Proof by contradiction

~50 min · WMA14 · 1.1

WMA14 · 1.1 · 50 min

A proof by contradiction is won or lost on four sentences, and only one of them is algebra. Assume the opposite of what you want to show. Derive, correctly, from that assumption. State — in words, with a reason — exactly why what you've reached is impossible. Conclude that the assumption must have been false. Three separate examiner-report series (Jan 2021, Oct 2021, Jan 2024) converge on the same finding: candidates who can do the algebra in the middle routinely lose the marks either side of it, not because the mathematics is wrong, but because the sentences that hold it together were never written down.

Key terms in this lesson

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What proof by contradiction is, and what this paper actually rewards

Spec 1.1 names the technique in one line, with guidance naming exactly two required worked examples and one further instruction that matters as much as either: "Proof by contradiction... including proof of the irrationality of √2 and the infinity of primes, and application to unfamiliar proofs." That last phrase is not decoration — it is the paper telling you directly that a proof-by-contradiction question can be built around a statement you have never seen before, so the two named proofs below are worth learning as *worked instances of a reusable structure*, not as two answers to memorise whole.

The structure itself, stripped to its shape: to prove a statement P, assume the opposite of P is true. Use only valid algebra or logic to see what that assumption forces to be true. If it forces something that cannot be true — however the impossibility shows up — then the assumption itself must have been the flaw, since everything built on top of it was done correctly. Conclude that the assumption is false, and therefore P is true. Four moves: assume, derive, name the impossibility and why, conclude.

This is genuinely new content when it first appears on this qualification, and the paper's own general commentary says so directly: "Students find new topic of 'proof' challenging and... candidates often omit questions on this topic or struggle to adopt a suitable strategy to complete the proof" (Jan 2021, general report). Two consequences follow directly from that sentence. First, attempting the question is itself worth something — a strategy that is only partly right still scores method marks a blank answer cannot. Second, "struggle to adopt a suitable strategy" points at the moves either side of the algebra, which is exactly where this lesson's own worked examples and trap list are built to focus.

One more reason this topic is worth the disproportionate attention this lesson gives it: rigorous argument and proof — precisely the skill this technique tests — is assessed under AO2, and AO2 carries 25–30 of this paper's 75 marks (the spec's own figure), the same weighting as AO1 (recall and technique) itself. A 'show that' or 'hence show' instruction anywhere else on the paper is drawing on the exact same skill this lesson teaches, not a separate one confined to the Proof section.

What "a contradiction" actually is

In plain terms

Pretend, just for the sake of the argument, that the thing you're trying to prove is false. Follow that pretend world through some completely correct steps. If you end up somewhere that genuinely cannot happen — a number that is both even and odd at once, a fraction that was supposed to be in its simplest form but turns out not to be, a real number whose square is negative — then the pretend world you built could never have existed in the first place. Nothing you did along the way was wrong, so the only thing that could be wrong is the very first pretend step. That means the "opposite" you assumed was false all along — which means the thing you actually wanted to prove was true the whole time.

In the vocabulary this lesson uses from here on: the "opposite" you assume is the negation of the statement you want to prove. The correct steps that follow from it are the derivation. The impossible thing you reach is the contradiction — and "impossible" here means genuinely impossible, not just unlikely or unusual: "even = odd" and "(a real number)² is negative" are both things that are false for every possible case, with no exceptions to check.

Formally

To prove a statement P, assume ¬P (the negation of P). If a chain of valid steps from ¬P produces some statement Q together with its own negation ¬Q — both true at once, which is never possible for any Q — then ¬P cannot itself be true, because a valid chain of reasoning cannot turn a true starting point into an impossible conclusion. Since ¬P is false, P is true. Nothing about this needs the specific content of P: it is a fact about valid reasoning itself, which is exactly why the same four-move structure applies whether the statement being proved is about irrational numbers, primes, or a cubic's stationary points — a statement this lesson has never shown you a worked example for.

