Chapter 1: Real Numbers

Overview

This page provides comprehensive Chapter 1: Real Numbers – Board Exam Notes aligned with the latest CBSE 2025–26 syllabus. Covers Fundamental Theorem of Arithmetic, HCF & LCM by Prime Factorisation, Proofs of Irrationality ($\sqrt{2}, \sqrt{3}, \sqrt{5}, 3+2\sqrt{5}$), Solved Board Questions, and Interactive Tools.

Fundamental Theorem of Arithmetic • Prime Factorisation • HCF & LCM • Irrationality Proofs

Exam Weightage & Blueprint

Total: ~6 Marks

Real Numbers falls under Unit I: Number Systems (6 marks). As per the latest syllabus, the board focus is on Fundamental Theorem of Arithmetic, HCF/LCM applications, and proving irrationality of numbers like $\sqrt{2}, \sqrt{3}, \sqrt{5}, 3+2\sqrt{5}$.

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 ($\sqrt{2}, \sqrt{3}, 3+2\sqrt{5}$)

Fundamental Theorem of Arithmetic

Theorem 1.1: 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.

Concept Check: Numbers ending with digit Zero
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.

$$ HCF(a, b) \times LCM(a, b) = a \times b $$

HCF-LCM Verifier

Enter two numbers to calculate and verify the formula.

⚠️ Common Mistake: The formula $ HCF \times LCM = a \times b \times c $ is NOT TRUE for three numbers.

Standard Board Question Types:

Find HCF using prime factorisation
Find LCM using prime factorisation
Find smallest number divisible by given numbers
Find greatest number dividing given numbers

Golden Rule:
Smallest number → use LCM
Greatest number → use HCF

📝 Solved Board Exam Questions

Step-by-step solutions in board exam format with marking scheme.

Q1. Find HCF and LCM of 306 and 657 by prime factorisation. Verify that HCF × LCM = Product of the two numbers. 3 Marks

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!

Q2. If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$, where $x, y$ are prime numbers, find $\text{HCF}(a, b)$ and $\text{LCM}(a, b)$. 1 Mark (Frequent Board MCQ)

Sol.

$\text{HCF}(a, b) =$ Product of the smallest power of each common prime factor $= x^{\min(3, 1)} \cdot y^{\min(2, 3)} = \mathbf{x y^2}$

$\text{LCM}(a, b) =$ Product of the highest power of each prime factor $= x^{\max(3, 1)} \cdot y^{\max(2, 3)} = \mathbf{x^3 y^3}$

Q3. Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers. 2 Marks

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.

Q4. Find the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively. 2 or 3 Marks

Sol. The required number divides $(70 - 5 = 65)$ and $(125 - 8 = 117)$ exactly.

∴ Required number = $\text{HCF}(65, 117)$.

$65 = 5 \times 13$

$117 = 3^2 \times 13$

$\text{HCF}(65, 117) = \mathbf{13}$. Hence, the largest number is 13.

Q5. Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 8:00 AM, when will they next ring together? 2 Marks

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.

Q6. Prove that $3 + 2\sqrt{5}$ is irrational. 3 Marks

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 $\text{HCF}(a, b) = 1$.

$\Rightarrow 2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b}$

$\Rightarrow \sqrt{5} = \frac{a - 3b}{2b}$

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. ■

Q7. Prove that $\frac{1}{\sqrt{2}}$ is irrational. 2 Marks

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. ■

🎯 Board Pattern Quick Rule:

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.

Step 1: Assumption (Contradiction Method) 0.5 Mark

Step 2: Squaring Both Sides 1 Mark

Step 3: Substitution 1 Mark

Step 4: Conclusion 0.5 Mark

📌 Board Pattern Alert:
Proof of irrationality of $\sqrt{3}$, $\sqrt{5}$ follows the exact same steps as $\sqrt{2}$. Simply replace 2 with the respective prime throughout.

📊 PYQ Trend (2018–2025):
CBSE alternates between $\sqrt{2}$ and $\sqrt{3}$ proofs. Compound irrationality ($3 + 2\sqrt{5}$ type) has appeared in 2023, 2024, and 2025 papers.

Competency Based Question (Case Study)

Scenario: A seminar is being conducted by an Educational Organisation. The number of participants in Hindi, English, and Mathematics are 60, 84, and 108 respectively.

Q1: Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject.

📋 One-Page Board Revision Checklist

✅ Every composite number = unique product of primes (FTA)
✅ HCF = product of smallest powers of common primes: $\text{HCF}(x^3y^2, xy^3) = xy^2$
✅ LCM = product of greatest powers of all primes: $\text{LCM}(x^3y^2, xy^3) = x^3y^3$
✅ 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 $r_1, r_2$ → $\text{HCF}(a - r_1, b - r_2)$
✅ $\sqrt{p}$ 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

💡 Exam Tip:
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: