Mini-Lesson

Proof

This mini-lesson covers the whole of Topic 1 — Proof in Edexcel 9MA0: the structure of a mathematical argument (⇒, ⇐, ⇔, ∴, ≡), proof by deduction, proof by exhaustion, disproof by counter-example, and — new at A-level — proof by contradiction, including the two classics: √2 is irrational and there are infinitely many primes.

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.

Proof · language

The language of proof

A proof is a watertight chain of reasoning from things already agreed (definitions, axioms, established results) to the statement you want. Every line must follow from the last — an example is not a proof.

P ⇒ Q · P ⇐ Q · P ⇔ Q⇒ “implies” · ⇐ “is implied by” · ⇔ “implies and is implied by” (if and only if)

x = 3 ⇒ x² = 9 is true, but the converse is false (x could be −3), so we may not write ⇔ here.
x − 1 = 0 ⇔ x = 1 — both directions hold, so ⇔ is correct.
An identity uses ≡ : it is true for every value, e.g. (x + 1)² ≡ x² + 2x + 1. An equation is true only for particular values.
∴ means “therefore” and closes the argument.

Exam trap: markers penalise the lazy ⇔. Only use it when both directions genuinely hold. Also: squaring is not reversible, so a ⇒ b does not give you b ⇒ a for free.

Proof · deduction

Proof by deduction

Deduction is the workhorse: take a general object, argue algebraically, land on the result. The trick is choosing the right general form.

even = 2k · odd = 2k + 1 · consecutive = n, n + 1k, n ∈ ℤ — always say what your letters are

Worked example — prove that the square of any odd number is odd

Let n be odd, so n = 2k + 1 for some integer k.

n² = (2k + 1)² = 4k² + 4k + 1 = 2(2k² + 2k) + 1

2k² + 2k is an integer, so n² is 2 × (integer) + 1, i.e. odd. ∎

Worked example — prove (n + 1)³ − n³ is odd for every integer n

(n + 1)³ − n³ = 3n² + 3n + 1 = 3n(n + 1) + 1

n and n + 1 are consecutive, so one of them is even ⇒ n(n + 1) is even ⇒ 3n(n + 1) is even.

Even + 1 is odd. ∎

Exam trap: writing “let n = 2k + 1 and m = 2k + 1” for two odd numbers forces them to be equal. Use different letters: 2k + 1 and 2m + 1.

Quick check

Is it a proof?

?A student writes: “3² = 9 is odd, 5² = 25 is odd, 7² = 49 is odd, so the square of any odd number is odd.” What is wrong with this?

Proof · exhaustion

Proof by exhaustion

Exhaustion splits the problem into a finite list of cases that between them cover every possibility, then checks each one. It is only valid if the cases really are exhaustive.

Worked example — prove that every prime p with 3 < p < 30 has the form 6k ± 1

Every integer is 6k, 6k ± 1, 6k ± 2 or 6k + 3.

6k, 6k ± 2 are even; 6k + 3 is a multiple of 3. So any prime bigger than 3 must be 6k ± 1.

Check: 5 = 6−1, 7 = 6+1, 11 = 12−1, 13 = 12+1, 17 = 18−1, 19 = 18+1, 23 = 24−1, 29 = 30−1. ∎

Case-splitting is exhaustion too: to prove n² + n is even for all integers n, split into n even and n odd — two cases that cover ℤ.

Exam trap: exhaustion with a huge or infinite number of cases is not a proof. If you cannot list the cases, deduce instead.

Calculate

Your turn — the famous near-miss

1Euler noticed that n² + n + 41 is prime for n = 1, 2, 3, … . It eventually fails. Find the smallest positive integer n for which n² + n + 41 is not prime.

Set-up

If n = 40: 40² + 40 + 41 = 1600 + 40 + 41 = 1681. And 1681 = 41 × 41, so it is composite.

