4. Proof by Contradiction

Video coming soon3 worked examples

Theory

Some statements are hard to prove directly but easy to prove by ruling out the alternative. In proof by contradiction you assume the statement is false, then reason until you reach something impossible. Since the reasoning was sound, the only faulty step must have been the assumption — so the original statement is true.

To prove “if $A$ then $B$ ”, assume $A$ is true and $B$ is false, and derive a contradiction. The classic results proved this way are that $\sqrt{2}$ is irrational and that there are infinitely many primes.

The most-used tool is the representation of integers: an even number is $2k$ and an odd number is $2k+1$ (or $2k-1$ ) for integer $k$ . Squaring these keeps the parity visible.

The Golden Rule: state the assumption explicitly — “Assume, for contradiction, that…” — carry it through until two statements directly conflict, then name the contradiction before concluding. The examiner is looking for that named clash.

⚠️ Common Examiner Traps

Negating the statement wrongly: the opposite of “ $n$ is even” is “ $n$ is odd”, and you must assume that alongside the given hypothesis, not instead of it.
Never actually reaching a contradiction: the proof only works once you point to two things that cannot both be true. Circular reasoning that never conflicts proves nothing.
Confusing it with contrapositive: contradiction assumes the hypothesis and the negated conclusion; contrapositive proves “not $B$ implies not $A$ ” directly, with no assumption of $A$ .
Rushing the conclusion: after the contradiction, state plainly that the assumption was false and therefore the original statement holds.

Worked examples

Example 1

Prove by contradiction that if $n^2$ is even, then $n$ is even (for $n \in \mathbb{N}$ ).

Step 1: Assume, for contradiction, that $n^2$ is even but $n$ is odd.

Step 2: If $n$ is odd, write $n = 2k + 1$ for some integer $k$ . Then:

n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1

Step 3: This has the form $2m + 1$ , so $n^2$ is odd.

Step 4: But we assumed $n^2$ was even. It cannot be both even and odd — a contradiction. Therefore the assumption that $n$ is odd is false, so $n$ must be even.

Example 2

Prove by contradiction that $\sqrt{2}$ is irrational.

Step 1: Assume, for contradiction, that $\sqrt{2}$ is rational. Then it can be written as a fraction in lowest terms:

\sqrt{2} = \frac{a}{b}, \qquad a, b \in \mathbb{Z}, \quad b \neq 0, \quad \text{with no common factor}

Step 2: Square both sides and rearrange:

2 = \frac{a^2}{b^2} \implies a^2 = 2b^2

Step 3: So $a^2$ is even, which (by the previous result) means $a$ is even. Write $a = 2c$ :

(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2

Step 4: Then $b^2$ is even, so $b$ is even too.

Step 5: But now $a$ and $b$ are both even, so they share a factor of $2$ — contradicting the assumption that the fraction was in lowest terms. Therefore $\sqrt{2}$ cannot be written as such a fraction, so it is irrational.

Example 3

Prove by contradiction that if $7n$ is even, then $n$ is even (for $n \in \mathbb{N}$ ).

Step 1: Assume, for contradiction, that $7n$ is even but $n$ is odd.

Step 2: If $n$ is odd, write $n = 2k + 1$ :

7n = 7(2k + 1) = 14k + 7 = 2(7k + 3) + 1

Step 3: This has the form $2m + 1$ , so $7n$ is odd.

Step 4: This contradicts the assumption that $7n$ is even. Hence $n$ cannot be odd, so $n$ is even.