Fundamental Theorem Of Arithmetic

Two friends are cutting a big block of chocolate: whichever way one of them snaps it into pieces, the smallest square blocks are always identical — and there is always the same number of them. Whole numbers behave exactly like that chocolate. Every composite number can be snapped apart into the same unchanging set of prime "square blocks", and that single fact quietly powers most of the Real Numbers chapter — HCF, LCM, decimal-expansion tests and irrationality proofs are all just this one idea in disguise.

Definition: A prime number is a natural number greater than 1 whose only divisors are 1 and itself (2, 3, 5, 7, 11, 13, …). A composite number is a natural number greater than 1 that has at least one divisor other than 1 and itself (4, 6, 8, 9, 10, 12, …).

Definition: The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors are written. In symbols, for every composite N there is exactly one collection of primes such that N = p₁ × p₂ × … × pₖ (the same primes, each appearing the same number of times).

Section 1 — What the theorem really says

The theorem carries two claims, and a board answer must respect both. The existence claim says: pick any composite number and you can break it into primes. The uniqueness claim says: no matter how you break it, the bag of primes you end up with is exactly the same — same primes, same number of copies of each. The only thing free to change is the order in which you write them.

Take 60. Split it as 60 = 6 × 10, or as 60 = 4 × 15, or as 60 = 2 × 30 — expand each fully and you always land on two 2s, one 3 and one 5. There is no secret alternative factorisation hiding anywhere. That is why we write the answer in exponent form with primes in ascending order:

60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.

This compact form captures the "bag of primes" cleanly and is exactly what an examiner expects to see — write 2² × 3 × 5, never 5 × 3 × 2 × 2.

Why 1 is deliberately not a prime. If 1 were allowed as a prime, then 60 = 2² × 3 × 5 = 1 × 2² × 3 × 5 = 1 × 1 × 2² × 3 × 5, and so on — infinitely many "different" factorisations. Uniqueness would collapse. Mathematicians exclude 1 from the primes precisely to keep the theorem clean; a prime needs two distinct divisors, and 1 has only one.

Section 2 — The factor-tree method, step by step

A factor tree is the standard tool to find the prime factorisation. Follow a fixed routine so you never miss a step.

Step 1: Write the number at the top.
Step 2: Split it into any product of two factors (grab a small prime like 2, 3 or 5 if you can spot one).
Step 3: For each branch ask, "Is this prime?" If yes, stop that branch. If no, split it further.
Step 4: Continue until every branch ends in a prime.
Step 5: Collect all the primes and write them with exponents, in ascending order.

The tree below factorises 540. Notice how every path — whichever split you begin with — funnels down to the same primes: 540 = 2² × 3³ × 5.

Before you put your pen down, run three checks: (1) Are all my factors prime — no 4, 6, 8, 9, 15, 21, 25 left hiding? (2) If I multiply them back, do I recover the original number? (3) Have I written the answer in exponent form with primes in ascending order? A common slip is leaving a composite factor inside the answer: writing "60 = 4 × 15" is a factorisation but not a prime factorisation, because 4 and 15 are still composite. Always push every branch down to a prime.

Section 3 — HCF and LCM from prime factorisation

Once numbers are in prime form, HCF (Highest Common Factor) and LCM (Lowest Common Multiple) become almost mechanical.

• HCF = product of the smallest power of each prime that is common to all the numbers.
• LCM = product of the greatest power of every prime that appears in any of the numbers.

The strip diagram below shows the shared and extra primes of 60 = 2² × 3 × 5 and 72 = 2³ × 3². The overlap (the smaller powers 2² and 3) is the HCF; the full union (largest of every prime) is the LCM.

For two positive integers a and b there is a golden identity:

HCF(a, b) × LCM(a, b) = a × b.

This gives instant cross-checks and shortcuts: LCM = (a × b) ÷ HCF, and HCF = (a × b) ÷ LCM. Important caution — this product rule holds only for two numbers. For three or more it fails in general; for 6, 72, 120 we get HCF = 6 and LCM = 360, but 6 × 360 = 2160 while 6 × 72 × 120 = 51840, nowhere near equal.

Section 4 — Why uniqueness is the bedrock (with the irrationality lemma)

Why is the uniqueness clause such a big deal? Because nearly every result in Real Numbers quietly leans on it. When you declare "HCF(60, 72) is definitely 12", you are relying on each number having one and only one prime skeleton; a second hidden factorisation could shift the answer.

