In short

Proof by contradiction works by assuming the opposite of what you want to prove and then showing that this assumption leads to something impossible — a logical contradiction. Since a false assumption cannot produce only true statements, the original claim must be true. The technique is especially powerful for proving that something cannot exist or that a number is not of a particular kind, such as proving that \sqrt{2} is irrational.

Some mathematical truths resist the direct approach. Try to prove directly that \sqrt{2} is irrational — that it cannot be written as a fraction p/q with p and q integers. Where would you even start? You cannot check every possible fraction (there are infinitely many), and there is no obvious algebraic path from "\sqrt{2} exists" to "\sqrt{2} is not a ratio of integers." The direct road has no clear first step.

But suppose, just for the sake of argument, that \sqrt{2} is rational. Suppose you can write it as p/q. What happens? You follow the algebra carefully, and you arrive at a statement that contradicts itself — something that is both true and false at the same time. That contradiction means your starting assumption was wrong. And if the assumption "\sqrt{2} is rational" is wrong, then \sqrt{2} must be irrational.

That is proof by contradiction: assume the opposite, follow the logic, and wait for it to break. The technique has been used for over two thousand years. The proof that \sqrt{2} is irrational is one of the oldest and most beautiful arguments in all of mathematics, and it reaches its conclusion not by building something up, but by watching a false assumption tear itself apart.

The logical foundation

In Logic and Propositions, you met the idea that a proposition is either true or false — there is no third option. This is the law of the excluded middle: for any proposition P, either P is true or \lnot P (not P) is true.

Proof by contradiction exploits this law. If you want to prove P is true, you can instead assume \lnot P and show that \lnot P leads to a contradiction — a statement that is both true and false, which is impossible. Since \lnot P leads to impossibility, \lnot P cannot be true. And since one of P or \lnot P must be true, it follows that P is true.

Proof by Contradiction (Reductio ad Absurdum)

To prove a statement P:

  1. Assume \lnot P (the negation of P).
  2. Using this assumption and known facts, derive a contradiction — a statement of the form "Q and \lnot Q" for some proposition Q.
  3. Conclude that \lnot P is false, so P must be true.

The Latin name reductio ad absurdum — "reduction to absurdity" — captures the idea perfectly. You reduce the negation of your claim to an absurdity, and the absurdity proves your claim.

Structure of a proof by contradictionA flow diagram with four boxes. The first box says Assume not-P. An arrow leads to a second box labelled logical steps. Another arrow leads to a third box saying Q and not-Q, a contradiction, coloured red. A final arrow curves back to a fourth box saying Therefore P is true. The contradiction box is marked with a cross symbol.Assume ¬PLogical stepsQ and ¬Qcontradiction!Therefore P is true
The skeleton of a proof by contradiction. You assume the negation of what you want to prove, follow the logic until you hit an impossibility, and then conclude that the original statement must be true. The contradiction (the red box) is the engine — it is what forces the conclusion.

The classic example: \sqrt{2} is irrational

This is the proof that every mathematics student should see at least once. It is short, it is clean, and it demonstrates the power of contradiction better than any abstract description.

Claim. \sqrt{2} is irrational — that is, there are no integers p and q (with q \neq 0) such that \sqrt{2} = p/q.

Proof. Assume, for the sake of contradiction, that \sqrt{2} is rational. Then \sqrt{2} = p/q for some integers p and q with q \neq 0. Choose p and q so that the fraction p/q is in lowest terms — that is, p and q share no common factor other than 1. (You can always reduce a fraction to lowest terms by dividing out common factors, so this is a valid choice.)

Step 1. Square both sides.

2 = \frac{p^2}{q^2}

so p^2 = 2q^2.

Step 2. Conclude that p^2 is even. Since p^2 = 2q^2, the right side is 2 times an integer, so p^2 is even.

Step 3. Conclude that p is even. If p were odd, then p^2 would be odd (because the square of an odd number is odd — proved in Mathematical Proof — Direct Proof). Since p^2 is even, p must be even. So p = 2k for some integer k.

Step 4. Substitute back. Since p = 2k:

p^2 = (2k)^2 = 4k^2

But p^2 = 2q^2, so 4k^2 = 2q^2, which gives q^2 = 2k^2.

Step 5. Conclude that q^2 is even, so q is even. The same argument from Steps 2 and 3 applies: q^2 = 2k^2 means q^2 is even, so q is even.

Step 6. Arrive at the contradiction. Both p and q are even, so they share a factor of 2. But we chose p/q to be in lowest terms — meaning p and q share no common factor other than 1. The statements "p and q share a factor of 2" and "p and q share no common factor other than 1" cannot both be true.

