Chapter 1: Real Numbers
Overview
Master Chapter 1 (Real Numbers) for CBSE Class 10 with concise revision notes on the Fundamental Theorem of Arithmetic, HCF-LCM relationships, irrationality proofs ($\sqrt{2}, \sqrt{3}, \sqrt{5}$), and board-pattern examples.
Fundamental Theorem of Arithmetic • Prime Factorisation • HCF & LCM • Irrationality Proofs
Exam Weightage & Blueprint
Total: ~6 MarksFor 2026–27, the CBSE syllabus assigns 6 marks to Unit I: Number Systems, which includes Real Numbers. The board focus is on Fundamental Theorem of Arithmetic and proofs of irrationality, such as $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, and $3+2\sqrt{5}$.
| Question Type | Marks | Frequency | Focus Topic |
|---|---|---|---|
| MCQ | 1 | High | HCF/LCM, Prime Factorisation |
| Short Answer | 2 | Medium | FTA Applications, Real-life Problems |
| Long Answer | 3 | Very High | Irrationality Proofs ($\sqrt{2}, \sqrt{3}, 3+2\sqrt{5}$) |
CBSE 2026–27 Syllabus Status: Real Numbers
Unit I: Number Systems • 6 MarksThis chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:
In 2026–27 Board Syllabus:
Deleted / Not in Board Exam:
Fundamental Theorem of Arithmetic
For any number like $a^n$ (e.g., $6^n$ or $4^n$) to end with the digit $0$, its prime factorisation must contain the primes 2 and 5. For example, $6^n = (2 \times 3)^n$ does not contain 5 as a prime factor, so $6^n$ can never end with the digit 0 for any natural number $n$.
Prime Factorization Engine
Enter a number to see its unique prime factors (Tree Structure).
HCF and LCM
For any two positive integers $a$ and $b$:
HCF (Highest Common Factor)
Product of the smallest power of each common prime factor.
LCM (Least Common Multiple)
Product of the greatest power of each prime factor involved.
HCF-LCM Verifier
Enter two numbers to calculate and verify the formula.
- Consecutive Numbers Rule: Any two consecutive natural numbers are always co-prime → $\text{HCF}(n, n+1) = 1$.
- Consecutive Even / Odd Numbers: $\text{HCF}(2n, 2n+2) = 2$ and $\text{HCF}(2n+1, 2n+3) = 1$.
- Smallest Prime vs Composite: Smallest prime is $2$, smallest composite is $4$. Therefore, $\text{HCF}(2, 4) = 2$ and $\text{LCM}(2, 4) = 4$.
- Total Factors Formula: If prime factorisation $N = p_1^{a} \cdot p_2^{b} \cdot p_3^{c} \dots$, total number of factors $= (a+1)(b+1)(c+1)\dots$
- Arithmetic of Radicals:
• $\text{Rational} \pm \text{Irrational} = \text{Irrational}$
• $\text{Non-zero Rational} \times \text{Irrational} = \text{Irrational}$
• Sum or product of two irrationals can be rational or irrational (e.g. $\sqrt{3} + (-\sqrt{3}) = 0$, $\sqrt{2} \times \sqrt{2} = 2$).
- Find HCF using prime factorisation
- Find LCM using prime factorisation
- Find smallest number divisible by given numbers
- Find greatest number dividing given numbers
Smallest number → use LCM
Greatest number → use HCF
📝 Solved Board Exam Questions
Step-by-step solutions in board exam format with marking scheme.
Sol. $306 = 2 \times 3^2 \times 17$
$657 = 3^2 \times 73$
HCF = Product of smallest powers of common primes = $3^2 = 9$
LCM = Product of greatest powers of all primes = $2 \times 3^2 \times 17 \times 73 = 22338$
Verification: HCF × LCM = $9 \times 22338 = 201042$
Product = $306 \times 657 = 201042$ Verified!
Sol.