Mechanism

Why one impossibility proves an infinite claim — and why a case split inside it still has to be exhaustive

This paper's WMA12 unit — read as this lesson's own prerequisite — derives a fact worth carrying over directly: a claim of the form "true for every n" is really an infinite AND, one statement per value of n, joined together. Proving that AND directly, case by case, is , and it only works when the list of cases is finite and short enough to write out. Contradiction sidesteps the infinite list a different way: instead of checking every case, it works with a single, general, unspecified case — an arbitrary integer n, or an arbitrary fraction p/q — that stands for every possible instance of the claim at once. If assuming that this one general case fails to satisfy P leads to something impossible, the argument never depended on which particular n, p or q was chosen — so no case, out of infinitely many, can be the exception. That is the entire reason contradiction can establish a TRUE claim over an infinite domain, which is precisely the one thing exhaustion (bounded to a finite list) and (which only ever disproves a FALSE claim) cannot do. There is one place this logic bites back inside a contradiction proof itself, and it is worth stating precisely because it is where marks are actually lost: if the derivation genuinely splits into more than one case — as it does below, where an assumed integer n is either even or odd — then "every integer is even or odd" is itself a two-case for-all claim, and by the exact same AND logic, BOTH branches have to be followed to their own impossibility before the proof is finished. Reaching a contradiction down one branch and treating the whole argument as settled is not a smaller version of a correct proof; it is an incomplete one, in exactly the way a real examiner report records candidates losing marks for reaching "the equation formed had no solutions" and stopping — the same incompleteness, whether it is a missing reason on one branch or a missing branch altogether.

Worked, in full

√2 is irrational — the spec's own named example, in full

01
Assume, for contradiction, that √2 is rational. Then √2 can be written as a fraction $\frac{p}{q}$ , where $p$ and $q$ are integers, $q \neq 0$ — and, one detail worth stating now and remembering for later: the fraction is written in its lowest terms, so $p$ and $q$ share no common factor other than 1.
Earns: M1 — states the assumption and, specifically, the lowest-terms condition. A real examiner report on exactly this proof shape records that most candidates who reached the correct destination still lost the final mark for never writing this exact sentence — stated here, deliberately, at the start, because the whole proof is built to need it again at the very end.
02
Square both sides and rearrange to remove the square root entirely: $\left(\sqrt{2}\right)^2 = \left(\frac{p}{q}\right)^2 \Rightarrow 2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2$ .
Earns: M1 — squares and rearranges to a square-root-free equation. This step is unconditional algebra; nothing about the lowest-terms assumption is used yet.
03
$p^2 = 2q^2$ means $p^2$ is even. To see what that forces about $p$ itself: if $p$ were odd, $p = 2k + 1$ for some integer $k$ , and $p^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ — odd. An odd number can never square to an even one, so by this contrapositive, $p$ itself must be even: $p = 2r$ for some integer $r$ .
Earns: A1 — establishes p is even WITH the reasoning shown (the odd-squares-to-odd argument), not asserted on its own. A near-identical claim, about a cube rather than a square, is exactly what a real examiner report on a two-part proof of this kind flags as commonly skipped: candidates "did correctly factorise the expression, but few made a comment to state that it was odd. Both of these aspects were required to show the contradiction" (Oct 2021, Q10).
04
Substitute $p = 2r$ back into $p^2 = 2q^2$ : $(2r)^2 = 2q^2 \Rightarrow 4r^2 = 2q^2 \Rightarrow q^2 = 2r^2$ .
Earns: A1 — correct substitution and simplification.
05
$q^2 = 2r^2$ is even, by the identical argument used for $p$ in stage 3: if $q$ were odd, $q^2$ would be odd, so $q$ must be even: $q = 2s$ for some integer $s$ .
Earns: A1 — establishes q is even, by repeating the same justification pattern as stage 3 rather than asserting it "by the same logic" with no working shown. Both applications of the lemma are marked identically, because both are doing identical work.
06
State what has actually been reached: $p$ and $q$ are both even, so they share a common factor of 2. This directly contradicts the condition set out in stage 1 — that $\frac{p}{q}$ was already in its lowest terms, with no common factor other than 1.
Earns: A1 — draws the contradiction against the specific, named condition from stage 1, rather than a vague "but this seems wrong." This is the exact line a real examiner report records candidates skipping: reaching "both even" and never stating which earlier condition it breaks.
07
The assumption that √2 is rational leads to something impossible, however $p$ and $q$ were chosen. So the assumption is false: √2 cannot be written as a fraction of integers in lowest terms. √2 is irrational.
Earns: B1 — the concluding statement, required whether or not it feels obvious by this point. A real examiner report on this precise style of question ends with the same lesson stated directly: "It is really important in a proof to include all necessary steps."