This is the contradiction. The assumption that \sqrt{2} is rational has led to an impossibility. Therefore \sqrt{2} is irrational. \square

Flow of the proof that root two is irrationalA vertical chain of six boxes showing each step of the contradiction proof. The first box assumes root two equals p over q in lowest terms. The second derives p squared equals 2 q squared. The third concludes p is even. The fourth substitutes p equals 2k and derives q squared equals 2 k squared. The fifth concludes q is even. The sixth box, highlighted in red, states the contradiction: both p and q are even, but the fraction was in lowest terms.Assume √2 = p/q in lowest termsp² = 2q²square both sidesp² is even → p is even → p = 2k4k² = 2q² → q² = 2k²substituteq² is even → q is evenp and q both even — contradicts lowest terms∴ √2 is irrational
The entire proof in six steps. The assumption at the top — $\sqrt{2} = p/q$ in lowest terms — leads, through clean algebra, to the conclusion that both $p$ and $q$ are even. But a fraction in lowest terms cannot have both numerator and denominator even. The contradiction at the bottom kills the assumption, and the irrationality of $\sqrt{2}$ is established.

The proof uses nothing beyond the definition of "even," the fact that the square of an odd number is odd, and the definition of "lowest terms." That is why it is so striking — a deep truth about the nature of \sqrt{2} follows from the simplest possible ingredients.

The geometry behind \sqrt{2}

The irrationality of \sqrt{2} is not just an algebraic curiosity — it has a geometric origin. Consider a square with side length 1. The diagonal of this square, by the Pythagorean theorem, has length \sqrt{1^2 + 1^2} = \sqrt{2}. So the proof above is really saying: there is no ruler, marked with equally spaced rational points, that can measure both the side and the diagonal of a unit square exactly. The side is 1 unit; the diagonal is \sqrt{2} units; and no common unit of measurement — no matter how small — divides both lengths evenly.

A unit square with its irrational diagonalA square with side length one unit drawn in the centre of the figure. The diagonal from the bottom-left to the top-right corner is drawn as a thick red line and labelled root 2. The two sides forming the right angle are labelled 1. A small square in the bottom-left corner marks the right angle.11√2diagonal² = 1² + 1² = 2
A unit square and its diagonal. The diagonal has length $\sqrt{2}$ by the Pythagorean theorem. The irrationality proof says this length cannot be expressed as any fraction $p/q$ — the diagonal and the side are fundamentally *incommensurable*. This discovery, made in ancient India and ancient Greece independently, was one of the first signs that the rational numbers do not fill up the entire number line.

This incommensurability was one of the earliest hints that the rational numbers are incomplete — that there are "gaps" on the number line that no fraction can reach. Those gaps are exactly the irrational numbers, and the proof by contradiction is the tool that revealed them.

The interactive figure below lets you explore this gap. Drag the point to choose a denominator q, and the readout shows the closest fraction p/q to \sqrt{2} and how far off it is. No matter how large you make q, the error never reaches zero — the fractions get closer and closer but never arrive.

Interactive rational approximations to root twoA number line from 1 to 20 with a draggable red point that sets a denominator q. Readouts above the line show q, the nearest fraction p over q to root two, and the approximation error. The error shrinks as q increases but never reaches zero.15101520↔ drag to choose denominator q
Drag the red point to choose a denominator $q$. The readouts compute the integer $p$ closest to $q\sqrt{2}$, the fraction $p/q$, and the approximation error $|\sqrt{2} - p/q|$. At $q = 7$, the best fraction is $10/7 \approx 1.4286$, with error $\approx 0.0128$. At $q = 12$, the best is $17/12 \approx 1.4167$, with error $\approx 0.0012$. The error shrinks but never hits zero — because $\sqrt{2}$ is irrational, proved by contradiction.

When to use proof by contradiction

Proof by contradiction is the right tool when:

The trade-off: proof by contradiction is powerful but sometimes indirect. When a direct proof exists, it is usually clearer and more informative, because it shows why the conclusion holds, not just that it holds. A contradiction proof shows that the opposite is impossible, but it does not always illuminate the mechanism. Use contradiction when the direct path is genuinely blocked, not as a lazy alternative to thinking directly.

The difference between contradiction and contrapositive

These two techniques are easy to confuse because both involve negation. Here is the difference.

Proof by contradiction assumes the negation of the entire conclusion and derives a contradiction — any contradiction at all.

Proof by contrapositive applies only to statements of the form "if P then Q" and instead proves "if \lnot Q then \lnot P" — the logically equivalent contrapositive — by a direct proof.

