The Timpson Lyceum · Discrete Math (2025)

Modular Arithmetic, Chinese Remainder Theorem, and Factoring - Lecture 20 (Discrete Math)

Modular Arithmetic, Chinese Remainder Theorem, and Factoring – Lecture 20

Overview

This lecture completes the study of modular arithmetic and opens the topic of factoring. We define division in $\mathbb{Z}_n$ via multiplicative inverses, discover that — unlike on the real numbers — most elements have no inverse, and prove the central fact that an element is invertible exactly when it is relatively prime to $n$. We then use the extended Euclidean algorithm to compute inverses (and hence divide) even in large moduli, solve single congruences and systems of congruences (the Chinese Remainder Theorem in action). Finally we introduce prime factorization, state the Fundamental Theorem of Arithmetic, and prove the prime-divisibility lemmas (Euclid's lemma and its generalization) that the uniqueness proof will rest on.

Key Concepts

The set $\mathbb{Z}_n$. $\mathbb{Z}_n = \{0, 1, 2, \dots, n-1\}$, the first $n$ natural numbers. Note the largest element is $n-1$, one less than $n$.

Addition and multiplication mod $n$. Both operations piggyback on ordinary arithmetic and then reduce:

• $a + b$ in $\mathbb{Z}_n$ means $(a + b) \bmod n$ — take the remainder after dividing by $n$.
• $a \cdot b$ in $\mathbb{Z}_n$ means $(a \cdot b) \bmod n$.

Algebraic properties (for $n \ge 2$, so $0,1 \in \mathbb{Z}_n$). These are inherited from the reals:

• Associativity: $(a+b)+c = a+(b+c)$ and $(ab)c = a(bc)$.
• Commutativity: $a+b = b+a$ and $ab = ba$.
• Identities: $a + 0 = a$ (0 is the additive identity); $a \cdot 1 = a$ (1 is the multiplicative identity); $a \cdot 0 = 0$.
• Distributivity: $a(b+c) = ab + ac$.

Subtraction. Defined from addition: $a - b$ is the unique $x$ solving $a + x = b$. It turns out $a - b = (a - b) \bmod n$, matching the way addition is defined.

Reciprocal / multiplicative inverse. For $a \in \mathbb{Z}_n$, we call $b$ the reciprocal (or inverse) of $a$ if $a \cdot b = 1$. If such a $b$ exists, $a$ is invertible and we write $b = a^{-1}$.

Division. For $a, b \in \mathbb{Z}_n$ with $b$ invertible, $a / b = a \cdot b^{-1}$. The requirement "$b$ invertible" is the modular analog of "$b \ne 0$" on the reals — you cannot divide by something with no inverse. Why an inverse is needed: cancellation $ab = ac \Rightarrow b = c$ is valid only when $a$ has an inverse. (Example in $\mathbb{Z}_{10}$: $2\cdot 3 = 6$ and $2 \cdot 8 = 16 \equiv 6$, so "$6/2$" is ambiguous — $2$ has no inverse.)

Inverses in $\mathbb{Z}_{10}$. From the multiplication table, the invertible elements are $1, 3, 7, 9$:

• $1^{-1} = 1$, $\quad 3^{-1} = 7$ (since $3\cdot 7 = 21 \equiv 1$), $\quad 7^{-1} = 3$, $\quad 9^{-1} = 9$ (since $9 \cdot 9 = 81 \equiv 1$).
• $2, 4, 5, 6, 8$ have no inverse — note even the prime $2$ fails. $0$ never has an inverse in any $\mathbb{Z}_n$.

Uniqueness of inverses. If $a$ has a reciprocal, it has exactly one. Proof: suppose $b$ and $c$ both satisfy $ab = 1$ and $ca = 1$. Then $b = b\cdot 1 = b(ac) = (ba)c = 1\cdot c = c,$ so $b = c$. (This is why a $1$ appears at most once in any row or column of a multiplication table.) Taking the inverse twice returns the original: $(a^{-1})^{-1} = a$.

Invertibility criterion (the key theorem). $a \in \mathbb{Z}_n$ is invertible if and only if $\gcd(a, n) = 1$ (i.e. $a$ and $n$ are relatively prime).

• Forward: if $a a^{-1} = 1$ in $\mathbb{Z}_n$, then $a\cdot a^{-1} = 1 + kn$ for some integer $k$, i.e. $a\cdot a^{-1} - kn = 1$. Bézout's identity says the smallest positive value of $ax + ny$ is $\gcd(a,n)$; getting $1$ forces $\gcd(a,n) = 1$.
• Backward: if $\gcd(a,n)=1$ then $ax + ny = 1$ for some integers $x, y$. Reducing mod $n$ kills $ny$, giving $a x \equiv 1$. Setting $x' = x \bmod n$ keeps $x' \in \mathbb{Z}_n$ and $a x' \equiv 1$, so $x' = a^{-1}$.

