Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic is a cornerstone of CBSE Class 10 Chapter 1: Real Numbers.
It states that every composite number can be broken down into prime building blocks in exactly one way — just as every molecule is made of atoms, every integer greater than 1 is made of primes.
The theorem has two parts:
- Existence: Every composite number CAN be factorised into primes.
- Uniqueness: There is only ONE way to do it (up to rearranging the factors).
This uniqueness guarantees that HCF, LCM, and divisibility results are well-defined and unambiguous.
What is Fundamental Theorem of Arithmetic?
Definition: 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.
n = p₁a₁ × p₂a₂ × ... × pₖaₖ
Where p₁ < p₂ < ... < pₖ are distinct primes and a₁, a₂, ..., aₖ are positive integers.
Examples:
- 36 = 2 × 2 × 3 × 3 = 2² × 3²
- 150 = 2 × 3 × 5 × 5 = 2 × 3 × 5²
- 1001 = 7 × 11 × 13
No matter how you factorise, the result is the same. For instance, 36 = 4 × 9 = (2 × 2) × (3 × 3) = 2² × 3². Or 36 = 6 × 6 = (2 × 3) × (2 × 3) = 2² × 3².
Key terms:
- Prime number: A natural number greater than 1 with no divisors other than 1 and itself. Examples: 2, 3, 5, 7, 11, 13.
- Composite number: A natural number greater than 1 with at least one divisor other than 1 and itself. Examples: 4, 6, 8, 9, 10, 12.
- Prime factorisation: The expression of a composite number as a product of primes.
Important:
- The number 1 is neither prime nor composite and is not covered by this theorem.
- The number 2 is the smallest prime and the only even prime.
- If a prime p divides the product ab, then p must divide a or b (or both). This property is used in irrationality proofs.
Fundamental Theorem of Arithmetic Formula
Key formulas from the Fundamental Theorem:
- Prime factorisation form: Every integer n > 1 can be written as n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ.
- HCF using prime factorisation: Product of the smallest powers of all common prime factors.
- LCM using prime factorisation: Product of the greatest powers of all prime factors.
- HCF-LCM relationship:
HCF(a, b) × LCM(a, b) = a × b
- Number of factors: If n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ, then total positive divisors = (a₁ + 1)(a₂ + 1)...(aₖ + 1).
- Decimal expansion test: A rational number p/q (in lowest terms) has a terminating decimal if and only if q contains only primes 2 and 5.
Derivation and Proof
Proof of the Fundamental Theorem of Arithmetic
Part 1: Existence of Prime Factorisation
- Let n be a composite number greater than 1. Since n is composite, n = a × b where 1 < a, b < n.
- If a is prime, stop. If composite, factorise a further into smaller factors.
- Each factorisation produces strictly smaller numbers. Since positive integers cannot decrease indefinitely, the process terminates.
- At termination, all factors are prime. So n is expressed as a product of primes.
Part 2: Uniqueness of Prime Factorisation
- Suppose n = p₁ × p₂ × ... × pₖ = q₁ × q₂ × ... × qₗ (two factorisations).
- p₁ divides n = q₁ × q₂ × ... × qₗ. Since p₁ is prime, p₁ must divide some qₙ.
- Since qₙ is also prime and p₁ | qₙ, we get p₁ = qₙ.
- Cancel p₁ = qₙ from both sides. Repeat for remaining factors.
- After cancelling all factors, every p₤ equals some qₙ and vice versa.
- The two factorisations are identical (up to order). QED.
Using the theorem to prove √2 is irrational:
The proof relies on: if 2 divides p², then 2 divides p.
- In the prime factorisation of p, the prime 2 appears k times.
- In p², it appears 2k times.
- If 2 | p², then 2k ≥ 1, so k ≥ 1, meaning 2 | p.
Types and Properties
Types of numbers based on prime factorisation:
| Type | Prime Factorisation | Examples |
|---|---|---|
| Prime | The number itself (single prime) | 2, 3, 5, 7, 11, 13, 17, 19, 23 |
| Composite | Product of 2 or more primes | 4 = 2², 12 = 2²×3, 30 = 2×3×5 |
| Perfect Square | All exponents are even | 36 = 2²×3², 100 = 2²×5² |
| Perfect Cube | All exponents are multiples of 3 | 8 = 2³, 216 = 2³×3³ |
| Square-free | All exponents are exactly 1 | 6 = 2×3, 30 = 2×3×5 |
Applications in the NCERT syllabus:
- Finding HCF and LCM: Decompose numbers into primes; pick smallest powers for HCF, greatest powers for LCM.
