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Class 10 Mathematics
Real Numbers - Euclid's division lemma, fundamental theorem of arithmetic, and irrationality proofs
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MathematicsClass 10Real Numbers

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

2026-08-289 min readRHS Academic Faculty
Overview & Key Summary:Class 10 Mathematics: Master Real Numbers Real Numbers is the very first chapter of your NCERT textbook, and it sets the foundation for algebra, geometry, and beyond. In this tu...

Class 10 Mathematics: Master Real Numbers

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}Dividend=(Divisor×Quotient)+Remainder

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

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

  • aaa: Dividend (the number being divided)
  • bbb: Divisor (the number dividing)
  • qqq: Quotient (how many times bbb goes into aaa)
  • rrr: Remainder (what is left over)

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

Writing this mathematically:
17=(5×3)+217 = (5 \times 3) + 217=(5×3)+2
Notice that the remaining chocolates (222) must always be less than the box size (555). That’s why 0≤r<b0 \le r < b0≤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 aaa and bbb (a>ba > ba>b):

  1. Apply Euclid's Division Lemma: a=bq+ra = bq + ra=bq+r.
  2. If r=0r = 0r=0, then bbb is the HCF!
  3. If r≠0r \neq 0r=0, apply the lemma again to bbb (new dividend) and rrr (new divisor).
  4. Repeat the process until the remainder becomes 000. 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 312=2×2×3=22×3 No matter how you factorize 121212, you will always get two 222's and one 333.


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 aaa and bbb: HCF(a,b)×LCM(a,b)=a×b\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times bHCF(a,b)×LCM(a,b)=a×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}qp​, where ppp and qqq are integers and q≠0q \neq 0q=0. Examples include 2,3,5,π\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi2​,3​,5​,π, etc.

The Key Theorem Needed for Proofs

Before proving irrationality, keep this critical theorem in mind:

Let ppp be a prime number. If ppp divides a2a^2a2 (where aaa is a positive integer), then ppp also divides aaa.

Example: If 333 divides 626^262 (363636), then 333 also divides 666.


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

To prove that a number like 2\sqrt{2}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}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}2​ is irrational

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

  2. Formulation: Therefore, we can find two co-prime integers aaa and bbb (b≠0b \neq 0b=0) such that: 2=ab\sqrt{2} = \frac{a}{b}2​=ba​ (Co-prime means aaa and bbb have no common factor other than 111.)

  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)}2=b2a2​⟹2b2=a2— (Equation 1)

  4. Deduction 1: Since 222 divides a2a^2a2, by our theorem, 222 must also divide aaa. So, we can write a=2ca = 2ca=2c for some integer ccc.

  5. Substitution: Substitute a=2ca = 2ca=2c into Equation 1: 2b2=(2c)2  ⟹  2b2=4c2  ⟹  b2=2c22b^2 = (2c)^2 \implies 2b^2 = 4c^2 \implies b^2 = 2c^22b2=(2c)2⟹2b2=4c2⟹b2=2c2

  6. Deduction 2: Since 222 divides b2b^2b2, 222 must also divide bbb.

  7. The Contradiction: From steps 4 and 6, both aaa and bbb have at least 222 as a common factor. But this contradicts our initial fact that aaa and bbb 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}2​ is rational. Hence, 2\sqrt{2}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 909090 and 144144144 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}HCF×LCM=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 5^190=2×3×3×5=21×32×51
  • 144=2×2×2×2×3×3=24×32144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2144=2×2×2×2×3×3=24×32

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

  • Common prime factors: 222 and 333.
  • Smallest power of 222 is 212^121.
  • Smallest power of 333 is 323^232. HCF(90,144)=21×32=2×9=18\text{HCF}(90, 144) = 2^1 \times 3^2 = 2 \times 9 = 18HCF(90,144)=21×32=2×9=18

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

  • Factors involved: 2,3,52, 3, 52,3,5.
  • Greatest power of 222 is 242^424.
  • Greatest power of 333 is 323^232.
  • Greatest power of 555 is 515^151. 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 = 720LCM(90,144)=24×32×51=16×9×5=720

Step 4: Verification.

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

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


Question 2

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

Solution:

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

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

Step 3: Rearrange the equation to isolate 3\sqrt{3}3​: 3=5−ab\sqrt{3} = 5 - \frac{a}{b}3​=5−ba​ 3=5b−ab\sqrt{3} = \frac{5b - a}{b}3​=b5b−a​

Step 4: Analyze the result: Since aaa and bbb are integers, (5b−a)(5b - a)(5b−a) and bbb are also integers. Therefore, 5b−ab\frac{5b - a}{b}b5b−a​ is a rational number.

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

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

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


Question 3

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

Solution:

A number is called composite if it has factors other than 111 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 + 13E=7×11×13+13

Take out 131313 as a common factor: E=13×(7×11+1)E = 13 \times (7 \times 11 + 1)E=13×(7×11+1) E=13×(77+1)E = 13 \times (77 + 1)E=13×(77+1) E=13×78E = 13 \times 78E=13×78

Now factorize 787878: 78=2×3×1378 = 2 \times 3 \times 1378=2×3×13

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

Since the given expression can be written as a product of prime factors (222, 333, and 131313), 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 + ra=bq+r where 0≤r<b0 \le r < b0≤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 bHCF(a,b)×LCM(a,b)=a×b.
  4. Irrationality Proofs: Always start by assuming the number is rational, use p/qp/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.

Verified NCERT & Board Exam Aligned Material
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