Indirect Proof (Proof by Contradiction)
Write indirect proofs by assuming the negation of the conclusion and deriving a contradiction. Apply this technique to prove geometric theorems where direct proof is impractical.
A lesson of Geometry, in Triangle Relationships, about 10 minutes.
By the end of this lesson, you will be able to write indirect proofs using the proof-by-contradiction technique, identify when indirect proof is the most appropriate strategy, and apply it to geometric theorems.
The Big Idea
Some statements are hard to prove directly — it's difficult to build a chain of reasoning from hypothesis to conclusion. Indirect proof turns this around: instead of proving "if $p$ then $q$," you assume $q$ is false and show this leads to a contradiction. Since mathematics cannot contain contradictions, $q$ must be true.
This is the same as proving the contrapositive ($\neg q \to \neg p$), but framed as reaching an absurd conclusion.
The Technique
Steps for an indirect proof:
- Assume the opposite (the negation) of what you want to prove
- Reason logically from that assumption
- Derive a contradiction — a statement that contradicts a given, a definition, a postulate, or a theorem
- Conclude the assumption was false, therefore the original statement is true
Example 1 — A Classic
Prove: $\sqrt{2}$ is irrational.
Assume the opposite: $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$ in lowest terms ($p, q$ have no common factor).
Squaring: $2 = \dfrac{p^2}{q^2} \implies p^2 = 2q^2$
So $p^2$ is even → $p$ is even → write $p = 2k$.
Then $4k^2 = 2q^2 \implies q^2 = 2k^2$, so $q^2$ is even → $q$ is even.
But then $p$ and $q$ are both even — contradicting "in lowest terms." ✗
Therefore $\sqrt{2}$ is irrational. ∎
Example 2 — Geometry: Triangle Has At Most One Right Angle
Prove: A triangle can have at most one right angle.
Assume the opposite: triangle $\triangle ABC$ has two right angles, say $\angle A = 90°$ and $\angle B = 90°$.
Then $\angle A + \angle B = 180°$.
By the Triangle Angle-Sum Theorem: $\angle A + \angle B + \angle C = 180°$.
Substituting: $180° + \angle C = 180° \implies \angle C = 0°$.
But an angle of a triangle must be positive → contradiction. ✗
Therefore a triangle can have at most one right angle. ∎
Example 3 — An Exterior Angle
Prove: The exterior angle of a triangle is greater than each remote interior angle.
Let exterior angle $= \angle 4$ at vertex $C$, remote interior angles $\angle A$ and $\angle B$.
Assume $\angle 4 \leq \angle A$ (the opposite of what we want to prove).
By the Exterior Angle Theorem: $\angle 4 = \angle A + \angle B$.