The Timpson Lyceum · Discrete Math (2025)

Proof by Smallest Counterexample and Induction - Lecture 11 (Discrete Math)

Overview

This lecture develops two closely related techniques for proving that every natural number has some property: proof by smallest counterexample and proof by induction. Both rest on the well-ordering principle — the fact that every non-empty set of natural numbers has a least element. We first set up smallest counterexample as a contradiction argument (assume the set of "bad" numbers is non-empty, extract its least element, and derive an impossibility), then prove the principle of induction as a theorem from well-ordering and use it to verify summation formulas. These methods are foundational: any later course built on mathematics will use induction constantly to prove statements that must hold across infinitely many cases.

Key Concepts

Counterexample vs. smallest counterexample. A single counterexample disproves a universal statement (e.g. $3$ disproves "every integer is even"). A smallest counterexample is the least value that falsifies the statement. Over the integers there is no smallest counterexample to "every integer is even" (the integers have no least element); over the naturals $\mathbb{N} = \{0,1,2,\dots\}$ the smallest counterexample is $1$. Smallest counterexamples exist for the naturals precisely because of well-ordering.

Well-ordering principle. Every non-empty subset of $\mathbb{N}$ has a least element. This is an axiom (it is provided by, and is logically equivalent to, the axiom of choice in ZFC); it is not something we prove. It supplies the notion of order that bare set theory does not.

Proof by contradiction (review). $\alpha \Rightarrow \beta$ is logically equivalent to $(\alpha \wedge \neg\beta) \Rightarrow \text{False}$. Intuitively: if having $\alpha$ without $\beta$ is impossible, then $\alpha$ forces $\beta$.

Property / predicate. A property $P$ is a one-place predicate: feed it a natural number, it returns true or false (e.g. "is even").

Smallest-counterexample template. To prove every $n \in \mathbb{N}$ has property $P$:

• Set up the bad set. Let $A = \{\, n \in \mathbb{N} : \neg P(n) \,\}$ — all naturals lacking the property. The goal is to show $A = \varnothing$, since an empty bad set means every natural has $P$.
• Assume for contradiction that $A \neq \varnothing$.
• Extract the least element. Since $A \subseteq \mathbb{N}$ and $A \neq \varnothing$, the well-ordering principle gives a least element $x \in A$.
• Derive a contradiction. Use the minimality of $x$ (typically: $x-1$ is not in $A$, so $x-1$ has the property) to show $x$ must in fact have the property, i.e. $x \notin A$ — contradicting $x \in A$.
• Conclude $A = \varnothing$, hence $P(n)$ holds for all $n \in \mathbb{N}$.

The opening and closing of every such proof are identical boilerplate; the real work is the middle step that turns "$x-1$ has the property" into "$x$ has the property."

Principle of induction (a theorem, proved from well-ordering). If $A \subseteq \mathbb{N}$ satisfies $0 \in A \quad\text{and}\quad \big(k \in A \Rightarrow k+1 \in A\big),$ then $A = \mathbb{N}$. Intuition: $0$ is in, so $1$ is in, so $2$ is in, and so on through every natural. The proof takes $B = \mathbb{N} \setminus A$, assumes $B \neq \varnothing$, lets $x$ be its least element; then $x \neq 0$ (since $0 \in A$), so $x-1 \in A$, and the closure property forces $x = (x-1)+1 \in A$ — contradicting $x \in B$.

Proof by induction (the technique). To show $P(n)$ for all $n$, let $A = \{\, n \in \mathbb{N} : P(n) \,\}$ (the set of values making the statement true) and verify the two hypotheses above:

• Base case — show the starting value (usually $0$, sometimes $1$) lies in $A$.
• Inductive step — assume $x \in A$ (the inductive hypothesis) and prove $x+1 \in A$.

Then $A = \mathbb{N}$ (or all naturals from the base value up), so $P$ holds everywhere. Unlike smallest counterexample, this is not a proof by contradiction, which makes it easier to follow; the two methods are logically equivalent.

