HookThe proof that took 358 years — and then a year to fix
In 1637 the French lawyer and amateur mathematician Pierre de Fermat scribbled a note in the margin of a textbook: the equation \(x^n+y^n=z^n\) has no solution in positive whole numbers when \(n\gt 2\). He added that he had found a truly marvellous proof, but the margin was too narrow to contain it. That single sentence tormented mathematicians for 358 years. It was finally settled in June 1993, when the British mathematician Andrew Wiles — who had worked almost in secret at Princeton for seven years — announced a proof to a stunned lecture room in Cambridge. Then the referees checking his manuscript found a gap. One step did not hold. It took Wiles another fourteen months, and help from his former student Richard Taylor, to repair the hole; the corrected proof appeared in 1995.
The episode is the perfect lesson in what a proof actually is. A proof is not a pile of supporting evidence, and it is not a statement that has been checked in a thousand separate cases. It is a chain of logical steps with no gap anywhere — and a single unhandled case, like the one the referees spotted, sinks the whole argument until it is mended. This section is the grammar of that certainty. You will meet the four moves the Edexcel specification names by name — proof by deduction, proof by exhaustion, disproof by counter-example and proof by contradiction — and finish with the two most celebrated arguments in all of mathematics: the proof that \(\sqrt2\) is irrational, and Euclid's proof that the prime numbers never run out.
ModelWhat a proof is — and what it is not
A mathematical proof is a watertight argument that a statement is true for every case it claims to cover, built only from definitions, agreed axioms and results already proved. The structure barely changes from one proof to the next: you state clearly what you are assuming, you make a sequence of deductions in which every line follows from the one before, and you arrive at the thing you set out to show. Many mathematicians close with the letters QED — Latin quod erat demonstrandum, 'which was to be demonstrated' — or a small filled square \(\square\), both meaning simply 'the proof is complete'.
The distinction that earns and loses marks is general versus specific. Showing that a statement holds for \(n=1,\ 2,\ 3\) proves it for \(n=1,\ 2,\ 3\) and nothing more. To prove it in general you must represent an arbitrary case with algebra — an even number as \(2k\), an odd number as \(2k+1\), a multiple of three as \(3k\) — and reason about that symbol, so a single argument covers every value at once. Notation carries weight too: use \(\Rightarrow\) for 'implies', and reserve the word 'therefore' for a genuine logical consequence, not a hopeful guess. Edexcel's four methods are four shapes this watertight argument can take, and choosing the right shape for the statement in front of you is half the skill.
MechanismProof by deduction — building straight from what you know
Proof by deduction is the workhorse: you start from known facts and definitions and move, step by watertight step, to the conclusion. It is also called direct proof because you never assume anything false and you never split into cases — you build straight from the general representation of the object to the property you want. Almost all the craft is in the algebra of that general object, which is why the opening line is nearly always a definition such as 'Let \(n\) be any integer'.
The most common trap is to prove the converse by accident, or to smuggle in the very thing you are trying to prove as though it were already known. Keep the logic flowing one way. Each line must be justified by the line above it or by a standard result, and the conclusion must appear only at the end, as the final consequence, never as an assumption planted at the start.
Prove that the product of any two odd numbers is odd.
Let the two odd numbers be \(2a+1\) and \(2b+1\), where \(a\) and \(b\) are any integers — this is the general form of an odd number, so it stands for every possible pair at once. Multiply them out: \((2a+1)(2b+1)=4ab+2a+2b+1\). Factor a \(2\) out of the first three terms: \(2(2ab+a+b)+1\). The part \(2(2ab+a+b)\) is a multiple of \(2\) and therefore even, and one more than an even number is always odd. Because \(a\) and \(b\) represented any integers, the product of every possible pair of odd numbers is odd. \(\;\square\)
MechanismProof by exhaustion — when the cases are finite
Proof by exhaustion works when a statement can be broken into a finite number of cases that between them cover every possibility. You prove the statement separately in each case; once no case is left standing, the statement is proved for all of them. The whole art is choosing cases that are genuinely exhaustive — leave a gap and the proof fails, in exactly the way Wiles' first draft failed.
The most famous exhaustion in mathematics is the Four Colour Theorem: that any flat map can be coloured with only four colours so that no two regions sharing a border have the same colour. In 1976 Kenneth Appel and Wolfgang Haken at the University of Illinois reduced the infinite problem to a finite list of unavoidable configurations — around \(1936\) of them — and checked every single one by computer. It was the first major theorem proved by exhaustion on a machine, and it caused a genuine row at the time about whether an argument no human could read by hand still counted as a proof. At A-level the case-lists are short enough to do by pen, but the logic is identical: split into cases, settle each, and leave none out.
Prove that the square of any integer leaves a remainder of \(0\) or \(1\) when divided by \(3\) — it is never \(2\).
