1. Chapter at a Glance
Every whole number greater than 1 is either a prime or can be built by multiplying primes — and in only one way. This simple-sounding fact, the Fundamental Theorem of Arithmetic, is the backbone of this chapter. With it we find HCF and LCM quickly, prove that numbers like √2, √3 and √5 are irrational, and understand which fractions have terminating decimals. This chapter carries about 6 marks in the CBSE board exam (Number Systems unit) and its proofs are frequently asked.
- The Fundamental Theorem of Arithmetic (FTA).
- HCF and LCM by prime factorisation; HCF × LCM = product of two numbers.
- Proving irrationality of √p (p prime) and of numbers like 3 + 2√5.
- Applying FTA to "can 6ⁿ end in 0?" type questions.
2. Fundamental Theorem of Arithmetic
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.
Example: 2520 = 2³ × 3² × 5 × 7. No other set of primes multiplies to 2520.
Why uniqueness matters: it lets us reason about which primes "can" or "cannot" appear in a number, which is the key to many proofs.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution: A number ending in 0 is divisible by 10 = 2 × 5, so 5 must be a prime factor. But 6ⁿ = 2ⁿ × 3ⁿ. By uniqueness of prime factorisation, 5 never appears. So 6ⁿ can never end in 0.
3. HCF and LCM by Prime Factorisation
- HCF = product of the smallest power of each common prime factor.
- LCM = product of the greatest power of each prime factor involved.
- For two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b.
Find the HCF and LCM of 96 and 404, and verify the product rule.
Solution: 96 = 2⁵ × 3; 404 = 2² × 101. HCF = 2² = 4. LCM = 2⁵ × 3 × 101 = 9696. Check: 4 × 9696 = 38784 = 96 × 404 ✓.
Find the HCF and LCM of 6, 72 and 120.
Solution: 6 = 2 × 3; 72 = 2³ × 3²; 120 = 2³ × 3 × 5. HCF = 2 × 3 = 6. LCM = 2³ × 3² × 5 = 360.
HCF × LCM = product is true for two numbers only. For 6, 72, 120: 6 × 360 = 2160, but 6 × 72 × 120 = 51840.
4. Proving Irrationality
Let p be a prime. If p divides a², then p divides a. (Follows from FTA: the primes of a² are exactly the primes of a, each doubled.)
- Assume √2 is rational: √2 = a/b, where a, b are coprime integers, b ≠ 0.
- Then 2b² = a², so 2 divides a², hence 2 divides a. Write a = 2c.
- 2b² = 4c² → b² = 2c², so 2 divides b², hence 2 divides b.
- Now 2 divides both a and b — contradicting that a and b are coprime.
- Hence √2 is irrational. ∎
The same proof works for √3, √5, √7 (any prime). In the board exam, write each step with the reason.
Proof: Suppose 5 − √3 = r, rational. Then √3 = 5 − r. Since 5 and r are rational, 5 − r is rational, so √3 would be rational — contradiction (√3 is irrational). Hence 5 − √3 is irrational. ∎
| Combination | Result | Example |
|---|---|---|
| rational + irrational | Irrational | 2 + √5 |
| non-zero rational × irrational | Irrational | 3√2 |
| irrational + irrational | Either | √2 + (−√2) = 0 |
| irrational × irrational | Either | √2 × √8 = 4; √2 × √3 = √6 |
5. Decimal Expansions (Link)
A rational number p/q in lowest terms has a terminating decimal if and only if q = 2ⁿ × 5ᵐ (n, m ≥ 0). Otherwise it is non-terminating recurring. Example: 13/3125 = 13/5⁵ terminates; 29/343 = 29/7³ recurs.
6. Solved Problems (Board Pattern)
Sonia takes 18 minutes to complete one round of a field; Ravi takes 12 minutes. They start together at the same point in the same direction. After how many minutes will they meet again at the start?
Solution: LCM(18, 12): 18 = 2 × 3², 12 = 2² × 3 → LCM = 2² × 3² = 36 minutes.
Explain why 7 × 11 × 13 + 13 is composite.
Solution: = 13(7 × 11 + 1) = 13 × 78. It has factors other than 1 and itself, so it is composite.
HCF(306, 657) = 9. Find LCM(306, 657).
Solution: LCM = 306 × 657 ÷ 9 = 22338.
