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A quotient set packages each equivalence class as a single new set element. It remembers which class an original set element belongs to and forgets which member of that class it was. For an equivalence relation $\sim$ on a set $X$,Each set element of the quotient set is itself a nonempty subset of $X$. The canonical projection of a quotient set sends an original set element to this entire equivalence class. ^[ref]

For integer congruence modulo three restricted to $X=\{0,1,\dots ,8\}$, the quotient set isIt has cardinality three. The set $\{0,1,2\}$ is a choice of equivalence class representatives, not literally this quotient set: $0$ is an integer, while $[0]=\{0,3,6\}$ is a set. The function sending each chosen equivalence class representative to its equivalence class is a bijection between these two different sets.

A quotient set alone specifies no arithmetic. To obtain the familiar modular arithmetic example with addition, use integer congruence on the whole set of integers. Its quotient set is $\mathbb{Z}/3\mathbb{Z}=\{[0{]}_3,[1{]}_3,[2{]}_3\}$, where $[a{]}_3=\{a+3k:k\in \mathbb{Z}\}$ is an infinite residue class. Define addition byThe answer must be independent of the equivalence class representatives. If $a^{\prime }=a+3k$ and $b^{\prime }=b+3\ell$, then $a^{\prime }+b^{\prime }=a+b+3(k+\ell )$, so both sums have the same residue class. This is what it means for the group operation to be well-defined. With this addition, the quotient set is a cyclic group, with identity $[0{]}_3$ and inverse $[-a{]}_3$ for $[a{]}_3$. ^[ref]

For example, $[2{]}_3+[2{]}_3=[4{]}_3=[1{]}_3$. Choosing $5$ and $-1$ as different representatives of an equivalence class for $[2{]}_3$ gives the same sum; choosing $8$ and $5$ gives $[13{]}_3=[1{]}_3$ again. These calculations use residue classes of all integers. Raw addition on the finite $X=\{0,\dots ,8\}$ does not stay in $X$: $8+8=16\notin X$. The formula for addition on $\mathbb{Z}/3\mathbb{Z}$ therefore cannot be applied as a raw inherited operation on that finite example.

For a positive modulus $n$, a reduced residue system contains exactly one representative of each residue class coprime to $n$. Its size is the Euler totient function $\phi (n)$. Multiplication by any unit modulo n permutes these classes, which proves the Fermat-Euler theorem after multiplying and cancelling their unit product.

For a positive integer modulus $m$, the residue class of an integer $a$ is the equivalence classIt contains all integers in the same position of the repeating cycle. For example,The residue class is this whole infinite collection; its canonical representative is the least nonnegative residue. Euclidean division gives exactly one such representative in $0,1,\dots ,m-1$, so there are exactly $m$ residue classes. When $m=1$, there is just one, $[0{]}_1=\mathbb{Z}$.

Define addition and multiplication of residue classes byThese definitions must give the same answer whichever integers represent the classes. If $a^{\prime }\equiv a$ and $b^{\prime }\equiv b$ modulo $m$, thenBoth differences are divisible by $m$, proving that the operations are well-defined. For example, $[9{]}_{12}+[5{]}_{12}=[2{]}_{12}$ and $[9{]}_{12}[5{]}_{12}=[9{]}_{12}$, since $45=3\cdot 12+9$. Replacing nine by $-3$ gives the same answers with smaller intermediate integers.

With these operations the residue classes form the finite commutative ring $\mathbb{Z}/m\mathbb{Z}$, a quotient ring. It need not be a field: $[6{]}_{12}[2{]}_{12}=[0{]}_{12}$ although neither factor is the zero class. These zero divisors explain why division requires a unit modulo n. Reducing after each multiplication is especially useful in modular exponentiation, where the unreduced powers can become enormous.

For $S(a)={\sum }_{u}e(a{u}^k/p)$, character orthogonality counts representations of $x$ by $p^{-1}{\sum }_{a}S(a{)}^{3}e(-ax/p)$. Its zero-frequency term is $p^2$. The power Gauss sum over a prime field bounds the other terms in total by $k^3{p}^{3/2}$, strictly less than $p^2$ when $p>k^6$. Every residue class therefore has a representation.

For $F(n)=n(n+2)$, every odd prime has two forbidden residue classes. Use Selberg sieve weights supported on odd squarefree integers $d\le R=x^{1/8}$. Their main quadratic form is $1/J$, where $J={\sum }_{d\le R}{\prod }_{p\mid d}2/(p-2)$. The truncated Euler-product lower bound applied to primes up to $R^{1/8}$ gives $J\gg {\log }^{2}R$: the mean logarithmic divisor size is $2{\sum }_{3\le p\le R^{1/8}}(\log p)/p<\frac{1}{2}\log R$, by the Mertens first theorem, and the full product is $\asymp {\log }^{2}R$, by the Mertens second theorem. The optimal Selberg weights have modulus at most one, so the remainder is at most $O({\sum }_{d,e\le R}[d,e])=O(R^4)$. The finitely many possible pairs with $p\le R$ contribute $O(R)$. Thus the count is $O(x/J+R^4+R)$, proving the bound.

For a positive integer modulus $n$, a residue class $[a{]}_n$ is a unit when it has a modular multiplicative inverse: some integer $u$ satisfies $au\equiv 1(\mathrm{mod}n)$. Modulo twelve, the units are exactlyIn particular, five is its own modular inverse, since $5\cdot 5=25\equiv 1(\mathrm{mod}12)$. Multiplying all twelve residue classes by five gives a permutation; multiplying the outputs by five recovers the inputs. By contrast, multiplying by six gives only zero and six, with six inputs for each output. It cannot be reversed.

In general, the exact criterion isFor the forward direction, $au-kn=1$ implies every common divisor of $a,n$ divides one. Conversely, if their greatest common divisor is one, Bézout's identity supplies $ua+vn=1$, so $u$ is a modular inverse. The extended Euclidean algorithm computes these coefficients. The units form the unit group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ under multiplication. ^[ref]

Multiplicative cancellation while keeping the same integer modulus is valid for a unit modulo n: from $ca\equiv cb(\mathrm{mod}n)$, multiply by a modular inverse of $c$ to get $a\equiv b(\mathrm{mod}n)$. A nonunit cannot be cancelled as a general rule. For example, $6\cdot 0\equiv 6\cdot 2(\mathrm{mod}12)$, yet $0\not\equiv 2(\mathrm{mod}12)$. Thus division by six cannot undo this multiplication; solving a linear congruence instead keeps track of all possible classes.

This distinction marks the boundary between commutative rings and fields. Modulo a prime number $p>1$, every nonzero residue class is a unit, so $\mathbb{Z}/p\mathbb{Z}$ is a finite field. Modulo twelve it is a commutative ring with zero divisors, and is not a field. For $n=1$, the sole class satisfies $[0{]}_1=[1{]}_1$ and is a unit; this one-element zero ring is not a field. ^[ref]

An ultrafilter on the positive integers selects exactly one residue class modulo a fixed positive integer $q$. Under addition on the Stone-Čech compactification of the natural numbers, selected residues add. An idempotent ultrafilter must therefore select a residue $r$ with $2r=r$ in the finite cyclic group, hence $r=0$. In particular, the multiples of every fixed positive integer belong to every additive idempotent ultrafilter.