(All of n = 1 … 39 do give primes — the pattern really does hold that far, which is exactly why one counter-example matters.)

n

Hint: think about when the 41 can be factored out — try n = 40 and factorise.

Proof · counter-example

Disproof by counter-example

To disprove a universal statement (“for all …”) you need exactly one case where it fails. One counter-example is a complete disproof — no more work needed.

“n² + n + 41 is prime for all positive integers n” — false: n = 40 gives 1681 = 41².
“x² > x for all real x” — false: x = 0.5 gives 0.25 > 0.5, which is untrue.
“If x² = 9 then x = 3” — false: x = −3.
“a > b ⇒ a² > b²” — false: a = 1, b = −5 gives 1 > −5 but 1 < 25.

Exam trap: a counter-example must be stated and checked. Write the numbers in and show the statement fails — quoting the value alone is usually 1 mark of 2. And a counter-example can never prove a statement, only disprove one.

Quick check

Kill the claim

?Which single value disproves the claim “for all real x, x² ≥ x”?

Proof · contradiction

Proof by contradiction — the method

New at A-level. To prove a statement P: assume P is false, reason correctly, and arrive at something impossible. Since the reasoning was sound, the assumption must have been wrong — so P is true.

Assume ¬P ⇒ … ⇒ contradiction ∴ Pthe contradiction is your proof — say clearly what it contradicts

Worked example — if n² is even then n is even

Assume the opposite: n² is even but n is odd. Then n = 2k + 1.

n² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is odd.

That contradicts n² being even. So n must be even. ∎

Exam trap: negate the statement properly. The negation of “for all x, P(x)” is “there exists an x with not P(x)” — not “for all x, not P(x)”.

Proof · contradiction

The two classics you must know

√2 is irrational

Assume √2 is rational: √2 = a/b with a, b integers, b ≠ 0, and the fraction in lowest terms (no common factor).

Then 2 = a²/b² ⇒ a² = 2b², so a² is even ⇒ a is even (previous screen) ⇒ a = 2c.

So 4c² = 2b² ⇒ b² = 2c² ⇒ b² is even ⇒ b is even.

But then a and b are both even — they share a factor 2, contradicting “lowest terms”. ∴ √2 is irrational. ∎

There are infinitely many primes

Assume there are finitely many: p₁, p₂, …, p_n is the complete list.

Let N = p₁ × p₂ × … × p_n + 1. Dividing N by any p_i leaves remainder 1, so no p_i is a factor of N.

But every integer > 1 has a prime factor. That prime is not on the list — contradiction. ∴ there are infinitely many primes. ∎

Subtlety worth a mark: N itself need not be prime. With the list 2, 3, 5, 7, 11, 13 you get N = 30031 = 59 × 509 — composite, but its prime factors (59, 509) are missing from the list. The contradiction still lands.

Calculate

Your turn — Euclid’s number

2Suppose (falsely) that the only primes are 2, 3, 5, 7, 11 and 13. Following Euclid, calculate N = 2 × 3 × 5 × 7 × 11 × 13 + 1.

Set-up

2 × 3 = 6; 6 × 5 = 30; 30 × 7 = 210; 210 × 11 = 2310; 2310 × 13 = 30030.

N = 30030 + 1 = 30031 = 59 × 509 — and 59 and 509 are primes that were not on the list.

Hint: 2 × 3 × 5 × 7 × 11 × 13 = 30030. Then add 1.

Quick check

Opening line

?You are asked to prove by contradiction that √2 is irrational. Which is the correct opening?

Sort it

Which method of proof?

Tap a card, then tap the method of proof it belongs to.

🧩 Deduction

📋 Exhaustion

💥 Counter-example

Proof · toolkit

The algebraic proof toolkit

Most deduction marks come down to writing the right general form and then factorising to reveal the structure.

Even: 2k · Odd: 2k + 1 · Consecutive integers: n, n + 1 · Consecutive odds: 2k + 1, 2k + 3
“Multiple of m” means you must end at m × (integer) — say the bracket is an integer.
Difference of two squares is your friend: a² − b² = (a − b)(a + b).