- Irrationality proofs: The consequence "if p | a² then p | a" proves √2, √3, √5 are irrational.
- Decimal expansion: The prime factorisation of the denominator determines terminating vs. repeating decimals.
- Perfect square/cube check: Examine whether all exponents in the factorisation are even (square) or multiples of 3 (cube).
Methods of prime factorisation:
- Factor tree method: Break the number step by step using any pair of factors until all leaves are prime.
- Repeated division (ladder method): Divide by the smallest prime repeatedly until the quotient is 1.
Properties from the theorem:
- If HCF(a, b) = 1, then a and b are coprime. Their factorisations share no common prime.
- Two consecutive integers are always coprime.
- If n = 2^a × 3^b × 5^c, the number of divisors is (a+1)(b+1)(c+1).
Solved Examples
Example 1: Finding prime factorisation using the factor tree method
Problem: Find the prime factorisation of 1764.
Solution:
- 1764 = 2 × 882
- 882 = 2 × 441
- 441 = 3 × 147
- 147 = 3 × 49
- 49 = 7 × 7
1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7²
Observation: All exponents are even, so 1764 is a perfect square. √1764 = 2 × 3 × 7 = 42.
Answer: 1764 = 2² × 3² × 7².
Example 2: Finding HCF and LCM using the Fundamental Theorem
Problem: Find the HCF and LCM of 510 and 92 using prime factorisation.
Solution:
Prime factorisations:
- 510 = 2 × 3 × 5 × 17
- 92 = 2² × 23
HCF: Common prime factor is only 2. Smallest power: 2¹. HCF = 2.
LCM: All primes with greatest powers: 2² × 3 × 5 × 17 × 23 = 23460.
Verification: HCF × LCM = 2 × 23460 = 46920. Product = 510 × 92 = 46920. ✓
Answer: HCF = 2, LCM = 23460.
Example 3: Verifying the HCF-LCM product relationship
Problem: Given HCF(306, 657) = 9, find LCM(306, 657).
Solution:
Using: HCF(a, b) × LCM(a, b) = a × b
- LCM = (a × b) / HCF = (306 × 657) / 9
- 306 × 657 = 201042
- LCM = 201042 / 9 = 22338
Verification: 306 = 2 × 3² × 17 and 657 = 3² × 73. HCF = 3² = 9 ✓. LCM = 2 × 3² × 17 × 73 = 22338 ✓.
Answer: LCM(306, 657) = 22338.
Example 4: Checking if 6^n can end with digit 0
Problem: Check whether 6n can end with the digit 0 for any natural number n.
Solution:
- A number ends with digit 0 if it is divisible by 10 (must have both 2 and 5 as prime factors).
- By the Fundamental Theorem: 6n = (2 × 3)n = 2n × 3n.
- The prime factorisation contains only 2 and 3. There is no factor of 5.
- Since 5 is not a factor of 6n, it is never divisible by 10.
Answer: No, 6n can never end with digit 0.
Example 5: Finding the smallest number divisible by given numbers
Problem: Find the smallest number that is divisible by each of 8, 15, and 20.
Solution:
Given: The smallest such number is LCM(8, 15, 20).
Prime factorisations:
- 8 = 2³
- 15 = 3 × 5
- 20 = 2² × 5
LCM: Greatest powers of all primes = 2³ × 3 × 5 = 8 × 3 × 5 = 120.
Verification: 120/8 = 15 ✓, 120/15 = 8 ✓, 120/20 = 6 ✓.
Answer: The smallest number divisible by 8, 15, and 20 is 120.
Example 6: Proving that √5 is irrational using the Fundamental Theorem
Problem: Prove that √5 is irrational.
Solution:
- Assume √5 is rational. Then √5 = a/b where a, b are coprime (HCF = 1) and b ≠ 0.
- Squaring: 5 = a²/b², so a² = 5b².
- 5 divides a². By the Fundamental Theorem, since 5 is prime and 5 | a², we get 5 | a. Let a = 5c.
- (5c)² = 5b² gives 25c² = 5b², so b² = 5c².
- Similarly, 5 | b.
