Published 2026-08-28
Chapter: Real Numbers

Real Numbers - Euclid's division lemma, fundamental theorem of arithmetic, and irrationality proofs

Real Numbers is the very first chapter of your NCERT textbook, and it sets the foundation for algebra, geometry, and beyond.

In this tutorial, we will break down the three pillar concepts of this chapter into simple, bite-sized, and logical steps:

  1. Euclid’s Division Lemma & Algorithm
  2. The Fundamental Theorem of Arithmetic (FTA)
  3. Proofs of Irrationality

Grab your notebook, a pen, and let's make Real Numbers easy and fun!


1. Euclid’s Division Lemma & Algorithm

What is a "Lemma"?

In mathematics, a lemma is a proven statement used as a stepping stone to prove another statement. Think of it as a trusted tool in your math toolbox!

The Concept: Division with a Twist

Do you remember how you checked your division problems in primary school? Dividend=(Divisor×Quotient)+Remainder\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}

Euclid’s Division Lemma is nothing more than this exact concept written in formal mathematical language!

Euclid's Division Lemma:
Given positive integers aa and bb, there exist unique integers qq and rr satisfying: a=bq+rwhere 0≤r<ba = bq + r \quad \text{where } 0 \le r < b

  • aa: Dividend (the number being divided)
  • bb: Divisor (the number dividing)
  • qq: Quotient (how many times bb goes into aa)
  • rr: Remainder (what is left over)

Analogy Time! 🍫
Imagine you have 17 chocolates (aa) and you want to pack them into boxes that hold 5 chocolates each (bb).
How many full boxes can you fill? 3 full boxes (qq).
How many chocolates remain unpacked? 2 chocolates (rr).

Writing this mathematically:
17=(5×3)+217 = (5 \times 3) + 2
Notice that the remaining chocolates (22) must always be less than the box size (55). That’s why 0≤r<b0 \le r < b!


Euclid's Division Algorithm: Finding the HCF

An algorithm is a step-by-step procedure for solving a problem. Euclid's Division Algorithm is used to calculate the Highest Common Factor (HCF) of two large numbers.

Step-by-step process to find HCF of two numbers aa and bb (a>ba > b):

  1. Apply Euclid's Division Lemma: a=bq+ra = bq + r.
  2. If r=0r = 0, then bb is the HCF!
  3. If r≠0r \neq 0, apply the lemma again to bb (new dividend) and rr (new divisor).
  4. Repeat the process until the remainder becomes 00. The divisor at this final stage is the HCF.

2. The Fundamental Theorem of Arithmetic (FTA)

The "Lego Bricks" of Mathematics 🧱

In chemistry, every complex substance is made up of fundamental elements (atoms). In mathematics, prime numbers are the atoms! Every composite number can be built by multiplying prime numbers together.

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

For example: 12=2×2×3=22×312 = 2 \times 2 \times 3 = 2^2 \times 3 No matter how you factorize 1212, you will always get two 22's and one 33.


Finding HCF and LCM using Prime Factorization

Using prime factorization, you can easily find the HCF and LCM of any set of positive integers:

  • HCF (Highest Common Factor): Product of the smallest power of each common prime factor involved in the numbers.
  • LCM (Lowest Common Multiple): Product of the greatest power of each prime factor involved in the numbers.

Important Formula:

For any two positive integers aa and bb: HCF(a,b)×LCM(a,b)=a×b\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b

(Note: This handy formula works only for two numbers, not three!)


3. Proofs of Irrationality

What is an Irrational Number?

A number is called irrational if it cannot be written in the form pq\frac{p}{q}, where pp and qq are integers and q≠0q \neq 0. Examples include 2,3,5,π\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi, etc.

The Key Theorem Needed for Proofs

Before proving irrationality, keep this critical theorem in mind:

Let pp be a prime number. If pp divides a2a^2 (where aa is a positive integer), then pp also divides aa.

Example: If 33 divides 626^2 (3636), then 33 also divides 66.


Proof by Contradiction: The "Math Detective" Method 🕵️‍♂️

To prove that a number like 2\sqrt{2} is irrational, we use a clever technique called Proof by Contradiction:

  1. We assume the opposite of what we want to prove (i.e., we assume 2\sqrt{2} is rational).
  2. We follow strict mathematical rules based on this assumption.
  3. We arrive at a logical absurdity (a contradiction).
  4. We conclude that our initial assumption was wrong, which means the number MUST be irrational!

Standard Proof: Prove that 2\sqrt{2} is irrational

  1. Assumption: Assume to the contrary that 2\sqrt{2} is rational.

  2. Formulation: Therefore, we can find two co-prime integers aa and bb (b≠0b \neq 0) such that: 2=ab\sqrt{2} = \frac{a}{b} (Co-prime means aa and bb have no common factor other than 11.)

  3. Squaring both sides: 2=a2b2  ⟹  2b2=a2— (Equation 1)2 = \frac{a^2}{b^2} \implies 2b^2 = a^2 \quad \text{--- (Equation 1)}

  4. Deduction 1: Since 22 divides a2a^2, by our theorem, 22 must also divide aa. So, we can write a=2ca = 2c for some integer cc.

  5. Substitution: Substitute a=2ca = 2c into Equation 1: 2b2=(2c)2  ⟹  2b2=4c2  ⟹  b2=2c22b^2 = (2c)^2 \implies 2b^2 = 4c^2 \implies b^2 = 2c^2

  6. Deduction 2: Since 22 divides b2b^2, 22 must also divide bb.

  7. The Contradiction: From steps 4 and 6, both aa and bb have at least 22 as a common factor. But this contradicts our initial fact that aa and bb are co-prime (have no common factors other than 1)!

