The Timpson Lyceum · Discrete Math (2025)

Fundamental Theorem of Arithmetic - Lecture 21 (Discrete Math)

Overview

This lecture proves the Fundamental Theorem of Arithmetic: every positive integer has a unique prime factorization. The argument has two halves — existence (every integer factors into primes) and uniqueness (it factors in only one way) — and both run by way of contradiction using the well-ordering principle and the smallest counterexample. With that theorem secured, the lecture closes with two classic "for fun" applications that depend on it: Euclid's proof that there are infinitely many primes, and the proof that $\sqrt{2}$ is irrational. Together these show how a single structural fact about the integers powers some of the oldest and most famous results in mathematics.

Key Concepts

Prime factorization. Writing a positive integer as a product of prime numbers, e.g. $60 = 2^2 \cdot 3 \cdot 5$ . A prime $p > 1$ has no positive divisors other than $1$ and $p$ ; a composite is an integer greater than $1$ that is not prime, so it can be written as $ab$ with $1 < a, b < n$ .

Edge cases.

$1$ has a prime factorization: the empty product, which equals $1$ (just as the empty sum equals $0$ ). So $1$ factors uniquely with no prime factors at all.
A prime $p$ is its own factorization: the single factor $p$ .

Fundamental Theorem of Arithmetic. Every positive integer $n$ has a prime factorization, and that factorization is unique up to the order of the factors.

Key lemma (proved previously, used in uniqueness). If a prime $p$ divides a product of primes $q_1 q_2 \cdots q_m$ , then $p$ must equal one of those primes $q_i$ . This is Euclid's lemma, and it is the hinge of the uniqueness argument.

Well-ordering principle. Every nonempty subset of the positive integers has a least element. This is what licenses the "smallest counterexample" technique: assume the set of counterexamples is nonempty, take its smallest member, and derive a contradiction — forcing the set to be empty.

Existence — proof strategy. Let $X = \{\, k \in \mathbb{Z}^+ : k \text{ has no prime factorization}\,\}$ . Note $1 \notin X$ and no prime is in $X$ . Suppose $X \neq \varnothing$ ; by well-ordering it has a least element $x > 1$ . Since $x$ is not prime it is composite, so $x = ab$ with $1 < a, b < x$ . Because $x$ is the smallest integer lacking a factorization, both $a$ and $b$ do factor into primes. Multiplying those factorizations gives a prime factorization of $x$ — contradicting $x \in X$ . Hence $X = \varnothing$ .

Uniqueness — proof strategy. Let $Y = \{\, k \in \mathbb{Z}^+ : k \text{ has two or more distinct prime factorizations}\,\}$ . Suppose $Y \neq \varnothing$ with least element $y$ , written two ways: $y = p_1 p_2 \cdots p_n = q_1 q_2 \cdots q_m$ . No $p_i$ can equal any $q_j$ — otherwise dividing out that common prime would produce a smaller integer with two factorizations, contradicting minimality of $y$ . But $p_1 \mid q_1 q_2 \cdots q_m$ , so by Euclid's lemma $p_1$ equals some $q_j$ — a shared prime. Contradiction. Hence $Y = \varnothing$ : every integer factors in exactly one way.

Infinitude of primes (Euclid). Suppose there are finitely many primes, with largest prime $p$ . Form $Q = (2 \cdot 3 \cdot 5 \cdots p) + 1$ . Dividing $Q$ by any prime in the list leaves remainder $1$ , so no known prime divides $Q$ . But $Q$ has a prime factorization, so $Q$ has a prime factor larger than $p$ — contradicting that $p$ was largest. This also gives a prime-generating procedure: multiply known primes and add $1$ to always uncover a new prime (it skips many, but never fails).

Irrationality of $\sqrt{2}$ . Suppose $\sqrt{2} = a/b$ with integers $a, b$ . Then $2b^2 = a^2$ . Counting the power of the prime $2$ : in $a^2$ it appears an even number of times, while in $2b^2$ it appears an odd number of times. Since the integer $a^2 = 2b^2$ has one unique factorization, the count of $2$ 's must agree — even cannot equal odd. Contradiction, so $\sqrt{2}$ is irrational. The Greeks found this shocking: a line segment they could construct (the diagonal of a unit square) corresponds to no ratio of integers, meaning no common unit ever measures both the side and the diagonal of a square.

Worked Example

Claim: $\sqrt{2}$ is irrational.

Assume for contradiction that $\sqrt{2} = a/b$ for integers $a, b$ . Squaring and clearing the denominator gives $2b^2 = a^2$ . Let $e_2(\cdot)$ denote the number of times the prime $2$ appears in a number's factorization. Squaring doubles every exponent, so $e_2(a^2) = 2\,e_2(a)$ (even) while $e_2(2b^2) = 2\,e_2(b) + 1$ (odd). Unique factorization forces these counts to be equal:

\begin{aligned} \sqrt{2} = \frac{a}{b} \;\Longrightarrow\; 2b^2 &= a^2 \\ e_2\!\left(2b^2\right) &= e_2\!\left(a^2\right) \\ 2\,e_2(b) + 1 &= 2\,e_2(a) \\ \underbrace{2\,e_2(b)+1}_{\text{odd}} &= \underbrace{2\,e_2(a)}_{\text{even}} \end{aligned}

No integer is both even and odd, so the equation is impossible. Therefore no such $a, b$ exist, and $\sqrt{2}$ is irrational. $\blacksquare$

Summary

The Fundamental Theorem of Arithmetic says every positive integer has a prime factorization, and it is unique up to order.
Both halves are proved by minimal counterexample: assume a nonempty set of bad integers, take its least element via well-ordering, and derive a contradiction.
Existence uses that a smallest non-factorable number would have to be composite, whose two smaller factors already factor — building the missing factorization.
Uniqueness relies on Euclid's lemma ( $p \mid q_1 \cdots q_m \Rightarrow p = q_i$ ) to force a shared prime between two supposedly disjoint factorizations of the smallest offender.
Euclid's proof of infinitely many primes: $Q = (\text{product of all primes}) + 1$ is divisible by no listed prime, yet must have a prime factor — so the list can never be complete.
$\sqrt{2}$ is irrational: counting the exponent of $2$ in $2b^2 = a^2$ gives odd $=$ even, an impossibility guaranteed by unique factorization.