Source — Examiner report, Oct 2021

"Most of these forgot, however, to add a statement that a/b was fully simplified, which would mean that the last mark in the question could not be awarded. It is really important in a proof to include all necessary steps."

Complete it yourself

Complete the chain — there are infinitely many prime numbers (the spec's other named example)

01
Assume, for contradiction, that there are only finitely many prime numbers. List every one of them: $p_1, p_2, \ldots, p_n$ — a complete, finite list, with nothing left out.
02
Construct one new number from the whole list at once: $N = p_1 \times p_2 \times \cdots \times p_n + 1$ . Since every $p_i \geq 2$ , $N$ is a genuine integer strictly greater than 1.

Marked, line by line

Prove, by contradiction, that n² − 2 is never divisible by 4, for n an integer. (4) — a REAL WMA14 question: the final question of the June 2022 paper (Q9), 4 marks. The real mark scheme (matched against wma14-01-rms-20220818.pdf) gives one default model answer plus three fully worked alternatives ("Alt 1/2/3"), every one scored M1 → A1 → dM1 → A1* — a dependent method mark ("dM1", the scheme's own notation for a mark that can only be earned once the M1 above it has) sits between the accuracy mark and the closing mark, not a second independent A1. This block is scored against "Alt 3," the one alternative that splits explicitly into the n-even and n-odd cases, the same shape this lesson's own mechanism block teaches. All four mark codes and all four criterion lines below are quoted directly from that real scheme; the specific algebra shown in each `working` field is VERIDIAN's own equivalent derivation (checked independently, mathematically equivalent to Alt 3's own working), not a transcription of Alt 3's working line-for-line.

4 marks available

“Show that” — the answer is already printed above

Matching the printed result isn’t the same as deriving it — real examiner reports describe scripts that adjust flawed working just to still land on it. Write your own full working below before checking it against the mark scheme.

Mechanism

A second route to "impossible": exhausting a target number's own factor pairs, nested inside a contradiction

Every impossibility this lesson has reached so far comes from one algebraic identity, read directly: an equation forces two sides of different parity, or forces a real number's square to be negative. A real WMA14 question (Jan 2026, Q8) does something structurally different. Reaching $(2p+q)(2p-q)=46$ doesn't itself say anything is impossible — it turns a claim about every possible pair of positive integers p, q (an infinite domain) into a claim about the finitely many ways the single number 46 factors as a product of two positive integers (a domain with exactly four members: 1, 2, 23, 46). Because p and q are positive, $2p+q-(2p-q)=2q>0$ , so $2p+q$ is always the larger factor — which rules out half the pairings outright and leaves exactly two to check: $46\times1$ and $23\times2$ . This is the WMA12 prerequisite lesson's own technique, exhaustion — a finite, listable set of cases, every one checked — but running INSIDE a proof by contradiction rather than as the whole proof: the outer shape is still assume, derive, contradiction, conclude, but here "derive" means deriving a short, provably complete list of candidates (provably complete because 46 itself has only four positive divisors, not because anyone chose to stop looking), and "contradiction" means showing every candidate on that list individually fails before the assumption as a whole can be called impossible. This is exactly why the real mark scheme's dependent mark for this route requires BOTH pairs, not one: by the same AND-logic this lesson's other mechanism block already derives for a case split, ruling out one of the two possibilities the assumption allows for is not the same as ruling out the assumption — a genuinely different way to fail a proof from missing the n-odd branch of a parity split, but the identical underlying reason it fails.

Same question, every valid method

Prove, by contradiction, that there are no positive integers p and q such that $4p^2-q^2=46$ . First, factorise $4p^2-q^2$ . (5) — a REAL WMA14 question: Q8 of the January 2026 paper (WMA14/01A, sat 22 January 2026; mark scheme Publications Code WMA14_01_2601_MS, Question Paper Log Number P87595A). Part (a), 1 mark, is the factorisation alone; part (b), 4 marks, is the proof. The real mark scheme credits three routes through part (b) on exactly equal footing — the identical M1 → A1 → dM1 → A1* mark ladder for every one — and adds its own note that any other credit-worthy proof found should be sent to review rather than assumed wrong. All three are shown below, quoting the scheme's own mark descriptions verbatim; the surrounding narration and the specific phrasing of each `working` line are VERIDIAN's own, checked independently against the real scheme and cross-verified with a computer-algebra system, not a line-for-line transcription.

