The Timpson Lyceum · Discrete Math (2025)

Proof by Contradiction and Contrapositive - Lecture 10 (Discrete Math)

Overview

This lecture introduces two new proof techniques to add to the direct proof you have used so far: proof by contrapositive and proof by contradiction. Both are built on logical equivalences you can verify with a truth table, and both give you an alternative route when a direct proof gets stuck. Proof by contrapositive proves $\alpha \Rightarrow \beta$ by instead proving $\lnot\beta \Rightarrow \lnot\alpha$, while proof by contradiction proves $\alpha \Rightarrow \beta$ by assuming $\alpha$ and $\lnot\beta$ together and deriving an impossible statement. These are not single-section tricks — they recur throughout this course and in every advanced math course beyond it, so it is worth getting comfortable with the logic now.

Key Concepts

Direct proof (review). To prove an implication $\alpha \Rightarrow \beta$ directly, you assume the antecedent $\alpha$ and show that the consequent $\beta$ must follow. This is the structure you have used up to now.

Proof by contrapositive. Instead of proving $\alpha \Rightarrow \beta$, you prove its contrapositive $\lnot\beta \Rightarrow \lnot\alpha$. These two statements are logically equivalent:

• $\alpha \Rightarrow \beta \;\equiv\; \lnot\beta \Rightarrow \lnot\alpha$
• This is useful when the assumption $\alpha$ is awkward to "unravel" (hard to turn the definition into something workable). Starting from $\lnot\beta$ is often easier.

Proof by contradiction. To prove $\beta$ (under hypothesis $\alpha$), assume $\alpha$ together with $\lnot\beta$ and show this leads to a contradiction — a statement that is always false. Since $\alpha \land \lnot\beta$ is impossible, $\alpha \Rightarrow \beta$ must hold:

• $\alpha \Rightarrow \beta \;\equiv\; (\alpha \land \lnot\beta) \Rightarrow \text{False}$
• The intuition: if it is impossible for a statement to be false, it must be true. Mark the impossible result with the contradiction symbol $\Rightarrow\!\Leftarrow$ (arrows pointing into each other).

Truth-table confirmation. Both equivalences can be checked column-by-column:

Supporting definitions used in the examples.

• Relation: a set of ordered pairs. A relation $R$ on a set $A$ has ordered pairs drawn from $A$.
• Five relation properties: reflexive, irreflexive, symmetric, antisymmetric, transitive.
• Equivalence relation: a relation that is reflexive, symmetric, and transitive (abbreviated E.R.).
• Equivalence class of $a$: the set of everything related to $a$, written $[a]$.
• Even / odd integers: $x$ is even iff $\exists\, a \in \mathbb{Z}$ with $x = 2a$; $x$ is odd iff $\exists\, b \in \mathbb{Z}$ with $x = 2b + 1$.

Choosing a style. You are never forced to use a particular proof style — pick whichever makes the reasoning easiest. Proof by contradiction is frequently the most natural choice. Other styles exist too (e.g. induction, the pumping lemma, pigeonhole), but contrapositive and contradiction are ubiquitous.

Worked Example

Claim: If $x \in \mathbb{Z}$, then $x$ is not both even and odd. (Proof by contradiction.)

Here $\alpha$ is "$x \in \mathbb{Z}$" and $\beta$ is "$x$ is not both even and odd." Assume by way of contradiction $\alpha \land \lnot\beta$: that $x$ is an integer and $x$ is both even and odd. By the definitions of even and odd there exist integers $a, b$ with $x = 2a$ and $x = 2b + 1$. Setting the two expressions equal and solving:

But $a - b$ is a difference of integers, so $a - b \in \mathbb{Z}$, while $\tfrac{1}{2} \notin \mathbb{Z}$. So $a - b$ is simultaneously an integer and not an integer — a contradiction $(\Rightarrow\!\Leftarrow)$. Therefore $x$ cannot be both even and odd. $\blacksquare$

Second example (proof by contrapositive). Let $R$ be an equivalence relation on $A$ and let $a, b \in A$. Claim: if $a \not\mathrel{R} b$ then $[a] \cap [b] = \varnothing$. Rather than unraveling "$a$ is not related to $b$," prove the contrapositive: assume $\lnot\beta$, i.e. $[a] \cap [b] \neq \varnothing$. Then there exists $x \in [a] \cap [b]$, so $x \mathrel{R} a$ and $x \mathrel{R} b$. By symmetry, $x \mathrel{R} a$ gives $a \mathrel{R} x$; combining $a \mathrel{R} x$ and $x \mathrel{R} b$ by transitivity gives $a \mathrel{R} b$ — which is $\lnot\alpha$. Having shown $\lnot\beta \Rightarrow \lnot\alpha$, the original statement $\alpha \Rightarrow \beta$ holds. $\blacksquare$

Summary

• Direct proof assumes $\alpha$ and derives $\beta$; the two new styles are alternatives for when that gets stuck.
• Proof by contrapositive proves $\alpha \Rightarrow \beta$ by proving the equivalent $\lnot\beta \Rightarrow \lnot\alpha$ — handy when $\alpha$ is hard to work with.
• Proof by contradiction assumes $\alpha \land \lnot\beta$ and derives an always-false statement, marked $\Rightarrow\!\Leftarrow$; impossibility of $\alpha \land \lnot\beta$ forces $\alpha \Rightarrow \beta$.
• Both styles rest on logical equivalences verifiable by truth table, and both lean on the negation $\lnot\beta$ as the starting point.
• You are never required to use a specific style — choose the one that makes the argument clearest (contradiction is often the most natural).
• These techniques recur throughout this course and all later math courses; mastering them now pays off repeatedly.