Assume the opposite of what you believe, follow the logic wherever it leads, and watch the assumption collapse under its own weight.
10 min read4 worked examples
The structure of the method
Every proof method so far has built forwards from known facts to a conclusion. Proof by contradiction works in reverse: to prove a statement is true, assume it is false, and show that this assumption leads somewhere impossible — a logical contradiction. Since a false assumption cannot logically produce a valid deduction chain ending in nonsense, the assumption itself must have been wrong, and so the original statement must be true.
The three-step skeleton
1. Assume the negation of the statement you want to prove — write it down explicitly. 2. Use logical, algebraic reasoning to deduce consequences of that assumption. 3. Arrive at a statement that is clearly false, or that contradicts the original assumption itself (or a known fact) — then conclude the original statement must be true.
Negating a statement correctly
The first step is often the one students get wrong: writing an accurate negation. "$\sqrt2$ is irrational" negates to "$\sqrt2$ is rational" — not "$\sqrt2$ is an integer" or some other unrelated claim. "There are infinitely many primes" negates to "there are finitely many primes" (a specific, countable list). Getting the negation exactly right matters, because the whole proof rests on correctly exhausting the only other possibility.
Strategy — know your contradiction before you start hunting for it
Common contradictions to aim for: reaching a fact that contradicts the assumption itself (e.g. showing a number assumed odd is actually even), producing a result that contradicts a known theorem or definition (e.g. a fraction in "lowest terms" that turns out to share a common factor), or finding a number smaller than a value already shown to be the smallest. Spotting which kind of contradiction the problem is steering towards helps structure the algebra.
Worked example 1
√2 is irrational
Prove, by contradiction, that $\sqrt2$ is irrational.
ProofAssume √2 is rational, in lowest terms, and derive a contradiction
1
Assume, for contradiction, that $\sqrt2$ is rational.
Negate the statement to be proved.
2
Then $\sqrt2 = \dfrac{a}{b}$ for integers $a,b$ with no common factor (the fraction is in its lowest terms), $b\neq0$.
Any rational number can be written as a fraction in lowest terms.
So $b^2$ is even, and hence $b$ is even too, by the same reasoning as step 4.
Same even-square argument applied to b.
7
But then $a$ and $b$ are both even, so they share a common factor of $2$ — contradicting the assumption that $\dfrac{a}{b}$ was in lowest terms.
This directly contradicts step 2.
8
This contradiction shows the assumption was false. Therefore $\sqrt2$ is irrational.
Conclude the original statement.
▪ QED
Examiner's eye
The phrase "no common factor" (or "in lowest terms" / "fully cancelled") is essential to state explicitly in step 2 — without it, there's nothing for step 7 to contradict, and the proof falls apart. This is the single most common mark lost on this classic proof.
Worked example 2
There are infinitely many primes
Prove, by contradiction, that there are infinitely many prime numbers.
ProofAssume a finite, complete list of primes, then construct a new one
1
Assume, for contradiction, that there are only finitely many primes: $p_1, p_2, \ldots, p_n$.
Negate "infinitely many" to "a finite, complete list".
2
Consider the number $N = p_1 \times p_2 \times \cdots \times p_n + 1$.
Multiply every prime in the list together, then add 1.
3
Dividing $N$ by any $p_i$ from the list leaves remainder $1$, since $p_1p_2\cdots p_n$ is exactly divisible by $p_i$.
N is one more than a multiple of every prime on the list.
4
So $N$ is not divisible by any prime in the list $p_1,\ldots,p_n$.
Direct consequence of step 3.
5
But every integer greater than $1$ is either prime itself or has a prime factor. So $N$ is either prime, or divisible by some prime not in the original list.
Every integer > 1 has at least one prime factor.
6
Either way, a prime exists that is not in the list $p_1,\ldots,p_n$ — contradicting the assumption that this list contained all the primes.
This contradicts step 1's claim of completeness.