7. Master Formula Table
| Result | Statement |
|---|---|
| FTA | n = p₁^a₁ × p₂^a₂ × … (unique) |
| HCF | Product of lowest powers of common primes |
| LCM | Product of highest powers of all primes |
| Two numbers | HCF × LCM = a × b |
| Prime divisibility | p | a² ⇒ p | a |
| Terminating test | q = 2ⁿ5ᵐ (lowest terms) |
| Ends in 0 | Needs both 2 and 5 as prime factors |
8. Master Summary Table
| Topic | Key idea | Typical board question |
|---|---|---|
| FTA | Unique prime factorisation | Express 3825 as primes; can 4ⁿ end in 0? |
| HCF/LCM | Lowest/highest powers | Bells ringing, laps, packets |
| Product rule | HCF × LCM = ab | Find LCM given HCF |
| Irrationality | Contradiction with coprime a, b | Prove √5 / 3 + 2√5 irrational |
| Composite numbers | Take common factor | 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 |
Chart — number of prime factors (with repetition):
9. Mind Map
- Unique primes
- Factor tree
- Lowest/highest powers
- HCF × LCM = ab
- √p proof
- a + b√p
- Ends in 0?
- Composite check
10. Tips and Tricks
- In irrationality proofs, always state "a and b are coprime" — the contradiction depends on it.
- Word problems with "together again", "minimum", "at the same time" → LCM; with "maximum size", "greatest number of rows" → HCF.
- To show a big expression is composite, factor out a common number.
- Using HCF × LCM = product for three numbers.
- Calling 1 a prime (it is neither prime nor composite).
- In the √p proof, forgetting to show that p divides b as well.
11. Important One-liners
- Every composite number has a unique prime factorisation.
- HCF ≤ each number ≤ LCM.
- If two numbers are coprime, their HCF is 1 and LCM is their product.
- √p is irrational for every prime p.
- 6ⁿ, 4ⁿ, 9ⁿ never end in 0; 10ⁿ always does.
- The sum of a rational and an irrational number is irrational.
12. Exam-pattern Questions with Detailed Answers
Express 156 as a product of primes.
Show answer
156 = 2 × 78 = 2 × 2 × 39 = 2² × 3 × 13.
Answer: 2² × 3 × 13
The HCF of two consecutive natural numbers is:
- 0
- 1
- 2
- Their product
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Consecutive numbers are coprime.
Answer: (B)
If HCF(a, b) = 12 and a × b = 1800, find LCM(a, b).
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1800 ÷ 12 = 150.
Answer: 150
Find the HCF and LCM of 12, 15 and 21.
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12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. HCF = 3; LCM = 2² × 3 × 5 × 7 = 420.
Answer: HCF 3, LCM 420
Explain why 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.
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= 5(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009. It has the factor 5, so it is composite.
Answer: 5 × 1009
Check whether 4ⁿ can end with the digit 0.
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4ⁿ = 2²ⁿ; the prime 5 is absent, so it can never be divisible by 10.
Answer: No
Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 8 a.m., when will they next ring together?
Show answer
LCM(9, 12, 15) = 2² × 3² × 5 = 180 min = 3 hours → 11 a.m.
Answer: 11 a.m.
An army contingent of 616 members marches behind a band of 32 members in the same number of columns. Find the maximum number of columns.
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616 = 2³ × 7 × 11, 32 = 2⁵ → HCF = 2³ = 8.
Answer: 8 columns
Prove that √5 is irrational.
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Assume √5 = a/b, a and b coprime, b ≠ 0. Then 5b² = a², so 5 | a², hence 5 | a. Let a = 5c: 5b² = 25c² → b² = 5c², so 5 | b. Then 5 divides both a and b, contradicting coprimality. Hence √5 is irrational. ∎
Answer: Proof by contradiction
Prove that 3 + 2√5 is irrational.
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Suppose 3 + 2√5 = r (rational). Then √5 = (r − 3)/2, which is rational — contradicting that √5 is irrational. Hence 3 + 2√5 is irrational. ∎
Answer: Proof by contradiction
Find the smallest number which, when divided by 28 and 32, leaves remainders 8 and 12 respectively.
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28 − 8 = 20 and 32 − 12 = 20, so the number is 20 less than a common multiple. LCM(28, 32) = 2⁵ × 7 = 224. Number = 224 − 20 = 204.
Answer: 204
Prove that √2 + √3 is irrational.
Show answer
Suppose √2 + √3 = r, rational. Then √3 = r − √2; squaring: 3 = r² − 2r√2 + 2 → √2 = (r² − 1)/(2r), rational — contradiction (r ≠ 0 since √2 + √3 > 0). Hence irrational. ∎
Answer: Proof by contradiction
13. Quick Revision Summary
- FTA: every composite number = unique product of primes.
- HCF: lowest powers of common primes; LCM: highest powers of all primes.
- HCF × LCM = product (two numbers only).
- √p irrational (p prime): assume a/b coprime → contradiction.
- rational ± irrational is irrational; used for 3 + 2√5 type proofs.