Worked example — the sum of the first n odd numbers is n²

1 + 3 + 5 + … + (2n − 1) is an arithmetic series with a = 1, d = 2.

S = n/2 [2a + (n − 1)d] = n/2 [2 + 2n − 2] = n/2 × 2n = n². ∎

Worked example — n³ − n is a multiple of 6

n³ − n = n(n² − 1) = (n − 1)n(n + 1): three consecutive integers.

Among any 3 consecutive integers one is a multiple of 3, and at least one is even. So the product is divisible by 2 × 3 = 6. ∎

Calculate

Your turn — sum of odd numbers

3Using the result above (the sum of the first n odd numbers is n²), find 1 + 3 + 5 + … + 49.

Set-up

The last term is 2n − 1 = 49 ⇒ n = 25 terms.

Sum = 25² = 625. (Check with the AP formula: 25/2 × (1 + 49) = 25/2 × 50 = 625. ✓)

Hint: 49 = 2n − 1 gives n = 25, and the sum is n².

Calculate

Your turn — difference of squares

4Prove-and-evaluate: it can be shown that (2n + 1)² − (2n − 1)² ≡ 8n. Evaluate the left-hand side when n = 7.

Set-up

Identity: (2n+1)² − (2n−1)² = (4n² + 4n + 1) − (4n² − 4n + 1) = 8n — always a multiple of 8.

At n = 7: 15² − 13² = 225 − 169 = 56, and 8 × 7 = 56. ✓

Hint: (15)² − (13)² = 225 − 169. It should equal 8 × 7.

Calculate

Your turn — divisible by 6

5You proved n³ − n is always a multiple of 6. For n = 10, calculate the value of (n³ − n) ÷ 6.

Set-up

n³ − n = 1000 − 10 = 990.

990 ÷ 6 = 165 — an integer, exactly as the proof guarantees. (990 = 9 × 10 × 11, three consecutive integers. ✓)

Hint: 10³ − 10 = 990, then divide by 6.

Match it

Notation and method

Tap a card on the left, then its partner on the right.

Situation

Correct description

Quick check

⇒ or ⇔?

?Let P be “n is a multiple of 4” and Q be “n is even” (n ∈ ℤ). Which is correct?

Proof · exam technique

Getting the marks

Define your letters. “Let n = 2k + 1, k ∈ ℤ” earns the first mark on countless questions.
Finish the sentence. After the algebra, write the conclusion in words: “which is 2 × integer + 1, hence odd”. The final mark is for the statement, not the algebra.
Contradiction: state the assumption, state the contradiction, state the conclusion. Three separate sentences, three marks.
Counter-example: give the value and the arithmetic showing it fails.
Never assume what you are proving. Starting from the answer and working back to 0 = 0 scores nothing unless every step is reversible and you say so.

Where this reappears: differentiation from first principles is a proof; the arithmetic-series sum formula is proved by reversal; compound-angle and R-form results are proved in Topic 5. Proof is not one topic — it is the whole paper.

Recap

The big ideas to take away

Deduction: general form (2k, 2k+1, n, n+1) ⇒ algebra ⇒ conclusion in words

Exhaustion: finite, genuinely complete list of cases — check every one

Counter-example: one failing case disproves a universal claim (n = 40 kills n² + n + 41)

Contradiction: assume the negation ⇒ absurdity. √2 irrational; infinitely many primes

Notation: ⇒ one way · ⇔ both ways · ≡ identity (true for all values)

That is the whole of Topic 1 — and the reasoning style the rest of 9MA0 is marked against. Press Finish to see your score.

🏆

Mini-lesson complete!

⭐⭐⭐

You have worked through Proof for Edexcel A-level Mathematics (9MA0). 🎉

Your stars: 0 / 0

Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to all subjects