$HCF(a, b) =$ Product of the smallest power of each common prime factor $= x^{\min(3, 1)} \cdot y^{\min(2, 3)} = \mathbf{x y^2}$
$LCM(a, b) =$ Product of the highest power of each prime factor $= x^{\max(3, 1)} \cdot y^{\max(2, 3)} = \mathbf{x^3 y^3}$
Sol. A composite number has more than two factors (factors other than 1 and the number itself).
1. $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 = 13 \times 2 \times 3 \times 13$
Since it has factors other than 1 and itself (namely 2, 3, 13), it is a composite number.
2. $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$
Since it has 5 as a factor in addition to 1 and itself, it is a composite number.
Sol. The required number divides $(70 - 5 = 65)$ and $(125 - 8 = 117)$ exactly.
∴ Required number = $HCF(65, 117)$.
$65 = 5 \times 13$
$117 = 3^2 \times 13$
$HCF(65, 117) = \mathbf{13}$. Hence, the largest number is 13.
Sol. We need LCM(9, 12, 15).
$9 = 3^2$, $12 = 2^2 \times 3$, $15 = 3 \times 5$
LCM = $2^2 \times 3^2 \times 5 = 180$ minutes = 3 hours
∴ They will ring together again at 11:00 AM.
Sol. Assume $3 + 2\sqrt{5}$ is rational.
Then $3 + 2\sqrt{5} = \frac{a}{b}$, where $a, b$ are integers, $b \neq 0$, and $HCF(a, b) = 1$.
$\Rightarrow 2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$
$\Rightarrow \sqrt{5} = \frac{a - 3b}{2b}$
Since $a, b$ are integers, $\frac{a - 3b}{2b}$ is rational.
∴ $\sqrt{5}$ is rational — contradiction (since $\sqrt{5}$ is irrational by Theorem 1.3).
Hence, $3 + 2\sqrt{5}$ is irrational. ■
Sol. Let us assume to the contrary that $\frac{1}{\sqrt{2}}$ is rational.
Then $\frac{1}{\sqrt{2}} = \frac{a}{b}$, where $a$ and $b$ are co-prime integers and $a \neq 0, b \neq 0$.
Rearranging, we get $\sqrt{2} = \frac{b}{a}$.
Since $a, b$ are integers, $\frac{b}{a}$ is rational, which means $\sqrt{2}$ is rational.
This contradicts the known fact that $\sqrt{2}$ is irrational. Hence, $\frac{1}{\sqrt{2}}$ is irrational. ■
- Sum/Difference: $a \pm \sqrt{b}$ (e.g., $5 - \sqrt{3}$) → isolate $\sqrt{b} = \frac{5b - a}{b}$ → contradiction.
- Quotient/Product: $\frac{a}{\sqrt{b}}$ or $a\sqrt{b}$ (e.g., $\frac{1}{\sqrt{2}}, 7\sqrt{5}$) → invert/divide to isolate $\sqrt{b}$ → contradiction.
- Sum of two radicals (Standard Paper): $\sqrt{2} + \sqrt{3} = x \Rightarrow 2 + 3 + 2\sqrt{6} = x^2 \Rightarrow \sqrt{6} = \frac{x^2 - 5}{2}$ → contradiction.
Revisiting Irrational Numbers
Theorem 1.3: Let \( p \) be a prime number. If \( p \) divides \( a^2 \), then \( p \) divides \( a \), where $a$ is a positive integer.
Proof Builder: $\sqrt{2}$ is Irrational
Click the steps to reveal the logic flow used in Board Exams.
Proof of irrationality of $\sqrt{3}$, $\sqrt{5}$ follows the exact same steps as $\sqrt{2}$. Simply replace 2 with the respective prime throughout.
CBSE alternates between $\sqrt{2}$ and $\sqrt{3}$ proofs. Compound irrationality ($3 + 2\sqrt{5}$ type) has appeared in 2023, 2024, and 2025 papers.