This rests on Bézout's identity — the lecture's "second most important theorem of number theory": for integers $a, b$ there exist integers $x, y$ with $ax + by = \gcd(a,b)$, and $\gcd(a,b)$ is the smallest positive value $ax+by$ can take. (The "most important" is the Division Algorithm: $a = qb + r$.)

Computing inverses with the extended Euclidean algorithm. To find $a^{-1}$ in $\mathbb{Z}_n$: run Euclid's algorithm on $n$ and $a$ to confirm $\gcd = 1$, then back-substitute to write $1 = xa + yn$. Reducing mod $n$ gives $a^{-1} \equiv x \pmod n$ (add a multiple of $n$ to land in $\mathbb{Z}_n$).

Solving linear congruences.

• $x \equiv c \pmod n$ has all solutions $x = c + kn$, $k \in \mathbb{Z}$ (infinitely many, spaced $n$ apart).
• $ax \equiv c \pmod n$: multiply both sides by $a^{-1}$ (exists when $\gcd(a,n)=1$), giving $x \equiv a^{-1}c$.

Systems of congruences (Chinese Remainder Theorem). To solve $x \equiv r_1 \pmod{m_1}$ and $x \equiv r_2 \pmod{m_2}$: substitute the general solution of the first ($x = r_1 + m_1 k$) into the second, solve the resulting congruence for $k$ using an inverse, then back-substitute. With coprime moduli the combined solution is unique mod $m_1 m_2$ — the final answer has the form $x = x_0 + (m_1 m_2)\,t$.

Factoring and the Fundamental Theorem of Arithmetic. Every positive integer $n$ factors into a unique product of primes, up to order. Examples: $30 = 2\cdot 3\cdot 5$, $\;12 = 2^2\cdot 3$, $\;60 = 2^2\cdot 3\cdot 5$. The number $1$ is the empty product of primes (its unique, prime-free factorization); $1$ itself is not prime. Prime factorizations let you cancel cleanly, e.g. $\frac{60}{12} = \frac{2^2\cdot 3\cdot 5}{2^2\cdot 3} = 5$.

Euclid's lemma. If $p$ is prime and $p \mid ab$, then $p \mid a$ or $p \mid b$. (False for composite divisors: $20 \mid 16\cdot 25 = 400$ but $20\nmid 16$ and $20 \nmid 25$.) Proof idea: if $p$ divides neither $a$ nor $b$, then since $p$'s only divisors are $1$ and $p$, we get $\gcd(p,a) = \gcd(p,b) = 1$. Bézout gives $px + ay = 1$ and $pu + bw = 1$; multiplying these, every term on the right is divisible by $p$ (each contains a factor of $p$ or the factor $ab$, which $p$ divides), so $p \mid 1$ — impossible. Hence $p \mid a$ or $p \mid b$.

Generalization to products of primes. If $p$ is prime and $p \mid q_1 q_2 \cdots q_n$ where each $q_i$ is prime, then $p = q_i$ for some $i$. Proved by induction on $n$: the base case $p \mid q_1$ forces $p = q_1$ (both have only trivial divisors); the inductive step splits $q_1\cdots q_{k+1}$ via Euclid's lemma into $p \mid (q_1\cdots q_k)$ — handled by the hypothesis — or $p \mid q_{k+1}$, giving $p = q_{k+1}$.

Worked Example

Compute $1/29$ in $\mathbb{Z}_{431}$ (i.e. find $29^{-1}$), then use it to evaluate $30/29$.

First run the Euclidean algorithm on $431$ and $29$, recording remainders:

The last nonzero remainder is $1$, so $\gcd(431,29)=1$ and $29$ is invertible. Back-substitute to write $1$ as a combination of $431$ and $29$:

Reducing mod $431$ kills the $7\cdot 431$ term:

So $1/29 = 327$. Now divide $30$ by $29$ using this inverse:

Therefore $30/29 = 328$ in $\mathbb{Z}_{431}$.

Summary

• $\mathbb{Z}_n = \{0,\dots,n-1\}$; addition and multiplication are ordinary arithmetic followed by reducing mod $n$, and they inherit associativity, commutativity, identities, and distributivity.
• Division is multiplication by a multiplicative inverse: $a/b = a\,b^{-1}$, defined only when $b$ is invertible — the modular stand-in for "nonzero."
• An element $a$ is invertible in $\mathbb{Z}_n$ iff $\gcd(a,n)=1$; when an inverse exists it is unique. Inverses are computed by back-substituting the extended Euclidean algorithm.
• Linear congruences $ax \equiv c \pmod n$ are solved by multiplying through by $a^{-1}$; systems are solved by substitution, yielding (for coprime moduli) a unique solution modulo the product — the Chinese Remainder Theorem.
• The Fundamental Theorem of Arithmetic asserts every positive integer has a unique prime factorization (with $1$ as the empty product); cancellation via factorizations simplifies fractions.
• Euclid's lemma ($p \mid ab \Rightarrow p\mid a$ or $p\mid b$ for prime $p$) and its induction-based extension to products of primes are the divisibility tools that the uniqueness of factorization will be built on.