# Is it mathematically correct to write $a \bmod n \equiv b$?

This is not a technical question, but a question on whether we can use a particular notation while doing modular arithmetic.

We write $a \equiv b \bmod n$, but is it right to write $a \bmod n \equiv b$?

• No, isn't right it. There is also the issue of potential confusion with the $a \bmod n=r$ of Computer Science. – André Nicolas Mar 16 '12 at 19:45
• What's the difference in the two notations? Should we not have $\equiv$ sign when relating an integer with an equivalence class? – Jalaj Mar 16 '12 at 19:49
• Lol Andre, you wrote "isn't right it" instead of "it isn't right". But I agree, it isn't right, see my answer. – Patrick Da Silva Mar 16 '12 at 19:54
• (Looks for Patrick's answer.) – Joe Mar 16 '12 at 19:56
• @Jay : Look again. =) (It wasn't there at the moment) – Patrick Da Silva Mar 16 '12 at 20:00

We have two different, but related, notions:

1. The equivalence relation "congruent modulo $n$".

Let $n$ be a fixed integer. If $a$ and $b$ are integers, we say that "$a$ and $b$ are congruent modulo $n$" if and only if $n|b-a$. We write it this way: $$a\equiv b\pmod{n}.$$ The symbol $\equiv$ is read "is congruent to" (as opposed to the symbol $=$ which is read "is equal to").

2. The binary operator $\bmod$.

Let $n$ be a positive integer. If $a$ is an integer, then $a\bmod n$ is the remainder (from a distinguished set, see below) of dividing $a$ by $n$. This is read "$a$ modulo $n$".

In mathematics, $a\bmod n$ is usually defined to be the unique integer $r$ such that $a=nq + r$ for some integer $q$ and $0\leq r \lt n$. In other areas, such as computer science (and sometimes in mathematics), one often requires that $a\bmod n$ be the unique integer $r$ such that $-\frac{n}{2}\lt r\lt \frac{n}{2}+1$ and $a-r$ is a multiple of $n$.

More generally, one may specify a "distinguished set of remainders modulo $n$", a set $R_n=\{a_0,\ldots,a_{n-1}\}$ such that every integer $x$ is congruent modulo $n$ to one and only one element of $R_n$, and define $\bmod$ as the operator such that $x\bmod n$ is the unique element of $a\in R_n$ such that $x\equiv a\pmod{n}$.

The operator $\bmod$ is like any other infix notation operator such as $+$; we write $2+3 = 5$, because the result of doing the operation $+$ to $2$ and $3$ is $5$. We write "$a\bmod n = b$" to signify that $b$ is the result of performing the operation "modulo $n$" to $a$.

The two notions are related in that if $a\bmod n = b$, then $a\equiv b\pmod{n}$. The converse does not hold in general, since we have, for example, $5\equiv 9\pmod{4}$, but $5\bmod 4 = 1\neq 9$.

Writing "$a\bmod n \equiv b$" confuses the two notions and is syntactically incorrect. You should use $=$, not $\equiv$. With $=$, it would be "mathematically correct" if and only if $b$ is the result of computing $a\bmod n$ (so $5\bmod 4 = 5$ would be wrong, but $5\bmod 4=1$ would be correct).

Writing $a\equiv b\bmod n$ also invites confusion of the two notions.

(Note, however, that "$a\bmod n \equiv b \pmod{n}$" would be syntactically correct, and would be mathematically correct if $a\equiv b\pmod{n}$.)

In computer science when we write $a \, \% \, n == b$, $\%$ is an operator/function/whatchamaycallit that acts on $a$ to return something, but in mathematics, writing "$\pmod n$" means that we are looking at an equality that works in some quotient of a group/ring/$(\dots)$. For instance, writing $$a \pmod n \equiv b$$ has no mathematical definition, because on the left "side" of the congruence we have $a \pmod n$ which is some equivalence class of integers (I assume), but on the right hand side we have an integer (I assume again), and then you want them to be "equiv" ($\equiv$ is \equiv in TeX) in some sense, but then again not defined. (It works in computer science, but then again... well.) The standard way of writing things in mathematics is that $$a \equiv b \pmod n$$ meaning that "$a$ and $b$ are equivalent up to an element of the ideal generated by $n$, that is, an $n$-multiple". You should read this as $$[[ a \equiv b ]] \pmod n$$ in the sense that $\pmod n$ is "something you apply to the equation", i.e. "you quotient".

Hope that helps,

• A few remarks. First, note that the OP did not write $a \pmod n \equiv b$, but something very different. Second, $a\equiv b \pmod n$ does not always imply that $a,b$ denote equivalence classes, e.g. it is often introduced in elementary number theory classes before quotient rings are studied (just as it was originally by Gauss). I cannot make any sense of your final displayed equation. – Bill Dubuque Mar 16 '12 at 20:31
• Minor comment: $a \% n$ is used in programming (C++ language and alike). In computer science, the algorithmic notation is often $a \bmod n$ (also $a \operatorname{rem} n$). – user2468 Mar 16 '12 at 20:55
• What I mean by the final equation is maybe not that clear... I don't like my answer that much after all. But what I tried to say was that when we use mod n in number theory it's usually something like $$[[a \equiv b \equiv c \equiv d \equiv e]] \pmod n$$ and we remove the brackets all the time, because it is clear from the context that putting the $\pmod n$ at the end of the equalities means "Read everything mod n". – Patrick Da Silva Mar 16 '12 at 22:23
• @Bill Dubuque : I know that $a \equiv b$ modulo $n$ does not mean that $a$ and $b$ are equivalence classes, but the "equality" stands between equivalence classes, that's what I mean. In what means do you distinguish $a \pmod n \equiv b$ and $a \mod n \equiv b$? I thought the parenthesis didn't mean much here. – Patrick Da Silva Mar 16 '12 at 22:25

It is often correct. $\TeX$ distinguishes the two usages: the \pmod control sequence is for "parenthesized" $\pmod n$ used to contextualize an equivalence, as in your first example, and the \bmod control sequence is for "binary operator" $\bmod$ when used like a binary operator (in your second example).

But in the latter case, you should use $=$, not $\equiv$. $7\bmod4 = 3$, and the relation here is a numeric equality, indicated by $=$, not a modular equivalence, which would be indicated by $\equiv$.