3 valid methods · every one reaches No positive integers p, q satisfy $4p^2-q^2=46$ — all three routes below reach that same conclusion for the same 5 marks. · 5 marks available

01
Factorise the left side: $4p^2 - q^2 = (2p+q)(2p-q)$ .
Quoted directly from the real mark scheme: "Correct factorisation" (Jan 2026, Q8(a)) — 1 of the question's 5 marks, earned in part (a) before the proof in part (b) even begins.
B1
02
Assume, for contradiction, that positive integers p and q exist with $4p^2-q^2=46$ , i.e. $(2p+q)(2p-q)=46$ . Since q is positive, $2p+q$ is always the larger of the two factors, so state one of the two valid pairs: $2p+q=46,\ 2p-q=1$ or $2p+q=23,\ 2p-q=2$ .
Quoted directly from the real mark scheme: "Deduces either of the valid pairs of equations. There is no requirement to justify that 2p + q must be greater than 2p − q and therefore solve the other possible equations" (Jan 2026, Q8(b)) — the scheme is explicit that the inequality doesn't need to be argued for this mark; stating either correct pair earns it outright.
M1
03
Solve the first pair: $2p+q=46,\ 2p-q=1 \Rightarrow p = 11.75$ — not an integer.
Quoted directly from the real mark scheme: "Correctly solves one of the two valid pairs of equations. No working is required, just the correct solution... it is acceptable to only go as far as finding either one of p or q as that is a non-integer" (Jan 2026, Q8(b)).
A1
04
State and solve the second pair too: $2p+q=23,\ 2p-q=2 \Rightarrow p = 6.25$ — also not an integer.
Quoted directly from the real mark scheme: "States and attempts to solve both valid pairs of equations" (Jan 2026, Q8(b)) — dependent on the M1 above, and it is specifically BOTH pairs that earn this mark, not either one alone.
dM1
05
Both valid pairs force a non-integer value of p, and — since $2p+q>2p-q>0$ — those are the only two ways 46 factors into a larger-times-smaller pair of positive integers. No positive integers p and q satisfy the original equation. The assumption is false.
Quoted directly from the real mark scheme: "Full proof by contradiction. This must follow M1A1dM1... a correct set up with words such as 'assume there is' or 'let there be'... correct solution of both valid pairs of equations... a (minimal) conclusion" (Jan 2026, Q8(b)).
A1*

The route this lesson's own new mechanism block is built around: turn the target number itself into the thing being exhausted. It needs no insight about parity, only the factorisation from part (a) and the discipline to find every valid pair before concluding — the exact discipline the dM1 mark above checks for.

In your own words

In one sentence: why does deriving a genuine impossibility from the negation of a statement prove that the original statement is true, rather than merely making it seem likely?

