Is it standard to write $ a \bmod m $?

Question

Does it make sense to write $ a \bmod n$? Most pages I see write $ a \equiv b \bmod n $, meaning two numbers leave same remainder when divided by $n$. I am trying to understand the meaning of the first, does it simply mean remainder of $a$ when divided by $n$?

If the above is correct, then which is the correct notation $ 3 \bmod 2= 1$ or $ 3 \bmod 2 = 1 \bmod 2$?

Nitpick: use bmod, not mod, in LaTeX for those things: $a \mod b$ vs. $a\bmod b$. — Clement C., Jun 15 '21 at 09:32
As far as I know, $a \equiv b \bmod n \Leftrightarrow a \bmod n = b \bmod n$. What I personally have never understood before is that why do many books on number theory directly start with congruence equivalence relation without talking about modular arithmetic. It just seems like a very important bit of information, at least to internalize the $\bmod n$ function. — William, Jun 15 '21 at 09:38

score 1 · Accepted Answer · answered Jun 15 '21 at 09:36

1

$a\bmod n$ means exactly what you suggest:

The unique value $b \in \{0,1,2,...n-1\}$ such that $a \equiv b\bmod n$.

Equivalently, it's the canonical representative of the equivalence class of $a$ modulo $n$.

answered Jun 15 '21 at 09:36

Clement C.

67,323

Thank you, to confirm $ 3 \mod 2 = 1 $ , right? – tryst with freedom Jun 15 '21 at 09:38
1

Yes, unless someone is using non-standard notation. @Buraian – Clement C. Jun 15 '21 at 09:39
Thank you so much for the speedy help – tryst with freedom Jun 15 '21 at 09:39
This is not "equivalently". $2 \mod 3 + 2 \mod 3 = 1 \mod 3$ is true for one interpretation but false for the other. – darij grinberg Jun 15 '21 at 09:43
Then is it standard to say something like "π mod e"? – Scratch---Cat Jul 07 '21 at 05:58

score 0 · Answer 2 · answered Jun 15 '21 at 09:46

Addition modulo $n$ is basically an equivalence relation on $\mathbb Z$ (the set of integers). Let me define the equivalence relation here:
Let $n\in \mathbb Z^+$. On $\mathbb Z$, define a relation $\sim$ such that for all $a$ and $b$ in $\mathbb Z, a\sim b$ if and only if $n$ divides $(a-b)$. Verify that this this relation $\sim$ is reflexive, transitive and symmetric. In more words, $\sim$ is equivalence relation on $\mathbb Z$.
What is the equivalence class of $a\in \mathbb Z$? It's precisely the set $\{a+kn:k\in \mathbb Z\}$, which is denoted by $[a]$ so called equivalence class of $a$ under addition modulo $n$.

So if $[a]=[b]$, then we mean that equivalence class of $a$ and $b$ are the same and it is possible if and only if $a\sim b$ that is when $n$ divides $(a-b)$. It is this fact that is usually denoted by: $a\equiv b \bmod n$.

Is it standard to write $ a \bmod m $?

2 Answers2