- Both a and b are divisible by 5, contradicting HCF(a, b) = 1.
Answer: √5 is irrational. ■
Example 7: Determining terminating or non-terminating decimals
Problem: Without performing division, determine whether the following have terminating decimal expansions: (i) 77/1400, (ii) 51/1500, (iii) 129/(2² × 5³ × 7²).
Solution:
(i) 77/1400:
- 77 = 7 × 11 and 1400 = 2³ × 5² × 7
- Simplify: 77/1400 = 11/(2³ × 5²) = 11/200
- Denominator = 2³ × 5² (only 2s and 5s) → Terminating
(ii) 51/1500:
- 51 = 3 × 17 and 1500 = 2² × 3 × 5³
- Simplify: 51/1500 = 17/(2² × 5³) = 17/500
- Denominator = 2² × 5³ (only 2s and 5s) → Terminating
(iii) 129/(2² × 5³ × 7²):
- 129 = 3 × 43. Since gcd(129, 7) = 1, 7² remains in denominator.
- Denominator has factor 7 (besides 2 and 5) → Non-terminating repeating
Answer: (i) Terminating, (ii) Terminating, (iii) Non-terminating repeating.
Example 8: Checking if a number is a perfect square or perfect cube
Problem: Using prime factorisation, determine if 5832 is a perfect cube. If yes, find its cube root.
Solution:
- 5832 = 2 × 2916 = 2 × 2 × 1458 = 2 × 2 × 2 × 729 = 2³ × 729
- 729 = 3 × 243 = 3 × 3 × 81 = 3 × 3 × 3 × 27 = 3 × 3 × 3 × 3 × 9 = 3⁶
- 5832 = 2³ × 3⁶
- All exponents (3 and 6) are multiples of 3 → perfect cube
- Cube root = 2¹ × 3² = 2 × 9 = 18
Answer: 5832 is a perfect cube. Cube root = 18.
Example 9: Word problem on LCM
Problem: In a school, Section A has 32 students and Section B has 36 students. Find the minimum number of books required so that the books can be distributed equally among students of either section.
Solution:
Given:
- Books must be divisible by both 32 and 36
- Minimum such number = LCM(32, 36)
Prime factorisations:
- 32 = 2⁵
- 36 = 2² × 3²
LCM: 2⁵ × 3² = 32 × 9 = 288
Verification: 288/32 = 9 ✓ and 288/36 = 8 ✓.
Answer: Minimum books required = 288.
Example 10: Finding the number of factors of a number
Problem: Find the total number of positive divisors of 2520.
Solution:
- 2520 = 2 × 1260 = 2 × 2 × 630 = 2 × 2 × 2 × 315 = 2³ × 315
- 315 = 3 × 105 = 3 × 3 × 35 = 3² × 5 × 7
- 2520 = 2³ × 3² × 5¹ × 7¹
- Number of divisors = (3+1)(2+1)(1+1)(1+1) = 4 × 3 × 2 × 2 = 48
Answer: 2520 has 48 positive divisors.
Real-World Applications
Applications of the Fundamental Theorem of Arithmetic:
- Cryptography: RSA encryption is based on the fact that multiplying two large primes is easy, but factorising their product is extremely hard. This asymmetry is a direct consequence of the theorem's uniqueness guarantee.
- Finding HCF and LCM: Every HCF/LCM problem can be solved systematically using prime factorisation. Used in simplifying fractions, ratio problems, and scheduling.
- Simplifying fractions: Divide numerator and denominator by their HCF (found via prime factorisation) to reduce to lowest terms.
- Decimal expansions: The theorem explains why 1/3 = 0.333... (repeating) while 1/4 = 0.25 (terminating). It depends on the prime factors of the denominator.
- Computer Science: Primality testing, integer factorisation, hash functions, and the Sieve of Eratosthenes all rely on the theorem.
- Chemistry: The concept of unique decomposition parallels the Law of Definite Proportions — a compound always has the same elements in the same ratios.
Key Points to Remember
- Every composite number has a unique prime factorisation (up to order).
- To find prime factorisation, use the factor tree method or repeated division (ladder) method.
- HCF = product of smallest powers of common prime factors.
- LCM = product of greatest powers of all prime factors.
- HCF(a, b) × LCM(a, b) = a × b. Use this to find one if the other is known.