Named traps

assumption-and-conclusion-omitted: The paper's own general commentary states this plainly: candidates "often omit questions on this topic or struggle to adopt a suitable strategy to complete the proof" (Jan 2021, general report), and a specific question the same series records the shape of that struggle directly: "The majority obtained the method mark for suggesting two appropriate odd numbers but full proofs with assumption, reason and conclusion were less common" (Jan 2021, Q3). On this evidence, the algebra in the middle is not where most marks are actually lost — the sentence before it and the sentence after it are.
plausible-but-false-assertion-in-place-of-derivationfudged-reverse-fit: Confirmed directly, and worth reading twice: "False reasoning was sometimes seen, for example: 'n is an integer so n² + 1 is odd'" (Jan 2021, Q3). Read as a claim about every integer n, this sentence is not merely unjustified — it is false: n² + 1 is odd exactly when n is even (n = 2 gives 5) and even when n is odd (n = 3 gives 10). A sentence that reads like a derivation but states something untrue is a worse failure than a visible gap in the working, precisely because it looks finished.
logical-step-skipped-before-the-conclusion-that-needs-it: On a two-part proof building toward the irrationality of √2: "A great many responses to part (a) did correctly factorise the expression, but few made a comment to state that it was odd. Both of these aspects were required to show the contradiction" (Oct 2021, Q10). The pattern here is an ordering failure, not an arithmetic one — reaching a correct intermediate expression and moving straight to what follows from it, without writing the sentence that actually licenses the move. This is exactly the gap the worked chain above forces open at stage 3 and stage 5, where "p is even" and "q is even" are each derived with their own stated reason, not asserted by analogy with each other.
lowest-terms-condition-not-stated: Confirmed on the same question, and specific to any proof built on a fraction assumed to be in simplest form: "Most of these forgot, however, to add a statement that a/b was fully simplified, which would mean that the last mark in the question could not be awarded. It is really important in a proof to include all necessary steps" (Oct 2021, Q10). This is the exact reason the √2 chain above states the lowest-terms condition in its first line rather than its last — it is needed again at the end, to say precisely what the final contradiction (both p and q even) actually contradicts.
impossible-claim-asserted-without-a-reason: On a proof by contradiction that a cubic has no stationary points: "The majority successfully set up their initial assumption... However, many then simply commented that the equation formed had no solutions, giving no explanation as to why, or they gave an inadequate justification, and so gained no further credit" (Jan 2024, Q8). Reaching the right equation is not the same as explaining why it cannot hold — "this has no solutions" is an assertion; "the left side is even and the right side is odd" is the reason that assertion actually needs.
non-algebraic-check-substituted-for-a-required-contradiction: On the same question, naming the specific wrong tools candidates reached for instead of the required derivation: "attempting the 'discriminant', using a graph only, testing values of x or attempting to use small angle approximations" (Jan 2024, Q8) — none of which scored, because a proof by contradiction on this paper is assessed on an algebraic derivation of an impossibility, not on evidence that a check was carried out. A graph or a handful of tested values can make a statement look true; only a derived contradiction proves it.
not-all-valid-factor-pairs-checked: A real WMA14 mark scheme names this exact gap directly, for the one mark that checks it: the dependent mark on the divisor-pair route to Q8 requires "States and attempts to solve both valid pairs of equations" (Jan 2026, Q8) — not one of the two pairs 46×1 and 23×2, however correctly solved, but both. This is a different shape of incompleteness from the n-even/n-odd split earlier in this lesson: there, the two branches come from every integer's own parity; here, they come from every way one specific target number (46) actually factors. The underlying reason both fail is identical, and it is this lesson's own mechanism block that derives it: a proof that rules out only some of the possibilities the assumption allows for has not yet ruled out the assumption itself.

Retrieval — with feedback on every choice

Question 1

3 marks

VERIDIAN-original scenario, in the style of a real question of this shape (Jan 2024, Q8): "Prove, by contradiction, that the curve $y = x^3 + 3x + 1$ has no stationary points." A candidate correctly assumes a stationary point exists, sets $\frac{dy}{dx} = 0$ , and reaches $3x^2 + 3 = 0$ , i.e. $x^2 = -1$ .

Which closing sentence earns full marks, based on real examiner-report guidance on exactly this question type?

Question 2

2 marks

A proof that √2 is irrational assumes √2 = p/q for integers p, q ≠ 0, but never states that p/q is in its lowest terms. What does the real record show about this specific omission?

Question 3

2 marks

A candidate assumes the negation of a statement, does some correct algebra, and reaches an equation that is merely unusual — true for very few integer pairs, but not shown to be impossible for every one. Does this complete a valid proof by contradiction?

Question 4

3 marks

Which of these statements is proof by contradiction the natural technique for — rather than exhaustion or a counter-example search?

Question 5

2 marks

In the proof that there are infinitely many primes, N = p₁ × p₂ × ⋯ × pₙ + 1 is built from an assumed complete list of primes. Dividing N by any prime pᵢ in that list always leaves remainder 1. Why does this matter for the proof?

Question 6

3 marks

A proof by contradiction reaches $(2p+q)(2p-q)=46$ for positive integers p and q. A candidate checks the pair $2p+q=46,\ 2p-q=1$ (giving $p=11.75$ , not an integer), and concludes the original statement is proved. Based on the real mark scheme for this exact question, what is missing?

