Can the HCF of two numbers ever be strictly greater than their LCM?

No. The HCF is always a factor of the LCM. For any two positive integers a and b, \text{HCF}(a,b) \le \min(a,b) \le \max(a,b) \le \text{LCM}(a,b), and their product satisfies \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b.

How do you prove that a number like $\sqrt{3}$ or $\sqrt{5}$ is irrational using contradiction?

Assume \sqrt{p} = \frac{a}{b} where a, b are coprime integers with b \neq 0. Squaring gives pb^2 = a^2, implying p divides a^2, and hence p divides a. Substituting a = pk gives b^2 = pk^2, implying p divides b. This shows p is a common factor of both a and b, contradicting their coprimality. Hence, \sqrt{p} is irrational.

Why is $\pi$ considered irrational when we commonly write $\pi = \frac{22}{7}$?

\frac{22}{7} and 3.14 are only convenient rational approximations used for practical calculations. The true value of \pi is a non-terminating, non-repeating decimal (3.14159265...), which cannot be expressed as a ratio of two integers.

How do you check if a simplified rational fraction $\frac{p}{q}$ terminates without long division?

First simplify \frac{p}{q} so that p and q are coprime. Then factorise the denominator q. If q = 2^n \times 5^m for non-negative integers n, m, the decimal expansion terminates after \max(n, m) decimal places. If any other prime factor (such as 3 or 7) is present, it is non-terminating repeating.

Is Euclid’s Division Lemma included in the CBSE 2026–27 examination syllabus?

No, Euclid's Division Lemma and Euclid's Division Algorithm have been rationalized and removed from the active CBSE syllabus. The active focus is on the Fundamental Theorem of Arithmetic, finding HCF/LCM by prime factorisation, and proving the irrationality of \sqrt{2}, \sqrt{3}, \sqrt{5}.

What is the relationship between HCF and LCM for three numbers $p, q, r$?

The simple product formula does NOT hold for three numbers: \text{HCF}(p,q,r) \times \text{LCM}(p,q,r) \neq p \times q \times r. For three numbers, you must find the prime factorisation of each number and take the lowest power of common primes for HCF and the highest power of all primes for LCM.

All Revision Notes

MathematicsCh-1 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 1

Real Numbers

Fundamental Theorem of Arithmetic, unique prime factorization, rigorous proofs of irrationality by contradiction, and LCM-HCF word problems.

Quick Key Takeaways:

Fundamental Theorem of Arithmetic: Every composite number can be uniquely expressed as a product of primes, up to the order of factors:

a = p_1^{k_1} p_2^{k_2} \dots p_n^{k_n}

Product Relationship:

\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b

(holds strictly for two positive integers only).

Irrationality by Contradiction: Proofs for

\sqrt{2}, \sqrt{3}, \sqrt{5}

and linear combinations

a + b\sqrt{p}

using coprime rational assumptions.

Terminating Decimal Criterion: Rational number

\frac{p}{q}

has a terminating decimal expansion if and only if

q = 2^m 5^n

for non-negative integers

m, n

Jump to Concept:1. 1. Fundamental Theorem of Arithmetic & Prime Factorization Architecture 2. 2. Rigorous Proofs of Irrationality by Contradiction 3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Circular Sports Track Synchronization 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

1. Fundamental Theorem of Arithmetic & Prime Factorization Architecture

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Real Numbers.

• The Fundamental Theorem of Arithmetic (Unique Factorization)

•

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

•

Canonical Form: Any integer

n > 1

is written as

n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}

, where

p_1 < p_2 < \dots < p_k

are distinct primes and

a_i \in \mathbb{N}

•

HCF and LCM Formulation via Primes:

•

\text{HCF}(a, b) =

Product of the smallest power of each common prime factor involved in the numbers.

•

\text{LCM}(a, b) =

Product of the greatest power of each prime factor involved in the numbers.

• Why $6^n$ or $4^n$ Cannot End with the Digit 0

•

Divisibility Rule for 0: For a number

x^n

to end with the digit 0, it must be divisible by 10, which means its prime factorization must contain both prime factors 2 and 5.