- If a prime p divides a², then p divides a. This is key for irrationality proofs.
- A number is a perfect square if all exponents are even. A perfect cube if all exponents are multiples of 3.
- A rational number p/q (lowest terms) has a terminating decimal if q = 2m × 5n.
- The HCF-LCM product formula works for two numbers only. For three numbers: LCM(a,b,c) = LCM(LCM(a,b), c).
- This topic is frequently tested in CBSE exams (2–5 marks): HCF/LCM problems, irrationality proofs, decimal expansion questions.
Practice Problems
- Find the prime factorisation of 17017 and determine whether it is a perfect square.
- Find the HCF and LCM of 26 and 91 using prime factorisation. Verify that HCF times LCM equals the product of the two numbers.
- Explain why 3 × 5 × 7 + 7 is a composite number.
- Check whether 12^n can end with the digit 5 for any natural number n.
- Two alarm clocks ring at intervals of 4 minutes and 6 minutes respectively. If both ring together at 9:00 AM, when will they next ring together?
- Find the smallest number which when divided by 15, 20, and 36 leaves no remainder.
- If LCM(91, 26) = 182, find HCF(91, 26) without prime factorisation.
- Prove that 5 + 3√2 is irrational using the Fundamental Theorem of Arithmetic.
Frequently Asked Questions
Q1. What is the Fundamental Theorem of Arithmetic in simple words?
Every whole number greater than 1 is either prime or can be written as a product of prime numbers in exactly one way (ignoring order). For example, 60 = 2 × 2 × 3 × 5 — no other combination of primes gives 60.
Q2. Why is the Fundamental Theorem of Arithmetic important?
It guarantees unique prime factorisation. This uniqueness is the foundation for computing HCF and LCM, proving irrationality of numbers like √2, determining decimal behaviour of fractions, and modern cryptography used in internet security.
Q3. How is the Fundamental Theorem of Arithmetic used to find HCF and LCM?
Find the prime factorisation of each number. For HCF, take the smallest powers of common primes and multiply. For LCM, take the greatest powers of all primes and multiply. Example: 72 = 2³ × 3² and 120 = 2³ × 3 × 5. HCF = 2³ × 3 = 24. LCM = 2³ × 3² × 5 = 360.
Q4. What is the difference between HCF and LCM?
HCF (Highest Common Factor) is the largest number that divides two or more numbers exactly. LCM (Least Common Multiple) is the smallest number divisible by two or more numbers. HCF uses smallest powers of common primes; LCM uses greatest powers of all primes. HCF ≤ both numbers; LCM ≥ both numbers.
Q5. Does the Fundamental Theorem of Arithmetic apply to 1?
No. The number 1 is neither prime nor composite and has no prime factorisation. Including 1 as a prime would break the uniqueness of the theorem, which is one reason 1 is excluded from the definition of prime numbers.
Q6. How do you prove a number is irrational using this theorem?
Use proof by contradiction. Assume √p (p prime) is rational, write it as a/b in lowest terms, square to get p | a². By the Fundamental Theorem, since p is prime and p | a², p must divide a. Substituting shows p also divides b, contradicting the lowest-terms assumption.
Q7. What is a factor tree and how do you draw one?
A factor tree is a diagram for finding prime factorisation. Start with the number at the top. Break it into any two factors (branches). Continue breaking composite factors into sub-factors. Stop when all end nodes (leaves) are prime. The product of all leaves is the prime factorisation. Different trees for the same number always produce the same prime factors.
Q8. Is the product of two prime numbers always a composite number?
Yes. If p and q are primes, p × q has factors 1, p, q, and pq — so it has factors other than 1 and itself, making it composite. Even p × p = p² is composite (factors: 1, p, p²).
Q9. Can two different numbers have the same prime factorisation?
No. The uniqueness part of the theorem guarantees each number has exactly one prime factorisation. If two numbers had the same factorisation, they would be the same number.
Q10. What are the Fundamental Theorem of Arithmetic questions for CBSE board exams?
Common questions: (1) Find HCF and LCM by prime factorisation (2 marks), (2) Prove √2, √3, or √5 is irrational (4 marks), (3) Check whether 6<sup>n</sup> can end with digit 0 (2 marks), (4) Determine terminating or non-terminating decimals (2 marks), (5) Verify HCF × LCM = product of numbers (2 marks).