7
This contradiction shows the assumption was false. Therefore there are infinitely many primes.
Conclude the original statement.
▪ QED
The key insight
The proof never claims $N$ itself is prime — only that some prime factor of $N$ must lie outside the assumed-complete list. This is a subtle but important distinction: the new prime found isn't necessarily $N$, just guaranteed to exist and be missing from the list.
Worked example 3
If n² is even, then n is even
Prove, by contradiction, that if $n^2$ is even then $n$ is even.
ProofAssume n is odd, and show n² must then be odd too
1
Assume, for contradiction, that $n^2$ is even but $n$ is not even, i.e. $n$ is odd.
Negate the conclusion while keeping the given condition fixed.
2
Since $n$ is odd, write $n = 2k+1$ for some integer $k$.
Standard form for an odd integer.
3
$n^2 = (2k+1)^2 = 4k^2+4k+1 = 2(2k^2+2k)+1$.
Expand and factor out 2 from the even part.
4
Since $2k^2+2k$ is an integer, $n^2 = 2(\text{integer})+1$ is odd.
This is exactly the form of an odd number.
5
But this contradicts the given fact that $n^2$ is even.
Direct contradiction with the premise in step 1.
6
This contradiction shows the assumption was false. Therefore, if $n^2$ is even, $n$ must be even.
Conclude the original statement.
▪ QED
Look deeper — this result underpins Example 1
This is precisely the fact quietly used at steps 4 and 6 of the $\sqrt2$ proof above ("$a^2$ even $\Rightarrow$ $a$ even"). Many A-level contradiction proofs lean on this small result as a building block — it's worth being able to reproduce it from scratch rather than just quoting it.
Worked example 4
No smallest positive rational number · Challenge
Prove, by contradiction, that there is no smallest positive rational number.
ProofAssume a smallest one exists, then construct a smaller one
1
Assume, for contradiction, that a smallest positive rational number exists, and call it $q$.
Negate "no smallest exists" to "a smallest one, q, exists".
2
Since $q$ is rational, so is $\dfrac{q}{2}$ (half of a rational number is rational).
Rationals are closed under division by the integer 2.
3
Since $q>0$, also $\dfrac{q}{2}>0$, so $\dfrac{q}{2}$ is a positive rational number.
Halving a positive number keeps it positive.
4
But $\dfrac{q}{2} < q$, since $q>0$.
Halving strictly decreases a positive number.
5
So $\dfrac{q}{2}$ is a positive rational number smaller than $q$ — contradicting the assumption that $q$ was the smallest positive rational.
Directly contradicts step 1.
6
This contradiction shows the assumption was false. Therefore, no smallest positive rational number exists.
Conclude the original statement.
▪ QED
Look deeper — applying the method to unfamiliar statements
This proof follows the identical skeleton as the first two, but the "construct something impossible" step looks different each time: an algebraic factor (Example 1), a new prime (Example 2), a smaller rational (here). When facing an unseen contradiction proof, the reliable starting move is always the same — write the precise negation, then ask "what can I build from this assumption that shouldn't be possible?"
Common mistake
Writing a sloppy or incorrect negation is the most frequent error — "not all $x$ are positive" does not negate to "all $x$ are negative" (it negates to "at least one $x$ is not positive", i.e. zero or negative). Always pause and check the negation is logically exhaustive with the original statement before building the rest of the proof on it.
Check yourself
Prove, by contradiction, that if $n$ is an integer and $n^3$ is even, then $n$ is even. (Hint: assume n is odd, and expand (2k+1)³.)
Proof
1
Assume $n^3$ is even but $n$ is odd: $n=2k+1$.
2
$n^3=(2k+1)^3=8k^3+12k^2+6k+1=2(4k^3+6k^2+3k)+1$, which is odd.
Expand and factor out 2 from the even terms.
3
This contradicts $n^3$ being even, so $n$ must be even.