Notation for "is divisible by"
Is there a standard way of writing $a$ is divisible by $b$ in mathematical notation?
From what I've search it seems that writing $a \equiv 0 \pmod b$ is one way? But also you can write $b \mid a$ as well (the middle character is a pipe)? And sometimes that pipe is replaced by $3$ vertical dots?
Or is there a way of writing $a$ is a multiple of $b$ which I think means the same thing?
EDIT: thanks for the answers, is there a way to extend this and write something like: $b \mid a$ when $a = k$
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$\begingroup$I have seen the following:
- $b \mid a$ that is with $\LaTeX$
\mid - $a = 0 \mod b$ that is with $\LaTeX$
\mod - $a = 0 \pmod b$ that is with $\LaTeX$
\pmod - $a \bmod b = 0$ that is with $\LaTeX$
\bmod - $a \equiv 0\ (b)$
- $a \equiv_{b} 0$
and of course there is
- $a = bk$ for some $k \in \mathbb{Z}$
Choose whatever suits you (and your friends or readers) best!
$\endgroup$ 2 $\begingroup$Alexander Merkurjev taught me a long time ago the ingenious Russian notation $6 \vdots 2$, which I immediately adopted .
It pleasantly "rhymes" with the equivalent $(6)\subset (2)$
There is also " $a \in b\mathbb Z$ ".
$\endgroup$ $\begingroup$I often write that as b divides a
Notation:
$$b \mid a$$
$\endgroup$ 1 $\begingroup$Definition: Integer $n$ is a divisible by an integer $d$, when $\exists k \in \mathbb{Z}, n=d\times k$.
Notation: $d \mid n$
Synonymous:
$n$ is a multiple of $d$.
$d$ is a factor of $n$
$d$ is a divisor of $n$
$d$ divides $n$
$a \equiv 0 \mod b$ and $b \mid a$ are both common, and their use depends on the context. Given a choice, I use the latter more than the former.
There are others such as $\text{lcm}(a,b)=a$ or $\text{hcf}(a,b)=b$ [or perhaps $\text{gcd}(a,b)=b$ if you prefer] which might also be used when more suitable for the context.
$\endgroup$ $\begingroup$COMMENT.- A notation forget in this page is $a\mid\mid b$ which is pertinent many times, very useful really, in number theory. It is used when a power of a prime, say $p^n$ is the maximal power that divides an integer $M$. In other words$$p^n\mid\mid M\iff p^n\mid M \text{ but } p^{n+1}\nmid M$$
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