², so 4c² = 2b², which gives 2c² = b².
5. So 2 divides b², and therefore 2 divides b.
6. Now 2 is a common factor of a and b. This contradicts the assumption that they are coprime.

So the assumption is false, and √2 is irrational.

√3 works the same way. Assume √3 = a/b (coprime). Then a² = 3b², so 3 divides a. Let a = 3c. Then 9c² = 3b², so b² = 3c², and 3 divides b. Both a and b have the factor 3, a contradiction.

In general, if p is a prime number, √p is irrational. The same steps prove that √5 is irrational.

Arithmetic with irrational numbers

• Rational + irrational, or rational − irrational, is always irrational.
• Non-zero rational × irrational, or irrational ÷ non-zero rational, is irrational.

To show 5 − √3 is irrational, assume 5 − √3 = a/b. Then √3 = 5 − a/b. The right side is rational, so √3 would be rational. That is a contradiction.

To show 3√2 is irrational, assume 3√2 = a/b. Then √2 = a/(3b), which is rational. That is a contradiction.

The non-zero condition matters: 0 × √2 = 0, which is rational. Also, two irrational numbers can give a rational answer: √2 × √2 = 2.

Remember this

• Every composite number has exactly one prime factorisation, apart from order.
• HCF: common primes, smallest powers. LCM: all primes, greatest powers.
• HCF × LCM = a × b works only for two positive integers.
• A number ends in 0 only if both 2 and 5 are among its prime factors.
• If a prime p divides a², then p divides a.
• √p is irrational for every prime p.

Important questions with answers

1. State the Fundamental Theorem of Arithmetic. Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order of the prime factors.

2. Find the HCF and LCM of 12 and 18 by prime factorisation and check the product rule. 12 = 2² × 3 and 18 = 2 × 3². HCF = 2 × 3 = 6. LCM = 2² × 3² = 36. Check: 6 × 36 = 216 and 12 × 18 = 216.

3. If HCF(306, 657) = 9, find LCM(306, 657). 9 × LCM = 306 × 657, so LCM = (306 × 657) / 9 = 34 × 657 = 22338.

4. Find the HCF and LCM of 12, 18 and 30. Is HCF × LCM equal to the product of the three numbers? HCF = 6 and LCM = 180. HCF × LCM = 1080, but 12 × 18 × 30 = 6480. So no; the product rule is only for two numbers.

5. Can 6ⁿ end with the digit 0 for any natural number n? No. 6ⁿ = 2ⁿ × 3ⁿ. To end in 0, a number must have 5 as a prime factor. By the Fundamental Theorem of Arithmetic, the factorisation of 6ⁿ is unique and contains no 5.

6. Prove that √2 is irrational. Use the six steps above: assume √2 = a/b with a, b coprime, get a² = 2b², show 2 divides a and then 2 divides b, and reach the contradiction.

7. Prove that 5 − √3 is irrational. Assume it is rational, equal to a/b. Then √3 = 5 − a/b, which is rational. This contradicts the fact that √3 is irrational.

8. Show that 3√2 is irrational. Assume 3√2 = a/b. Then √2 = a/(3b), a rational number. This contradicts the irrationality of √2.

9. If r is rational and s is irrational, what is r + s? r + s is always irrational.

Common mistakes to avoid

• Using HCF × LCM = product for three numbers. It is true only for two.
• Including a prime in the HCF that does not appear in every number.
• Forgetting to say "a and b are coprime" at the start of an irrationality proof. The contradiction depends on it.
• Applying "p divides a², so p divides a" when p is not prime.
• Saying "rational × irrational is always irrational" without the word non-zero.

Practise these proofs and factorisations until each step feels natural, and study this chapter with Joy whenever you want a patient guide.