  8. Conclusion: This contradiction has arisen because of our incorrect assumption that 2\sqrt{2} is rational. Hence, 2\sqrt{2} is irrational. ■\blacksquare


💡 Practice Questions with Solutions

Let’s test your understanding! Try solving these on your own first before reading the solutions.


Question 1

Find the HCF and LCM of 9090 and 144144 using the Prime Factorization method. Verify that HCF×LCM=Product of the two numbers\text{HCF} \times \text{LCM} = \text{Product of the two numbers}.

Solution:

Step 1: Write down the prime factorization of both numbers.

  • 90=2×3×3×5=21×32×5190 = 2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1
  • 144=2×2×2×2×3×3=24×32144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2

Step 2: Calculate HCF (smallest powers of common factors).

  • Common prime factors: 22 and 33.
  • Smallest power of 22 is 212^1.
  • Smallest power of 33 is 323^2. HCF(90,144)=21×32=2×9=18\text{HCF}(90, 144) = 2^1 \times 3^2 = 2 \times 9 = 18

Step 3: Calculate LCM (greatest powers of all factors involved).

  • Factors involved: 2,3,52, 3, 5.
  • Greatest power of 22 is 242^4.
  • Greatest power of 33 is 323^2.
  • Greatest power of 55 is 515^1. LCM(90,144)=24×32×51=16×9×5=720\text{LCM}(90, 144) = 2^4 \times 3^2 \times 5^1 = 16 \times 9 \times 5 = 720

Step 4: Verification.

  • LHS=HCF×LCM=18×720=12,960\text{LHS} = \text{HCF} \times \text{LCM} = 18 \times 720 = 12,960
  • RHS=Product of numbers=90×144=12,960\text{RHS} = \text{Product of numbers} = 90 \times 144 = 12,960

Since LHS=RHS\text{LHS} = \text{RHS}, the relationship HCF×LCM=a×b\text{HCF} \times \text{LCM} = a \times b is verified!


Question 2

Prove that 5−35 - \sqrt{3} is an irrational number, given that 3\sqrt{3} is irrational.

Solution:

Step 1: Let us assume to the contrary that 5−35 - \sqrt{3} is rational.

Step 2: Therefore, we can find co-prime integers aa and bb (b≠0b \neq 0) such that: 5−3=ab5 - \sqrt{3} = \frac{a}{b}

Step 3: Rearrange the equation to isolate 3\sqrt{3}: 3=5−ab\sqrt{3} = 5 - \frac{a}{b} 3=5b−ab\sqrt{3} = \frac{5b - a}{b}

Step 4: Analyze the result: Since aa and bb are integers, (5b−a)(5b - a) and bb are also integers. Therefore, 5b−ab\frac{5b - a}{b} is a rational number.

This implies that 3\sqrt{3} must also be a rational number!

Step 5: Conclusion: This contradicts the given fact that 3\sqrt{3} is irrational.

This contradiction arose because of our wrong assumption that 5−35 - \sqrt{3} is rational. Therefore, 5−35 - \sqrt{3} is irrational. ■\blacksquare


Question 3

Explain why 7×11×13+137 \times 11 \times 13 + 13 is a composite number.

Solution:

A number is called composite if it has factors other than 11 and itself (i.e., it can be expressed as a product of prime factors).

Given expression: E=7×11×13+13E = 7 \times 11 \times 13 + 13

Take out 1313 as a common factor: E=13×(7×11+1)E = 13 \times (7 \times 11 + 1) E=13×(77+1)E = 13 \times (77 + 1) E=13×78E = 13 \times 78

Now factorize 7878: 78=2×3×1378 = 2 \times 3 \times 13

Substitute back: E=13×(2×3×13)=2×3×132E = 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2

Since the given expression can be written as a product of prime factors (22, 33, and 1313), according to the Fundamental Theorem of Arithmetic, it must be a composite number. ■\blacksquare


🌟 Quick Recap & Revision Tips

  1. Euclid's Division Lemma: a=bq+ra = bq + r where 0≤r<b0 \le r < b.
  2. Fundamental Theorem of Arithmetic: Every composite number has a unique prime factorization.
  3. HCF & LCM Relation: HCF(a,b)×LCM(a,b)=a×b\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b.
  4. Irrationality Proofs: Always start by assuming the number is rational, use p/qp/q in simplest form, and find a logical contradiction!

Keep practicing your textbook exercises regularly. Math isn't about memorization—it's about building logical confidence step by step. You've got this!

Common Student Mistakes to Avoid

  1. Sign Errors in Algebraic Calculations: Mistakes in distributing negative signs across brackets or when transferring terms across the equals sign.
  2. Formula Misapplication: Memorizing formulas without checking required units or conditions (e.g. using diameter instead of radius).
  3. Skipping Intermediate Steps: Jumping directly to final numerical answers without showing step-by-step mathematical working, leading to partial credit loss in board exams.
  4. Incorrect Unit Conversions: Forgetting to convert parameters into uniform SI units (e.g., cm to meters or minutes to seconds) before computing.

Exam Preparation & Frequently Asked Questions (FAQ)

Q1. How should I revise Real Numbers for the Class 10 Mathematics examination?

Focus on mastering core textbook definitions, practicing 3-4 numerical problems daily with pen and paper, and reviewing previous year CBSE/NCERT board exam questions.

Q2. What are the key concepts that carry maximum marks in this chapter?

Pay special attention to core definitions, step-by-step derivations, solved textbook examples, and practical real-world applications outlined in your NCERT curriculum.

Q3. How can I avoid losing marks in long answer questions?

Always structure your answers with clear subheadings, write step-by-step working for numerical problems, state given values clearly, and highlight your final answers with correct SI units.

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