The clearest payoff is the proof that √2 is irrational, which NCERT builds on a single lemma:

Lemma: If a prime p divides a², then p divides a. (This is a direct consequence of the Fundamental Theorem: the prime p must appear in the unique factorisation of a².)

Proof that √2 is irrational (by contradiction).
Assume, to the contrary, that √2 is rational. Then √2 = p/q for some integers p, q with q ≠ 0 and HCF(p, q) = 1 (lowest terms).
Squaring: 2 = p²/q², so p² = 2q². Hence 2 divides p². By the lemma (with p = 2), 2 divides p. Write p = 2m.
Substitute: (2m)² = 2q² ⇒ 4m² = 2q² ⇒ q² = 2m². So 2 divides q², and again by the lemma 2 divides q.
But now 2 divides both p and q, contradicting HCF(p, q) = 1. The assumption is false, so √2 is irrational. ∎

The identical argument proves √3, √5, √7, … irrational (replace the prime 2 by 3, 5, 7). Two corollaries you must remember:

• Rational ± Irrational = Irrational (e.g., 5 + √2 is irrational).
• Non-zero rational × Irrational = Irrational (e.g., 3√2 is irrational).

Beware the trap "sum of two irrationals is always irrational" — false, since √2 + (−√2) = 0 is rational.

Section 5 — Everyday power of the theorem

The theorem is not just exam theory; it is the reason repeating events sync up. Imagine two LED signs outside a shop: one blinks every 60 seconds, the other every 72 seconds, both starting together at 12:00:00. They next blink together after LCM(60, 72) = 360 seconds = 6 minutes, i.e. at 12:06:00. RFID tags, traffic-signal timers and railway announcement cycles all ride on this LCM logic — and it returns a single definite answer only because the prime skeleton is unique.

The same skeleton explains a neat divisibility fact: a number ending in 0 must have both 2 and 5 in its factorisation (since 10 = 2 × 5). So 6ⁿ can never end in 0, because 6ⁿ = 2ⁿ × 3ⁿ contains no factor 5 — no amount of raising to powers can conjure a prime that was not there to begin with. This "you cannot invent a missing prime" idea is the engine behind a whole family of board HOTS questions.

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Work through these examples. They begin with plain factorisation and HCF/LCM, then build to the twists NCERT boards enjoy — verification, three-number cases, irrationality proofs, and "invent-a-missing-prime" reasoning. Keep handy: HCF = smallest powers of common primes, LCM = greatest powers of all primes, and HCF × LCM = a × b (two numbers only).

Example 1 — Factorise using a factor tree (basic)

Q: Express 96 as a product of its prime factors.
Solve: 96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3.
Collect the primes: five 2s and one 3.
Answer: 96 = 2⁵ × 3. Check: 2⁵ × 3 = 32 × 3 = 96 ✓

Example 2 — HCF and LCM by prime factorisation

Q: Find the HCF and LCM of 96 and 404, and verify HCF × LCM = product.
Given: 96 = 2⁵ × 3 and 404 = 2² × 101 (101 is prime).
Solve:
HCF = smallest power of each common prime. Only common prime is 2, smallest power 2² ⇒ HCF = 2² = 4.
LCM = greatest power of every prime = 2⁵ × 3 × 101 = 32 × 3 × 101 = 9696.
Verify: HCF × LCM = 4 × 9696 = 38784, and 96 × 404 = 38784.
Answer: HCF = 4, LCM = 9696, and both products equal 38784 ✓

Example 3 — Find the missing number from HCF and LCM

Q: The HCF of two numbers is 9 and their LCM is 360. If one number is 45, find the other.
Given: HCF = 9, LCM = 360, one number a = 45.
Formula: HCF × LCM = a × b (valid for two numbers).
Solve: 9 × 360 = 45 × b ⇒ 3240 = 45b ⇒ b = 3240 ÷ 45 = 72.
Answer: The other number is 72. Check: 45 = 3² × 5 and 72 = 2³ × 3² ⇒ HCF = 3² = 9 and LCM = 2³ × 3² × 5 = 360 ✓

Example 4 — Tricky: three numbers, and why the product rule breaks

Q: Find HCF and LCM of 6, 72 and 120 by prime factorisation. Does HCF × LCM = product of the three numbers?
Given: 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5.
Solve:
HCF = smallest powers of primes common to all three: 2¹ and 3¹ ⇒ HCF = 2 × 3 = 6.
LCM = greatest powers of every prime: 2³ × 3² × 5 = 8 × 9 × 5 = 360.
Test the product rule: HCF × LCM = 6 × 360 = 2160, but 6 × 72 × 120 = 51840. Not equal.
Answer: HCF = 6, LCM = 360. The identity HCF × LCM = product holds only for two numbers, so it correctly fails here. Trap: never apply the two-number shortcut to three or more numbers.