Reference — not a study method, a lookup

Structure: assume the OPPOSITE. Derive correctly. State WHY what you reached is impossible. Conclude the assumption is false, so the original is true.
M1 (real Jun 2022 MS): the opening mark is generous on wording — a correct equation earns it, no fixed phrase required.
The closing mark needs a REASON the contradiction is impossible (even ≠ odd; a square can't be negative) — not just 'no solutions.'
Spec-named examples: √2 is irrational; there are infinitely many primes. Also applies to statements you've never seen before.
Contradiction proves a TRUE, unbounded 'for all' claim. Exhaustion needs a finite domain; counter-example needs a FALSE claim.
A case split inside a proof (e.g. n even / n odd) is itself a for-all claim — both branches must reach their own contradiction.
A target number's OWN divisors can turn an infinite search into a short, finite list to exhaust (e.g. 4p² − q² = 46 → (2p+q)(2p−q) = 46 has exactly 2 valid factor pairs) — exhaustion nested inside a contradiction. Every pair on the list must be checked, not just one.
A real Jan 2026 mark scheme credits three separate routes on equal footing for one question — divisor-pair testing, odd/even parity, and a q = 2m substitution: proof by contradiction often has more than one legitimate path to the same contradiction.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against a primary Pearson document — WMA14-verified-facts.md for the trap-taxonomy and chain content, and the actual mark-scheme PDF (wma14-01-rms-20220818.pdf) fetched and checked directly for the marked-solution's mark codes and criteria — not carried over from prior course material. The marked-solution's real question ("prove n² − 2 is never divisible by 4," June 2022 Q9) is genuine and worth 4 marks; the real scheme scores it M1 → A1 → dM1 → A1* across one default solution and three labelled alternatives, and this block's M1, A1, dM1 and A1* criterion lines are quoted verbatim from the alternative ("Alt 3") that splits into the n-even/n-odd cases this lesson's own mechanism block teaches. The specific algebra in each working line is VERIDIAN's own equivalent derivation, mathematically checked, not a transcription of Alt 3's own working line-for-line (stated in full where the block itself introduces the question). Every other worked example in this lesson — all three prequestions (including the one built around the real M1 quote above, itself a VERIDIAN-original comprehension question, not a reproduction of the real exam question that quote comes from), the √2 chain, the primes chain-drill, the cubic in the MCQ block, and every other MCQ scenario — is VERIDIAN-original, built around the spec's own two named guidance examples or a fresh statement, never a reproduction of a real Pearson question; where mark codes are attached to VERIDIAN-original working, they are modelled on verified mark-scheme conventions (what M, A and B marks mean, what an ag/A1* final mark requires) rather than transcribed from a real scheme, which for an original question does not exist. A second real WMA14 question was added in a later verification pass: Q8 of the January 2026 paper (Publications Code WMA14_01_2601_MS, Question Paper Log Number P87595A), genuine and worth 5 marks (1 for the factorisation in part (a), 4 for the proof in part (b)), independently re-verified against both PDFs — question paper and mark scheme, fetched and pdftotext'd directly, cover pages checked — rather than carried over from any prior transcription. The method-comparison block's three routes and every one of their M1/A1/dM1/A1* criterion lines are quoted verbatim from that real scheme; the surrounding mechanism-block prose, the trap-taxonomy item and the MCQ item built around this question are VERIDIAN-original commentary, not reproductions of the real question's own wording beyond the question statement itself.

Question 13 marks

Which closing sentence earns full marks, based on real examiner-report guidance on exactly this question type?

" $x^2 = -1$ has no real solutions, since a real number squared can never be negative — so no such x exists, contradicting the assumption that a stationary point exists at some x. The curve has no stationary points."
Correct — this states the specific reason the equation is impossible (a real square cannot be negative), not just that it "has no solutions." That reason is exactly what a real examiner report on this question type records as the thing most often left out.
B" $x^2 = -1$ has no solutions."
This is the documented failure itself. An examiner report on this exact question type records that "many then simply commented that the equation formed had no solutions, giving no explanation as to why, or they gave an inadequate justification, and so gained no further credit" (Jan 2024, Q8) — this sentence is precisely that comment.
C"The graph of $y = 3x^2 + 3$ stays above the x-axis, so there is no stationary point."
A real examiner report on this question type names "using a graph only" directly among the non-credited approaches — a proof by contradiction on this paper requires the impossibility to be derived algebraically, not illustrated.
D"Testing $x = -2, -1, 0, 1, 2$ in $3x^2 + 3$ gives no zero, so there is no stationary point."
The same real report names "testing values of x" directly among the approaches that do not score. A handful of tested values, however many, never rules out every real x the way a derived impossibility does.