The contrapositive is a special case of contradiction. When you assume P is true and Q is false and derive that P is false, you have used contradiction — but the specific contradiction you found was "P and \lnot P," which is exactly what the contrapositive technique predicts. The article on proof by contrapositive explores this relationship in detail.

Comparison of direct proof, contrapositive, and contradictionThree rows each showing a proof strategy. Row one labelled Direct shows an arrow from P to Q. Row two labelled Contrapositive shows an arrow from not-Q to not-P. Row three labelled Contradiction shows Assume not-P leading to an explosion symbol labelled contradiction, followed by therefore P.Direct:PQContrapositive:¬Q¬PContradiction:¬P⊥P
Three proof strategies at a glance. Direct proof walks forward from $P$ to $Q$. Contrapositive walks forward from $\lnot Q$ to $\lnot P$ (which is logically the same as $P \Rightarrow Q$). Contradiction assumes $\lnot P$ (or more generally, the negation of the conclusion) and derives an impossibility, symbolised by $\bot$. All three are valid; the choice depends on which path is clearest.

Two worked examples

Example 1: Prove that there is no largest integer

Claim. There is no integer N that is greater than or equal to every other integer.

This is a negative existential statement — "there does not exist an N such that..." — and it is a natural candidate for contradiction.

Step 1. Assume the opposite: there is a largest integer. Call it N.

So N is an integer, and for every integer m, m \le N.

Why: this is what "largest integer" means — every other integer is at most N. The assumption gives you a concrete object to work with.

Step 2. Construct N + 1.

Since N is an integer, N + 1 is also an integer (the integers are closed under addition — see Operations and Properties).

Why: you need another integer to compare against N. The simplest candidate is N + 1, and closure guarantees it is a valid integer.

Step 3. Compare N + 1 with N.

N + 1 > N.

Why: adding 1 to any number makes it strictly larger. This is a basic property of the ordering on \mathbb{Z}.

Step 4. Identify the contradiction.

The assumption says every integer m satisfies m \le N. But N + 1 is an integer and N + 1 > N. These two statements contradict each other.

Why: the assumption says N is at least as large as everything, but N + 1 is something strictly larger. One of these must be false, and the ordering fact N + 1 > N is certainly true — so it is the assumption that fails.

Step 5. Conclude.

The assumption that a largest integer exists leads to a contradiction. Therefore, no largest integer exists. \square

Result. There is no largest integer.

Number line showing that N plus 1 always exceeds any supposed largest integer NA horizontal number line with a point labelled N marked by a filled red circle. To its right, another point labelled N plus 1 is marked with a filled red circle. An arrow extends further right, indicating the integers continue forever. A bracket below connects N to N plus 1, labelled plus 1.···NN + 1···+1N + 1 > N always — no integer is largest
No matter where you place $N$ on the number line, $N + 1$ sits to its right — strictly larger. The integers extend forever in both directions, and no endpoint exists. The proof by contradiction merely formalises this: if a largest integer existed, then adding $1$ would break the claim.

The proof is three lines of real work (construct N + 1, observe N + 1 > N, note the contradiction), but those three lines settle an infinite question. That is the leverage that contradiction gives you.

Example 2: Prove that $\sqrt{3}$ is irrational

Claim. \sqrt{3} is irrational.

The structure mirrors the \sqrt{2} proof, but the divisibility argument adapts to the prime 3.

Step 1. Assume, for contradiction, that \sqrt{3} is rational.

Then \sqrt{3} = p/q for some integers p, q with q \neq 0 and p/q in lowest terms (so p and q share no common factor other than 1).

Why: "rational" means "expressible as a fraction." Choosing lowest terms is a free move that gives you an extra constraint to eventually contradict.

Step 2. Square both sides and rearrange.

3 = \frac{p^2}{q^2} \implies p^2 = 3q^2

Why: squaring removes the square root, turning the equation into a statement about integers — which is where divisibility arguments can operate.

Step 3. Conclude that 3 \mid p.

Since p^2 = 3q^2, the left side is divisible by 3. Now use the fact that if 3 \mid p^2 then 3 \mid p. (This is a property of primes: if a prime divides a product, it divides at least one factor. Since p^2 = p \cdot p and 3 is prime, 3 \mid p.) So p = 3k for some integer k.

Why: the key algebraic fact is that 3 is prime. For a prime r, r \mid n^2 implies r \mid n. This is what drives the contradiction — the same argument would not work for a non-prime like 4.