•

Prime Factorization of $6^n$ :

6^n = (2 \times 3)^n = 2^n \times 3^n

. The only prime factors are 2 and 3.

•

Conclusion by Uniqueness: By the Fundamental Theorem of Arithmetic, 5 is not a factor of

6^n

. Hence, there is no natural number

n

for which

6^n

ends with the digit zero.

📊 Real Numbers: Prime Factorization & HCF-LCM HierarchyVisual Model

Visual schematic mapping unique prime factorization, factor trees, coprime divisibility laws, and the two-number LCM-HCF relationship.

2. Rigorous Proofs of Irrationality by Contradiction

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Real Numbers.

• Theorem Proof: Prove that $\sqrt{5}$ is an Irrational Number

•

Step 1 (Assumption of Rationality): Let us assume to the contrary that

\sqrt{5}

is rational. Then there exist coprime integers

a

and

b

(

b \neq 0, \gcd(a, b) = 1

) such that:

\sqrt{5} = \frac{a}{b} \implies a = b\sqrt{5}

•

Step 2 (Squaring Both Sides): Squaring both sides:

a^2 = 5b^2 \quad \text{--- (Equation 1)}

This implies that

5

divides

a^2

. By Theorem (if a prime

p

divides

a^2

, then

p

divides

a

), it follows that $5$ divides $a$ .

•

Step 3 (Substitution): Since

5

divides

a

, we can write

a = 5c

for some integer

c

. Substituting

a = 5c

into Equation 1:

(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2

•

Step 4 (Deduction of Common Factor): This shows that

5

divides

b^2

, which implies $5$ divides $b$ .

•

Step 5 (Contradiction): From Steps 2 and 4,

5

is a common factor of both

a

and

b

. But this directly contradicts our initial hypothesis that

a

and

b

are coprime (

\gcd(a, b) = 1

•

Conclusion: Our assumption was false. Therefore, $\sqrt{5}$ is irrational.

• Proof for Linear Combination: Prove that $3 + 2\sqrt{5}$ is Irrational

•

Let

3 + 2\sqrt{5} = \frac{a}{b}

, where

a, b \in \mathbb{Z}, b \neq 0, \gcd(a, b) = 1

•

Rearranging:

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

•

Since

a

and

b

are integers,

\frac{a - 3b}{2b}

is a rational number. This implies that

\sqrt{5}

must be rational.

•

But this contradicts the proven fact that

\sqrt{5}

is irrational. Hence, $3 + 2\sqrt{5}$ is irrational.

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: Given that $\text{HCF}(306, 657) = 9$ , find $\text{LCM}(306, 657)$ .

•

Formula: For any two positive integers

a

and

b

\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b

•

Given:

a = 306, b = 657, \text{HCF} = 9

•

Calculation:

\text{LCM}(306, 657) = \frac{a \times b}{\text{HCF}(a, b)} = \frac{306 \times 657}{9}

\text{LCM} = 34 \times 657 = 22,338

•

Final Boxed Answer:

\mathbf{\text{LCM}(306, 657) = 22,338}

• 5-Mark Heavyweight Board Problem / Rider: An army contingent of 616 members is to march behind an army band of 32 members in a Republic Day parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

•

Mathematical Formulation: The maximum number of columns is the Highest Common Factor of

616

and

32

, i.e.,

\text{HCF}(616, 32)

•

Prime Factorization Method:
-

616 = 2^3 \times 7 \times 11 = 8 \times 77

32 = 2^5 = 32

•

Finding HCF: Common prime factor with lowest power is

2^3 = 8

\text{HCF}(616, 32) = 8

•

Verification by Division:
-

616 = 32 \times 19 + 8

32 = 8 \times 4 + 0 \implies \text{HCF} = 8

•

Final Conclusion: The maximum number of columns in which the army contingent and band can march is 8 columns.

4. CBSE Case-Study Modeling: Circular Sports Track Synchronization

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Circular Sports Track Synchronization

Sonia and Ravi start driving around a circular sports track from the same starting point at the same time and in the same direction. Sonia takes 18 minutes to complete one round, while Ravi takes 12 minutes for the same.

•

Q1: After how many minutes will they meet again at the starting point?