Case Study: National Mathematics Seminar Seating
Scenario: An educational council is organizing an inter-school academic symposium. The registered participants are 60 students in Hindi, 84 students in English, and 108 students in Mathematics. To maintain equal academic standard and discipline, the organizers decide that in each lecture hall, the exact same number of participants must be seated, all being in the same subject.
Q1: Find the maximum capacity (number of participants) that can be seated in each hall under the given condition. 1 Mark
Q2: What is the minimum total number of lecture halls required to accommodate all participants? 1 Mark
Q3: If the registration fee is ₹250 per participant, find the total revenue collected from the seminar. (OR) If 12 additional participants register in Mathematics, how many extra rooms will be needed? 2 Marks
📋 One-Page Board Revision Checklist
- ✅ Every composite number = unique product of primes (FTA)
- ✅ HCF = product of smallest powers of common primes: $HCF(x^3y^2, xy^3) = xy^2$
- ✅ LCM = product of greatest powers of all primes: $LCM(x^3y^2, xy^3) = x^3y^3$
- ✅ HCF(a, b) × LCM(a, b) = a × b (valid for two numbers only)
- ✅ Composite number proof → factor out common prime to show $>2$ factors
- ✅ Largest number dividing $a, b$ leaving remainders $r_1, r_2$ → $HCF(a - r_1, b - r_2)$
- ✅ $\sqrt{p}$ is irrational when $p$ is prime (use Theorem 1.3 + contradiction)
- ✅ $a + b\sqrt{p}$ and $\frac{a}{\sqrt{b}}$ are irrational → isolate radical, show contradiction
- ✅ "Smallest number" type → use LCM
- ✅ "Greatest number" type → use HCF
- ✅ $\text{HCF}(n, n+1) = 1$ (Consecutive natural numbers are co-prime)
- ✅ Smallest prime = 2, Smallest composite = 4 → $\text{HCF}(2, 4) = 2, \text{LCM}(2, 4) = 4$
- ✅ Non-zero Rational × Irrational = always Irrational
- ✅ Total factors of $p_1^a p_2^b$ = $(a+1)(b+1)$
Always state Theorem 1.3 explicitly in irrationality proofs — writing "If $p$ divides $a^2$, then $p$ divides $a$" fetches marks even if the rest of the proof has minor errors.
Concept Mastery Quiz 🎯
Test your readiness for the board exam.
1. The HCF of two consecutive even numbers is always:
2. If LCM(a, b) = 1050 and HCF(a, b) = 30, and a = 150, then b = ?
3. The product of a non-zero rational and an irrational number is:
4. If $p$ and $q$ are co-prime, then $p^2$ and $q^2$ are:
5. $5 - 2\sqrt{3}$ is:
Frequently Asked Questions: Class 10 Maths Chapter 1
What is the weightage of Chapter 1 Real Numbers in CBSE Class 10 Board Exam?
Real Numbers carries 6 marks in Unit 1: Number Systems in the CBSE Class 10 Board Examination, typically featuring 1 MCQ/assertion-reason question and one 2-mark or 3-mark problem on irrationality proofs.
Is Euclid's Division Lemma included in the CBSE Class 10 syllabus?
No, Euclid's Division Lemma (EDL) has been removed from the rationalized CBSE Class 10 syllabus. The current syllabus focuses strictly on the Fundamental Theorem of Arithmetic and proving irrationality of square roots like sqrt(2), sqrt(3), and sqrt(5).
What is the relationship between HCF and LCM of two positive integers?
For any two positive integers a and b, HCF(a, b) * LCM(a, b) = a * b. Note that this property holds true strictly for two numbers and does not generally apply to three numbers.
How to write the proof of irrationality of sqrt(2) or sqrt(3) for full marks?
Assume the contradiction that the number is rational (a/b where a and b are co-prime integers). Square both sides, show that both a and b share a common prime factor (violating co-primality), and conclude that the initial assumption was incorrect.