The Timpson Lyceum · Discrete Math (2025)

Proof, Counterexample, and Boolean Algebra – Lecture 3 (Discrete Math)

Overview

This lecture moves from definitions to the central activity of discrete math: writing proofs. It explains what it actually means to prove a statement (showing it is impossible to be false, not merely collecting confirming examples), lays out a repeatable procedure for constructing direct proofs, and works through several representative proofs about even numbers, divisibility, and composites. It then introduces proof by counterexample as the way to disprove a universal statement, and closes by reframing the logical connectives from earlier as Boolean algebra, reviewing the standard identities and how to evaluate Boolean expressions.

Key Concepts

Core definitions (recalled). For integers $a, b$:

• Divides: $a \mid b$ means there exists an integer $c$ with $b = a \cdot c$.
• Even: $a$ is even means $\exists x \in \mathbb{Z}$ such that $a = 2x$.
• Odd: $a$ is odd means $\exists x \in \mathbb{Z}$ such that $a = 2x + 1$.
• Prime: $p > 1$ whose only positive divisors are $1$ and $p$.
• Composite: $c > 0$ for which there exists an integer $a$ with $1 < a < c$ and $a \mid c$.

What a proof is.

• To prove a statement is to show it is impossible to be false, not just true in many cases.
• Most claims can be cast as an implication $\alpha \Rightarrow \beta$. The backbone of a direct proof is the tautology: if $\alpha$ is true and $\alpha \Rightarrow \beta$ is true, then $\beta$ is true.
• An equivalent route is the contrapositive $\lnot\beta \Rightarrow \lnot\alpha$, which has the same truth table.

Examples are not proof.

• Finding many—even infinitely many—confirming cases does not establish a universal claim. "All primes are odd" survives $3, 5, 7, 11, \dots$ yet fails at $2$.
• The Goldbach conjecture (every even integer greater than $2$ is a sum of two primes, e.g. $8 = 5 + 3$, $10 = 7 + 3$) has been checked by computer to enormous values but remains unproven—verification is not proof.

Procedure for a direct proof.

1. Restate the proposition in if–then form: $\alpha$ (the antecedent) implies $\beta$ (the conclusion).
2. Write the first line = assume the antecedent, and the last line = the conclusion you must reach.
3. Unravel the definitions—replace each defined term with its meaning, working top-down from the antecedent and bottom-up from the conclusion.
4. Bridge the gap. There is no algorithm for this middle step; it requires insight specific to the problem.

• Closure of $\mathbb{Z}$ under addition and multiplication (sums/products of integers are integers) may be assumed.

If-and-only-if proofs. To prove $\alpha \iff \beta$, prove both $\alpha \Rightarrow \beta$ and $\beta \Rightarrow \alpha$—two proofs in one, often marked with directional arrows $(\Rightarrow)$ and $(\Leftarrow)$.

Proof by counterexample. To prove a universal statement is false, exhibit a single case where the antecedent is true but the conclusion is false. This works because $\alpha \Rightarrow \beta$ is false only in that one row of the truth table. (Contrast with proof by contradiction, which shows that assuming the antecedent forces an impossibility, i.e. $\alpha \Rightarrow \text{False}$.)

Boolean algebra. A Boolean is a value that is either true (T) or false (F)—exactly two values. The logical connectives are the Boolean operators: $\land$ (and), $\lor$ (or), $\Rightarrow$ (implies), $\iff$ (iff), $\lnot$ (not). Key identities, each provable by truth table:

• Commutative: $x \land y = y \land x$, $\quad x \lor y = y \lor x$.
• Associative: $(x \land y) \land z = x \land (y \land z)$, and likewise for $\lor$.
• Identity: $x \land \text{T} = x$, $\quad x \lor \text{F} = x$.
• Distributive: $x \land (y \lor z) = (x \land y) \lor (x \land z)$, and $x \lor (y \land z) = (x \lor y) \land (x \lor z)$.
• Complement: $x \land \lnot x = \text{F}$, $\quad x \lor \lnot x = \text{T}$.
• De Morgan: $\lnot(x \land y) = \lnot x \lor \lnot y$, $\quad \lnot(x \lor y) = \lnot x \land \lnot y$.
• Implication as disjunction: $x \Rightarrow y = \lnot x \lor y$.

Worked Example

Claim: The sum of two even integers is even.

Restate as implication: If $a$ and $b$ are even integers, then $a + b$ is even.

Proof. Assume $a$ and $b$ are even. Unraveling the definition of even, there exist integers $x, y$ with $a = 2x$ and $b = 2y$. We must reach $a + b = 2c$ for some integer $c$:

Since $x$ and $y$ are integers, $c = x + y$ is an integer (closure of $\mathbb{Z}$ under addition). Thus $a + b = 2c$, which is exactly the definition of even. Therefore $a + b$ is even. $\blacksquare$

The same template proves transitivity of divisibility: if $a \mid b$ and $b \mid c$, then $a \mid c$. Assuming the antecedent gives integers $m, n$ with $b = a m$ and $c = b n$; substituting, $c = (a m) n = a(mn)$, so with the integer $p = mn$ we have $c = a p$, i.e. $a \mid c$. $\blacksquare$

Summary

• To prove a statement is to show it is impossible to be false; no quantity of confirming examples (Goldbach being the cautionary case) ever suffices.
• The direct-proof recipe: restate as $\alpha \Rightarrow \beta$, assume the antecedent on line one, target the conclusion on the last line, unravel definitions from both ends, then bridge with problem-specific insight.
• Even/odd/divides/prime/composite are all defined by an existence statement ($\exists$ an integer such that …); unraveling these definitions is what turns a claim into algebra.
• An iff proof requires proving both directions; closure of the integers under $+$ and $\times$ may be assumed.
• A single counterexample disproves a universal claim—precisely the truth-table row where the antecedent is true and the conclusion is false.
• The logical connectives form a Boolean algebra with commutative, associative, identity, distributive, complement, and De Morgan laws, all verifiable by truth table; in particular $x \Rightarrow y = \lnot x \lor y$.