Why a composite number has one prime factorisation and no other

What to assume they know

• Prime and composite numbers, and why 1 counts as neither
• Divisibility tests for 2, 3, 5, and trial division by primes up to the square root to decide whether a number is prime
• Index notation: writing repeated factors as powers, and multiplying powers of the same base
• Building a factor tree from earlier classes, and reading its leaves
• That a number ending in 0 is divisible by both 2 and 5

What they should be able to do

• Generate composite numbers by multiplying a chosen set of primes with repetition, and explain why that process alone never proves the reverse direction
• Build the factor tree of a large number and collect its leaves into a product of prime powers
• Build a second, differently branched tree for the same number and observe that the collection of leaves is unchanged
• State what Theorem 1.1 asserts, separating the existence claim from the uniqueness claim, and say which of the two is doing the work in a given argument
• Write a factorisation in the chapter's ascending-prime, combined-powers form, and explain why fixing the order removes the theorem's only escape clause
• Argue that a stated number cannot end in the digit zero by showing that the prime 5 is missing from its factorisation
• Explain why an expression such as a product of primes plus a shared factor is composite, without computing the whole value first

Where it usually goes wrong

• "The theorem just says numbers can be factorised." Half of it does, and that half is visible on any tree. The clause that matters says there is no second, disagreeing factorisation hiding anywhere. Split the statement into its two claims and keep them apart.
• "Different factor trees give different answers." Students who start 32760 at 360 × 91 rather than 2 × 16380 often expect a different result and are surprised into thinking they made an error. Run both trees side by side so the agreement is something they watched happen.
• "3803 × 3607 is a prime factorisation." Not until someone has checked that neither factor splits further. The chapter is unusually explicit about handing this check to the reader.
• "A factorisation is unique, full stop." Only once order is pinned down. 2 × 3 × 5 × 7 and 7 × 5 × 3 × 2 are two different strings and one factorisation, which is exactly why the chapter fixes ascending order.
• "4ⁿ never ends in 0 because I checked the first six powers." A pattern in a table is a reason to suspect, not a reason to assert. The proof is that the prime 5 is not available, and it covers every n at once.
• "1 is prime, so it can go in the factorisation." Allowing 1 would let you pad any factorisation with as many extra factors as you liked, and uniqueness would die immediately. That is the structural reason 1 is excluded.

Questions to check understanding

• Split a given number entirely into primes and present it in index form (Exercise 1.1 Q1 sets five such numbers)
• Decide whether a given expression such as 6ⁿ or 15ⁿ can end in a stated digit, with the reason
• Explain why a sum of the form (product of primes) + (one of those primes) is composite
• Given two different factor trees for one number, state what the theorem predicts about their leaves
• State the Fundamental Theorem of Arithmetic and identify which clause is used in a given piece of reasoning

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data; the chapter prints no answers to its exercises, and no answer key was consulted.

• Small factorisations the chapter opens with (§1.2, p. 2): 2 = 2, 4 = 2 × 2, 253 = 11 × 23. The first is the reminder that a prime is already its own factorisation.
• Numbers built from the primes 2, 3, 7, 11, 23 (§1.2, p. 2). The five products printed, with their values: 7 × 11 × 23 = 1771; 3 × 7 × 11 × 23 = 5313; 2 × 3 × 7 × 11 × 23 = 10626; 2³ × 3 × 7³ = 8232; 2² × 3 × 7 × 11 × 23 = 21252. Verified: all five products are correct as printed. Note that 8232 draws on only three of the five primes and repeats two of them — the collection is a supply, not a checklist.
• The factor tree of 32760 (§1.2, p. 2, drawn as artwork; the numbers live inside the boxes and I read them off the printed page). The tree splits 32760 into 2 and 16380; 16380 into 2 and 8190; 8190 into 2 and 4095; 4095 into 3 and 1365; 1365 into 3 and 455; 455 into 5 and 91; 91 into 7 and 13. Leaves: 2, 2, 2, 3, 3, 5, 7, 13. Collected: 2³ × 3² × 5 × 7 × 13, which the chapter states on p. 3. Verified: 8 × 9 × 5 × 7 × 13 = 32760.
• A second tree for the same number (not in the book, not the chapter's). Split 32760 as 360 × 91 instead. Then 360 → 8 × 45 → 2³ and 3² × 5, and 91 → 7 × 13. Verified: the leaves are again 2, 2, 2, 3, 3, 5, 7, 13. This is the panel that makes uniqueness visible rather than asserted; the chapter draws only one tree, so the second must be built.
• 123456789 (§1.2, p. 3). The chapter gives 3² × 3803 × 3607 and explicitly leaves the primality of the two large factors for the reader to check. Verified: 3803 × 3607 = 13 717 421, and 9 × 13 717 421 = 123 456 789. Also verified by trial division up to 61: 3607 and 3803 are both prime — 61² is 3721, so testing primes to 61 settles both. This check is the section's real content: without it the line is a factorisation into three numbers, not into primes.
• Example 1 — powers of 4 and the digit zero (§1.2, p. 4). Input: 4ⁿ for natural n. The chapter's route is that ending in 0 forces divisibility by 5, while 4ⁿ is 2²ⁿ and so offers only the prime 2. Verified: 4, 16, 64, 256, 1024, 4096 — the last digit cycles 4, 6, 4, 6 and never reaches 0. The cycle is the observation, the missing prime is the proof.
• Exercise 1.1 Q1 — numbers to factorise (§1.2, p. 5): 140, 156, 3825, 5005, 7429. Verified: 140 = 2² × 5 × 7; 156 = 2² × 3 × 13; 3825 = 3² × 5² × 17; 5005 = 5 × 7 × 11 × 13; 7429 = 17 × 19 × 23. The last one has no small prime factor at all and is the item that forces genuine trial division.
• Exercise 1.1 Q5 (p. 5): whether 6ⁿ can end in 0. Verified: 6ⁿ = 2ⁿ3ⁿ, which contains no 5, so it cannot — the same argument as Example 1 with a different pair of primes, and worth running as the audience's turn.
• Exercise 1.1 Q6 (p. 5): why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite. The intended reading is that each is a sum of two terms sharing a factor, so that factor survives the addition. Verified: the first is 13 × 78 = 1014 = 2 × 3 × 13²; the second is 5040 + 5 = 5045 = 5 × 1009, and 1009 is prime.

Figures to have open

• The factor tree of 32760, redrawn as a schematic. It is the chapter's own figure (§1.2, p. 2) and the numbers sit inside the drawn boxes, so it must be rebuilt rather than lifted.
• A second factor tree of 32760 branching differently, with both leaf rows sorted and aligned underneath. This does not exist in the chapter and is an addition made here; it is the strongest single image in the topic.
• A row of "prime tiles" for 2³ × 3² × 5 × 7 × 13 that can be shuffled to show order changing while the tiles do not.
• The Gauss portrait is printed in a boxed aside on p. 3 with his dates. A portrait is not required; a dated label on a timeline carries the same point. Do not reproduce the printed image.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, printed Chapter 1 "Real Numbers", §1.2 "The Fundamental Theorem of Arithmetic", pp. 2–5. Theorem 1.1 is stated on p. 3; the general ascending-order form and Example 1 are on p. 4; Exercise 1.1 begins on p. 5.
• §1.1 "Introduction", p. 1, describes what the chapter intends to do with the theorem.
• The historical aside on p. 3 points at Proposition 14 of Book IX of Euclid's Elements and at Gauss's Disquisitiones Arithmeticae for the first correct proof. Both are named in the book; neither is reproduced here.
• §1.4 "Summary", p. 9, restates the theorem as the chapter's first summary point.