\rightarrow

They will meet at the starting point at a time that is the Lowest Common Multiple of 18 and 12, i.e.,

\text{LCM}(18, 12)

. Prime factors:

18 = 2 \times 3^2, 12 = 2^2 \times 3

\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = \mathbf{36\text{ minutes}}

•

Q2: How many rounds will Sonia and Ravi have completed by that time?

\rightarrow

Sonia completes

\frac{36}{18} = \mathbf{2\text{ rounds}}

. Ravi completes

\frac{36}{12} = \mathbf{3\text{ rounds}}

•

Q3: If a third runner Amit takes 24 minutes, when will all three meet at the start?

\rightarrow

\text{LCM}(18, 12, 24) = 2^3 \times 3^2 = 8 \times 9 = \mathbf{72\text{ minutes}}

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines

•

1 Mark: State coprime assumption (

\gcd(a, b) = 1

) and algebraic setup in contradiction proofs.

•

1 Mark: Prove 5 divides

a^2 \implies 5

divides

a

•

1 Mark: Substitute

a = 5c

and prove 5 divides

b

•

1 Mark: State the contradiction explicitly and write the final conclusion.

• Common Calculation Traps & Verification Checklist

•

Trap 1: Never apply

\text{HCF}(a, b, c) \times \text{LCM}(a, b, c) = a \times b \times c

for three numbers; this formula is strictly valid for two numbers only.

•

Trap 2: Forgetting to state that

a

and

b

are coprime integers in irrationality proofs leads to a mandatory 0.5-mark deduction.

•

Trap 3: In word problems, distinguish clearly between minimum time/interval (requires LCM) versus maximum capacity/size/columns (requires HCF).

Authentic Board Question (3 Marks)Topic: Fundamental Theorem of Arithmetic & Contradiction Method

Prove that

\sqrt{5}

is an irrational number.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Assumption & Coprime Setup: Assume

\sqrt{5} = \frac{p}{q}

where

p, q

are coprime integers and

q \neq 0

½ Mark

Step 2: Squaring & Divisibility of $p$: Square both sides (

5q^2 = p^2

), conclude

5

divides

p^2 \implies 5

divides

p

1 Mark

Step 3: Substitution ($p = 5k$) & Divisibility of $q$: Substitute

p = 5k \implies 5q^2 = 25k^2 \implies q^2 = 5k^2 \implies 5

divides

q

1 Mark

Step 4: Contradiction & Final Conclusion: State that

p

and

q

having common factor

5

contradicts coprimality

\implies \sqrt{5}

is irrational.

½ Mark

Model Student Answer (Target: Full 3/3 Marks):

Let us assume to the contrary that

\sqrt{5}

is rational.
Therefore,

\sqrt{5} = \frac{a}{b}

, where

a

and

b

are coprime integers and

b \neq 0

.
Squaring both sides:

5 = \frac{a^2}{b^2} \implies 5b^2 = a^2

...(1)
Since

5

divides

a^2

, by Theorem, $5$ divides $a$ .
Let

a = 5c

for some integer

c

.
Substituting into (1):

5b^2 = (5c)^2 = 25c^2 \implies b^2 = 5c^2

Since

5

divides

b^2

, $5$ divides $b$ .
Thus,

a

and

b

have at least

5

as a common factor. But this contradicts the fact that

a

and

b

are coprime.
Conclusion: This contradiction has arisen because of our incorrect assumption. Hence, $\sqrt{5}$ is irrational.

Examiner Mark Deduction Traps:

•Writing the word "coprime integers" is mandatory for full marks in Step 1.

•Never skip the concluding contradiction sentence—examiners deduct ½ mark if omitted.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

The Fundamental Theorem of Arithmetic guarantees that every composite number can be expressed as a product of primes in a way that is unique, except for the order in which the prime factors occur. Prime factorisation is the actual computational process of finding those prime factors (e.g.,

140 = 2^2 \times 5 \times 7

Real Numbers

1. Fundamental Theorem of Arithmetic & Prime Factorization Architecture

2. Rigorous Proofs of Irrationality by Contradiction

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

4. CBSE Case-Study Modeling: Circular Sports Track Synchronization

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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