Step 4. Substitute p = 3k back.

p^2 = 9k^2 = 3q^2 \implies q^2 = 3k^2

So 3 \mid q^2, and by the same prime argument, 3 \mid q.

Why: the same move as Step 3, applied to q instead of p. The factor of 3 bounces from p to q — exactly the pattern that creates the contradiction.

Step 5. Derive the contradiction.

Both p and q are divisible by 3, so they share 3 as a common factor. But p/q was in lowest terms — no common factor other than 1. Contradiction.

Therefore \sqrt{3} is irrational. \square

Result. \sqrt{3} is irrational.

The irrationality proof for root 3 as a ping-pong between p and qA diagram showing two columns labelled p and q. An arrow from p squared equals 3 q squared leads to 3 divides p. Then p equals 3k is substituted, leading to an arrow showing 3 divides q. At the bottom, a red box states the contradiction: both p and q divisible by 3 but the fraction was in lowest terms.pqp² = 3q²3 | p → p = 3ksubstituteq² = 3k² → 3 | q3 | p and 3 | q — contradicts lowest terms∴ √3 is irrational
The proof bounces the divisor $3$ from $p$ to $q$ like a ping-pong ball. First, $p^2 = 3q^2$ forces $3 \mid p$. Then substituting $p = 3k$ forces $3 \mid q$. Both being divisible by $3$ contradicts the lowest-terms assumption. The same pattern works for any prime: $\sqrt{5}$, $\sqrt{7}$, $\sqrt{11}$, and so on — each one is irrational by the same argument with the prime changed.

The pattern of this proof is reusable. For any prime r, the same argument shows \sqrt{r} is irrational: assume \sqrt{r} = p/q in lowest terms, derive r \mid p and r \mid q, contradiction. The prime's role is essential — the argument breaks for perfect squares like \sqrt{4} = 2, because 4 \mid p^2 does not force 2 \mid p in the same way (it forces 2 \mid p, but the factor only bounces once, not twice).

Common confusions

Going deeper

If you came here to learn the technique and see it applied to the classic irrationality proofs, you have everything you need. The rest of this section is for readers who want to see the broader landscape — where contradiction fits in the hierarchy of proof methods, and what it means for the foundations of logic.

Constructive mathematics and the status of contradiction

Proof by contradiction rests on the law of the excluded middle: every proposition is either true or false. This seems obvious, but there is a school of mathematics — constructive mathematics — that rejects this law. Constructivists accept proof by contradiction only for deriving negative statements ("there is no x such that...") but not for positive existence claims ("there exists an x such that..."). The reason is philosophical: a constructive proof of "there exists x" must actually produce the x, not merely show that its non-existence leads to trouble.

In standard (classical) mathematics, the law of the excluded middle is accepted, and proof by contradiction is fully valid for all statements. This is the framework used in Indian school and competitive mathematics. But it is worth knowing that the framework is a choice, not a necessity — and that serious mathematicians have explored the alternative.

Infinite descent

A powerful variation of proof by contradiction is infinite descent, associated with the French mathematician Pierre de Fermat. The idea: assume a solution exists, and show that it implies the existence of a smaller solution, which implies a still smaller solution, and so on forever. But an infinite descending sequence of positive integers is impossible (every non-empty set of positive integers has a least element). So the original assumption fails.

The irrationality proof for \sqrt{2} is secretly an infinite descent argument. If p/q is in lowest terms and both p and q are even, you can divide both by 2 to get a "more reduced" fraction — but the original was already in lowest terms, so there is nowhere left to descend. The same idea — reducing a solution to a smaller solution until you run out of room — shows up throughout number theory and is one of the most elegant proof strategies in all of mathematics.

The irrationality of e and \pi

The irrationality of \sqrt{2} and \sqrt{3} are entry-level results. Much deeper results, also proved by contradiction, include the irrationality of e (the base of natural logarithms) and \pi (the ratio of a circle's circumference to its diameter). These proofs are beyond introductory level, but they use the same core move: assume the number is rational, derive consequences, and find a contradiction. The proofs are longer and the algebra is harder, but the logical skeleton is identical to the one you learned in this article.

Ramanujan, one of the greatest mathematicians India has produced, discovered extraordinary formulas for \pi as infinite series, including:

\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^{\infty} \frac{(4k)!(1103 + 26390k)}{(k!)^4 396^{4k}}

Each term of this series gives roughly eight more correct digits of \pi. The series converges astonishingly fast — and the irrationality of \pi means it never terminates or repeats.

Where this leads next

Proof by contradiction opens a wide landscape of results that the direct method alone cannot reach.