Every integer is exactly one of three forms: \(3k\), \(3k+1\) or \(3k+2\). These three cases are exhaustive, so it is enough to square each. Case \(3k\): \((3k)^2=9k^2=3(3k^2)\), remainder \(0\). Case \(3k+1\): \((3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1\), remainder \(1\). Case \(3k+2\): \((3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1\), remainder \(1\). In all three cases the remainder is \(0\) or \(1\) and never \(2\); because the three cases cover every integer, the statement is proved. \(\;\square\)
MechanismDisproof by counter-example — one failure is enough
To disprove a universal statement — one that claims something holds for all values — you need only a single case where it fails. That one case is a counter-example, and exhibiting it is a complete disproof: no further argument is required. This is the deep asymmetry at the heart of proof. Confirming a hundred cases proves nothing about the general claim, yet one failure destroys it outright.
The classic warning against trusting a pattern is an expression studied by Leonhard Euler in the eighteenth century. The formula \(n^2+n+41\) produces a prime number for every integer \(n\) from \(0\) all the way to \(39\) — forty primes in a row, which looks overwhelmingly like a rule. It is not. The pattern is a trap, and the counter-example lands the moment you push one step further. This is precisely why 'it works for the first forty cases' can never be a proof — and why examiners love to hand you a plausible-looking claim and ask you to test it.
Disprove the claim: '\(n^2+n+41\) is prime for every non-negative integer \(n\)'.
You do not argue in general here — you hunt for one failure. Try \(n=40\): \(40^2+40+41=1600+40+41=1681\). Is \(1681\) prime? Notice that \(41^2=1681\), so \(1681=41\times 41\), which is composite. The single value \(n=40\) makes the expression non-prime, so the universal claim is false — and that one counter-example, with no further working, is the entire disproof. It is worth seeing why it breaks: at \(n=41\) the collapse is even plainer, because \(41^2+41+41=41(41+1+1)=41\times 43\) is obviously a multiple of \(41\).
ModelProof by contradiction — assume the opposite and watch it break
Proof by contradiction — Latin reductio ad absurdum — is the boldest move of the four. To prove a statement true, you assume it is false, then show that this assumption forces a logical impossibility, something that simply cannot be. Since the false assumption led to nonsense, the assumption must itself have been wrong, so the original statement is true. It is the natural weapon for claims that are awkward to build directly — especially claims that something is impossible, or that a list never ends.
Two arguments in exactly this shape sit on the Edexcel specification by name, and both are more than two thousand years old. The first proves that \(\sqrt2\) is irrational: that it cannot be written as a fraction of two whole numbers. The second is Euclid's proof, from around 300 BC, that there are infinitely many prime numbers. The engine is the same in each: assume the opposite (a fraction exists; the primes are finite), follow the logic without a single error, and crash into a contradiction that can only be escaped by throwing the assumption away.
Prove that \(\sqrt2\) is irrational.
Assume the opposite: that \(\sqrt2\) is rational, so \(\sqrt2=\dfrac{p}{q}\) for integers \(p\) and \(q\) sharing no common factor — the fraction is in its lowest terms. Squaring both sides gives \(2=\dfrac{p^2}{q^2}\), so \(p^2=2q^2\). Then \(p^2\) is even, which forces \(p\) itself to be even, so write \(p=2r\). Substituting, \((2r)^2=2q^2\Rightarrow 4r^2=2q^2\Rightarrow q^2=2r^2\). Now \(q^2\) is even, so \(q\) is even as well. But if \(p\) and \(q\) are both even they share the factor \(2\) — contradicting the assumption that the fraction was already in lowest terms. The assumption has produced an impossibility, so it was false: \(\sqrt2\) cannot be written as such a fraction, and is therefore irrational. \(\;\square\)
Euclid's proof that there are infinitely many primes runs the same way. Assume there are only finitely many, \(p_1,\ p_2,\ \dots,\ p_n\). Form the number \(N=p_1p_2\cdots p_n+1\). Dividing \(N\) by any prime on the list leaves remainder \(1\), so no listed prime divides it — yet \(N\gt 1\), so it must have some prime factor, one that is not on the list. That contradicts the list being complete, so the primes cannot be finite.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Proof questions are marked for logical structure as much as for the final line, so signpost the shape you are using. Open a deduction with a definition line ('Let \(n\) be any integer'), never a worked example — a single example is the classic zero-mark answer because it proves only that one case. When you write a general odd number as \(2k+1\) or a multiple of three as \(3k\), make the algebra land on a form the conclusion can be read off directly: a factor of \(2\) for even, a leftover \(+1\) for odd, a factor of \(3\) for divisibility.
For exhaustion, state that your cases are exhaustive and say why (for example, 'every integer is \(3k\), \(3k+1\) or \(3k+2\)') before working through them — the mark for completeness is separate from the marks for the individual cases. For a disproof, do not argue in general at all: simply present one counter-example, show the arithmetic that makes it fail, and state that the claim is therefore false. For contradiction, write 'Assume, for contradiction, that…' as your first line, keep every step reversible and correct, and finish by naming the contradiction explicitly ('but this contradicts \(p\) and \(q\) having no common factor'). The examiner is looking for that named clash — it is the moment the proof actually works.