Traps tested: Impossible claim asserted without reason · Graphical check substituted for algebra · Numerical testing substituted for proof

Question 22 marks

A proof that √2 is irrational assumes √2 = p/q for integers p, q ≠ 0, but never states that p/q is in its lowest terms. What does the real record show about this specific omission?

It is a documented, specifically costly omission — the final contradiction (that p and q are both even) only contradicts something if the proof stated at the start that p and q could share no common factor
Correct. An examiner report on exactly this proof shape records that most candidates who reached "p and q both even" still lost the final mark, because they "forgot ... to add a statement that a/b was fully simplified" (Oct 2021, Q10) — without that stated condition, "both even" is just an observation, not a contradiction of anything.
BIt doesn't matter — the proof works identically whether or not p/q is stated to be in simplest form
It does matter, structurally: the entire contradiction at the end of the chain (p and q share a factor of 2) depends on having assumed they shared no factor in the first place. Remove that opening condition and "both even" contradicts nothing at all.
CIt only matters for the infinitude-of-primes proof, not for the √2 proof
The reverse is true — the lowest-terms condition is specific to the √2 proof's own fraction-based structure. The primes proof doesn't involve a fraction at all; its contradiction comes from a prime factor of N not being on an assumed complete list.
DIt only matters if q happens to be negative
The sign of q plays no role in this contradiction. What matters is whether p and q share a common factor — a property completely independent of whether either number is positive or negative.

Traps tested: Lowest terms condition treated as optional · Lowest terms condition misattributed · Irrelevant property checked

Question 32 marks

No — a valid contradiction must be impossible without exception (like an even number equalling an odd one, or a real number's square being negative), not merely rare. A surviving case, however unlikely, is enough to leave the original claim unproved
Correct. The whole justification for the technique (see this lesson's own mechanism block) depends on the derived statement being impossible for every possible value, not just unusual for typical ones — an argument that only shows something is rare has not shown the assumption cannot hold.
BYes — mathematical proofs on this paper only need to establish that a result is very likely true
This paper's own AO2 weighting is specifically for "rigorous mathematical arguments and proofs through precise statements, logical deduction and inference" — a likelihood, however high, is not a logical certainty, and proof by contradiction specifically requires the latter.
CYes, provided several specific values were checked and none of them worked
Checking specific values is the method for disproving a claim by counter-example, not for completing a contradiction. A contradiction needs a derived, general impossibility — true for every case at once, not merely absent from the ones tested.
DIt depends on how unusual the equation is
Degree of rarity is not the standard being applied here at all. The only question that matters is whether the derived statement is impossible for every case or merely uncommon for some of them — there is no partial credit for "very unlikely."

Traps tested: Probabilistic standard substituted for logical necessity · Numerical testing substituted for proof · Impossibility treated as a matter of degree

Question 43 marks

Which of these statements is proof by contradiction the natural technique for — rather than exhaustion or a counter-example search?

"√2 is irrational" — an infinite, unbounded claim (about every possible pair of integers p, q) that is TRUE, with no way to check every case directly
Correct. Exhaustion needs a finite, listable domain, which this claim does not have. A counter-example search only settles a claim you suspect is FALSE. Contradiction is the one technique of the three that can establish a true, unbounded claim — exactly the mechanism this lesson's own mechanism block derives.
B"pq is even, for every pair of positive integers p, q with p + q = 7 and p < q"
This domain is finite and small — p can only be 1, 2 or 3 — which is exactly the shape exhaustion is built for (this is, in fact, the WMA12 prerequisite lesson's own worked example). Reaching for contradiction here would mean deriving a general impossibility for a claim that can simply be checked case by case, faster.
C"n² + n + 1 is prime, for every positive integer n"
This claim is false — n = 4 gives 4² + 4 + 1 = 21 = 3 × 7, not prime (the WMA12 prerequisite lesson's own worked disproof). A false claim needs a counter-example search, not a proof by contradiction: there is nothing true here to establish.
D"the sum of any two even numbers is even"
This has a direct, one-line constructive proof: two even numbers can be written as 2a and 2b for integers a, b, and their sum is 2a + 2b = 2(a + b), which is even by definition, for any a and b at all. No assumption of the opposite is needed when a direct route is already this short.