Example 5 — Tricky: word problem (LCM of blinking / bells)

Q: Three temple bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 6:00 a.m., at what time will they next ring together?
Given: Intervals 9 = 3², 12 = 2² × 3, 15 = 3 × 5.
Formula: Next common ring after LCM(9, 12, 15) minutes.
Solve: LCM = greatest powers = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes = 3 hours.
6:00 a.m. + 3 hours = 9:00 a.m.
Answer: They next ring together at 9:00 a.m. Sanity check: 180 ÷ 9 = 20, 180 ÷ 12 = 15, 180 ÷ 15 = 12 — all whole numbers, so 180 is indeed a common multiple ✓

Example 6 — Tricky: can a power end in zero? (HOTS)

Q: Check whether 6ⁿ can end with the digit 0 for any natural number n. Justify using the Fundamental Theorem of Arithmetic.
Solve: A number ends in 0 only if it is divisible by 10 = 2 × 5, i.e. its prime factorisation contains both 2 and 5. Now 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. By the uniqueness of prime factorisation, the only primes in 6ⁿ are 2 and 3 — the prime 5 never appears. Since 5 is absent, 6ⁿ is not divisible by 10.
Answer: No, 6ⁿ can never end with 0, because its unique prime factorisation contains no factor 5. Key idea: you cannot invent a prime that was never in the number.

Example 7 — Tricky: prove √5 is irrational

Q: Prove that √5 is irrational.
Solve (contradiction): Suppose √5 is rational, so √5 = p/q with integers p, q, q ≠ 0 and HCF(p, q) = 1.
Squaring: 5 = p²/q² ⇒ p² = 5q². So 5 divides p². By the lemma "prime p ∣ a² ⇒ p ∣ a", 5 divides p. Write p = 5m.
Then (5m)² = 5q² ⇒ 25m² = 5q² ⇒ q² = 5m². So 5 divides q², hence 5 divides q.
Now 5 divides both p and q, contradicting HCF(p, q) = 1.
Answer: The assumption is false, so √5 is irrational. ∎ (The same steps work for √2, √3, √7 — swap in the relevant prime.)

Example 8 — Tricky: assertion–reason on primes and 1

Q: Assertion (A): The number 1 is neither prime nor composite.
Reason (R): A prime has exactly two distinct factors and a composite has more than two; 1 has only one factor. Choose: (a) both true and R explains A; (b) both true but R does not explain A; (c) A true, R false; (d) A false.
Solve: 1 has exactly one factor (itself), so it fails the "two distinct factors" test for a prime and the "more than two factors" test for a composite — hence 1 is neither. That is precisely what R states, so R is the correct reason for A. Both statements are true and R explains A.
Answer: (a) Both A and R are true, and R is the correct explanation of A. (This is also why 1 is excluded from primes — to keep prime factorisation unique.)

Cover the answer, recall, then check. 8 cards on prime factorisation.

Q1. State the Fundamental Theorem of Arithmetic.
A1. Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of the factors.

Q2. Write 140 as a product of primes.
A2. 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7.

Q3. How do you find HCF using prime factorisation?
A3. Take the product of the smallest powers of each common prime factor.

Q4. How do you find LCM using prime factorisation?
A4. Take the product of the greatest powers of all prime factors that appear.

Q5. State the relation between HCF, LCM and two numbers.
A5. HCF(a,b) × LCM(a,b) = a × b.

Q6. If HCF(a, b) = 9 and LCM = 90, and a = 18, find b.
A6. b = (HCF × LCM)/a = (9 × 90)/18 = 45.

Q7. Find the HCF of 96 and 404 using prime factors.
A7. 96 = 2⁵×3, 404 = 2²×101; common = 2² = 4, so HCF = 4.

Q8. Why can the product of two odd numbers never be even?
A8. Neither has 2 as a prime factor, so their product has no factor of 2 and stays odd.