Definitions used in the examples.

• Even / odd. An integer $m$ is even if $m = 2a$ for some integer $a$, and odd if $m = 2a+1$ for some integer $a$.
• Summation notation. $\sum_{k=1}^{n} f(k)$ means add $f(1), f(2), \dots, f(n)$. A "sum" from a larger lower index to a smaller upper index (e.g. $\sum_{k=1}^{0}$) is the empty sum, equal to $0$ — distinct from $\sum_{k=0}^{0}$, which has one term.
• Fibonacci numbers. $F_0 = 1,\; F_1 = 1,\; F_n = F_{n-1} + F_{n-2}$ for $n \ge 2$: $1,1,2,3,5,8,13,21,\dots$

Worked Example

Claim (proved by induction): for every $n \in \mathbb{N}$, $\displaystyle \sum_{k=0}^{n} k = \frac{n(n+1)}{2}$.

Let $A = \big\{\, n \in \mathbb{N} : \sum_{k=0}^{n} k = \tfrac{n(n+1)}{2} \,\big\}$.

Base case ($n=0$): $\sum_{k=0}^{0} k = 0 = \dfrac{0(0+1)}{2}$, so $0 \in A$.

Inductive step: assume $x \in A$, i.e. $\sum_{k=0}^{x} k = \dfrac{x(x+1)}{2}$. Add the next term $x+1$ to both sides:

This is exactly the formula with $n$ replaced by $x+1$, so $x+1 \in A$. By the principle of induction $A = \mathbb{N}$, hence the formula holds for every natural number. (For example, $\sum_{k=0}^{1000} k = \frac{1000 \cdot 1001}{2} = 500{,}500$.)

Representative smallest-counterexample proof (sketch): to prove $\sum_{k=1}^{n}(2k-1) = n^2$ for all $n \ge 1$ — "the sum of the first $n$ odd numbers is $n^2$" — let $A$ be the naturals where it fails, assume $A \neq \varnothing$, take its least element $x$. Since $n=1$ works ($\sum_{k=1}^{1}(2k-1) = 1 = 1^2$), we have $x \ge 2$, so $x-1 \ge 1$ makes the statement true: $\sum_{k=1}^{x-1}(2k-1) = (x-1)^2$. Adding the $x$-th odd term $2x-1$ gives $\sum_{k=1}^{x}(2k-1) = (x-1)^2 + (2x-1) = x^2 - 2x + 1 + 2x - 1 = x^2$, so $x$ makes the statement true — contradicting $x \in A$. Thus $A = \varnothing$ and the formula holds for all $n \ge 1$.

Summary

• The well-ordering principle (every non-empty subset of $\mathbb{N}$ has a least element) is the axiom underlying both proof techniques.
• Proof by smallest counterexample is a contradiction argument: let $A$ be the set of naturals failing the property, assume $A \neq \varnothing$, take its least element $x$, and use that $x-1$ has the property to show $x$ does too — a contradiction forcing $A = \varnothing$.
• The principle of induction ($0 \in A$ and $k \in A \Rightarrow k+1 \in A$ imply $A = \mathbb{N}$) is a theorem proved from well-ordering, not a separate axiom.
• Proof by induction verifies a base case and an inductive step ($x \Rightarrow x+1$); it avoids contradiction and is the preferred everyday tool. The base value need not be $0$ — start wherever the statement begins (e.g. $n \ge 1$).
• The two methods are logically equivalent; anything provable one way is provable the other.
• Induction excels at summation formulas — e.g. $\sum_{k=0}^{n} k = \frac{n(n+1)}{2}$, $\sum_{k=1}^{n}(2k-1) = n^2$, $\sum_{k=0}^{n} k\cdot k! = (n+1)! - 1$, and $\sum_{k=0}^{n-1} 2^k = 2^n - 1$ — letting you prove a single identity holds for infinitely many cases.