Traps tested: Contradiction reached for when exhaustion would do · Contradiction attempted on a false claim · Contradiction reached for when a direct proof is shorter

Question 52 marks

It shows N is not divisible by any prime on the assumed list, so any prime factor of N (which must exist, since N > 1) is a prime not on the list — contradicting the assumption that the list was complete
Correct. The remainder-1 fact rules out every pᵢ on the list as a factor of N, all at once — which is exactly what forces N's own prime factor to be new, and it is that new prime, not N itself, that breaks the assumption of completeness.
BIt shows N itself must be prime
N is only guaranteed to have SOME prime factor (since every integer greater than 1 does) — N itself need not be prime. The proof never needs to know whether N is prime, only that whichever prime divides it cannot be on the original list.
CIt shows N is even
Nothing about the remainder-1 argument says anything about N's own parity, and N need not be even — if 2 is one of the primes in the list, N is odd (one more than an even product). This isn't the property the proof relies on.
DIt has no real bearing on the proof — the contradiction comes from N simply being larger than every pᵢ on the list
Size alone proves nothing: a larger number could still, in principle, be built entirely from primes already on the list (e.g. by multiplying some of them together again). What actually forces a NEW prime to exist is the remainder-1 fact ruling out every existing one as a factor.

Traps tested: N assumed to be prime itself · Irrelevant property inferred · Wrong quantity identified as the contradiction

Question 63 marks

The second valid pair, $2p+q=23,\ 2p-q=2$ , must also be checked and shown to fail — since $2p+q>2p-q>0$ , these are the only two ways 46 factors into two positive integers, and both possibilities the assumption allows for have to be ruled out before it is contradicted
Correct. The real mark scheme's dependent mark for this route is explicit: "States and attempts to solve both valid pairs of equations" (Jan 2026, Q8) — one failing pair shows one possibility fails, not that every possibility does.
BNothing — the assumption has already been shown impossible once, so the proof is finished
This is the exact gap the real mark scheme's dependent mark checks for. Showing one of the two valid factor pairs fails only rules out one of the two ways the assumption could hold — the other pair could, in principle, have worked.
CThe candidate should instead test every integer from 1 to 46 individually to be thorough
Unnecessary, and not what the scheme credits. Factorising in part (a) already reduces an unbounded search over all p, q to exactly two divisor pairs of 46 — testing individual integers one at a time abandons that reduction for a slower, uncredited method.
DThe candidate should have used the odd/even parity argument instead, since testing factor pairs directly isn't a valid method for this question
Divisor-pair testing is explicitly credited by the real scheme on equal footing with the parity route and the substitution route (all three score M1 → A1 → dM1 → A1*). The problem here isn't the choice of method — it's that only half of the chosen method was carried out.

Traps tested: Single case treated as exhaustive · Unbounded search substituted for a derived finite list · Valid method mistaken for invalid

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Oct 2021 · Q10 — cited directly in this lesson

Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA14.

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Up next

Vectors — scalar product, angle-finding, and skew/parallel/intersecting lines

Every "show that these lines are skew" question is really two separate proofs wearing one sentence — not parallel, and not intersecting — and the mark scheme will not credit the second one for showing something else entirely, however true it happens to be. The scalar-product half of this topic is one formula, \cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}, but the paper is almost never really asking whether you can evaluate it — it is asking whether you fed it the right two vectors in the first place, and real scripts get exactly that wrong often enough that it is the single most-documented error anywhere in this unit.

70 min

Proof by contradiction

What proof by contradiction is, and what this paper actually rewards

What "a contradiction" actually is

Why one impossibility proves an infinite claim — and why a case split inside it still has to be exhaustive

√2 is irrational — the spec's own named example, in full

Complete the chain — there are infinitely many prime numbers (the spec's other named example)

A second route to "impossible": exhausting a target number's own factor pairs, nested inside a contradiction

Deep dive

Practice this for real