Class 10 Maths Chapter 1: Real Numbers - Complete Guide with Solved Examples

Real Numbers is the first chapter in CBSE Class 10 Mathematics, and it sets the tone for the entire year. This chapter introduces concepts that are fundamental to higher mathematics, and it carries significant weightage in board exams. Many students rush through this chapter thinking it is "easy" and then lose marks on questions they should have gotten right.

This guide covers every concept in the chapter with solved examples and board exam tips.

Euclid's Division Lemma

Euclid's Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 is less than or equal to r, and r is less than b.

In simple terms, when you divide any positive integer a by another positive integer b, you get a quotient q and a remainder r, and the remainder is always between 0 and b-1.

This lemma is the foundation of Euclid's Division Algorithm, which is used to find the HCF (Highest Common Factor) of two numbers.

Euclid's Division Algorithm: Step by Step

To find the HCF of two numbers using Euclid's Division Algorithm, follow these steps:

Step 1: Apply the division lemma to the larger number (a) and the smaller number (b). Find q and r such that a = bq + r.

Step 2: If r = 0, then b is the HCF. Stop here.

Step 3: If r is not 0, apply the division lemma again with b and r. Now b becomes the new a, and r becomes the new b.

Step 4: Repeat until the remainder becomes 0. The divisor at this stage is the HCF.

Solved Example: Find the HCF of 455 and 42.

Step 1: 455 = 42 x 10 + 35 (here a = 455, b = 42, q = 10, r = 35) Step 2: r is not 0, so continue. Now a = 42, b = 35. Step 3: 42 = 35 x 1 + 7 (q = 1, r = 7) Step 4: r is not 0, so continue. Now a = 35, b = 7. Step 5: 35 = 7 x 5 + 0 (q = 5, r = 0) Step 6: r = 0, so HCF = 7.

Therefore, HCF(455, 42) = 7.

Solved Example: Find the HCF of 196 and 38220.

Step 1: 38220 = 196 x 195 + 0 Since the remainder is 0 in the very first step, HCF(196, 38220) = 196.

Board Exam Tip: Always show each step clearly. Write the division lemma equation for every step. Marks are awarded for the method, not just the answer.

The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of prime numbers, and this factorization is unique (apart from the order of the factors).

For example: 36 = 2 x 2 x 3 x 3 = 2 squared x 3 squared. No matter how you factorize 36, you will always get the same prime factors.

This theorem is used to find HCF and LCM of numbers using prime factorization.

Finding HCF and LCM Using Prime Factorization

To find HCF: Take the product of the smallest power of each common prime factor.

To find LCM: Take the product of the greatest power of each prime factor (common and uncommon).

Solved Example: Find HCF and LCM of 12, 15, and 21.

Prime factorizations: 12 = 2 squared x 3 15 = 3 x 5 21 = 3 x 7

HCF: The only common prime factor is 3 (with smallest power 1). HCF = 3

LCM: Take highest powers of all prime factors: 2 squared x 3 x 5 x 7 = 4 x 3 x 5 x 7 = 420. LCM = 420

Important Property: For any two positive integers a and b, HCF(a, b) x LCM(a, b) = a x b.

Solved Example: The HCF of two numbers is 4 and their LCM is 9696. If one number is 96, find the other.

Using the property: HCF x LCM = Product of two numbers 4 x 9696 = 96 x other number 38784 = 96 x other number Other number = 38784 / 96 = 404.

Board Exam Tip: This property only works for exactly two numbers. Do not apply it to three or more numbers.

Proving Numbers Irrational

This is one of the most frequently asked board exam questions. The method involves proof by contradiction.

To prove that the square root of 2 is irrational:

Assume the square root of 2 is rational. Then it can be expressed as p/q where p and q are co-prime integers (no common factor other than 1) and q is not 0.

So, square root of 2 = p/q Squaring both sides: 2 = p squared / q squared Therefore: p squared = 2 x q squared

This means p squared is divisible by 2, which means p is also divisible by 2. Let p = 2m for some integer m.

Substituting: (2m) squared = 2 x q squared 4m squared = 2 x q squared 2m squared = q squared

This means q squared is divisible by 2, which means q is also divisible by 2.

But if both p and q are divisible by 2, they have a common factor of 2. This contradicts our assumption that p and q are co-prime.

Therefore, our assumption was wrong. The square root of 2 is irrational.

Solved Example: Prove that 3 + 2 times the square root of 5 is irrational.

Assume 3 + 2 times the square root of 5 is rational. Then 3 + 2 times the square root of 5 = a/b where a and b are co-prime integers.

2 times the square root of 5 = a/b - 3 = (a - 3b)/b Square root of 5 = (a - 3b)/2b

Since a and b are integers, (a - 3b)/2b is rational. This means the square root of 5 is rational.

But we know the square root of 5 is irrational (can be proved similarly to the square root of 2). This is a contradiction.

Therefore, 3 + 2 times the square root of 5 is irrational.

Board Exam Tip: These proofs follow a standard pattern. Practice the proof for the square root of 2, square root of 3, and square root of 5. Most exam questions are variations of these.

Decimal Expansions of Rational Numbers

A key concept in this chapter is understanding when a rational number has a terminating decimal expansion and when it has a non-terminating repeating expansion.

A rational number p/q (in its simplest form) has a terminating decimal expansion if and only if the prime factorization of q has no prime factors other than 2 and 5.

In other words, q must be of the form 2 raised to n times 5 raised to m, where n and m are non-negative integers.

Solved Example: Without performing division, determine whether the following have terminating or non-terminating decimal expansions.

13/3125: 3125 = 5 raised to 5. Only factor is 5. Terminating.

17/8: 8 = 2 cubed. Only factor is 2. Terminating.

7/75: 75 = 3 x 5 squared. Has factor 3 (other than 2 and 5). Non-terminating repeating.

To convert a rational number with a terminating decimal expansion:

13/3125 = 13/5 raised to 5 = 13 x 2 raised to 5 / (5 raised to 5 x 2 raised to 5) = 13 x 32 / 10 raised to 5 = 416/100000 = 0.00416.

Important Board Exam Questions

Type 1: Find HCF and LCM using Euclid's Algorithm or prime factorization. These are almost guaranteed in every board paper.

Type 2: Prove that a given number is irrational. Practice the standard proofs and their variations.

Type 3: Determine whether a decimal expansion terminates or repeats.

Type 4: Application problems involving HCF and LCM (e.g., finding the largest tile size for a room, or the time when two events coincide again).

Solved Example (Application): Three bells ring at intervals of 9, 12, and 15 minutes. If they all ring together at 8:00 AM, when will they next ring together?

Find LCM of 9, 12, and 15. 9 = 3 squared, 12 = 2 squared x 3, 15 = 3 x 5. LCM = 2 squared x 3 squared x 5 = 4 x 9 x 5 = 180 minutes = 3 hours.

They will ring together at 11:00 AM.

Tips for Scoring Full Marks in This Chapter

Practice Euclid's Division Algorithm with different number pairs until the steps become automatic. Memorize the irrationality proof structure and practice at least five variations. Know the condition for terminating decimals by heart. For application problems, identify whether you need HCF or LCM: HCF for "largest" or "maximum" scenarios, LCM for "smallest" or "minimum" scenarios.

Padhaao's chapter-wise practice mode for Class 10 Maths covers all these question types with explanations, making it easy to drill this chapter until you are fully confident.

Real Numbers may seem straightforward, but it tests mathematical reasoning and proof-writing skills that carry over to other chapters. Master it thoroughly, and you build a strong foundation for the rest of Class 10 Mathematics.