What are the rules for using the divides operator aka "$\mid$"? Is it false to say $2\mid5$ since $5/2$ = $2.5$ and $2.5\notin\mathbb{Z}$? Or does my question imply a misunderstanding?
I am seeing this for the first time in the 6.042J, Lecture 2: Induction on MIT OCW.
Thanks for the help.
$\endgroup$ 15 Answers
$\begingroup$That's correct. We use the divide operator to denote the following:
We say for two integers $a$, $b$, that $a$ divides $b$ (or $a|b$) if $b = ka$ for some integer $k$.
$\endgroup$ $\begingroup$To say that $a\ \mid\ b$, or "a divides b", is to say that $r = 0$ in the division algorithm: $$b = qa + r.$$
$\endgroup$ 2 $\begingroup$$n\mid m$ means that $n$ divides $m$; $2\mid5$ is false since it leaves a remainder of $1$, thus $2\nmid5$. ($x\nmid y$ stands for that $x$ does not divide $y$)
More examples:
$2\mid6$ since remainder when $6$ is divided by $2$ is zero.
$3\mid 9^n+27n, \forall n\in\mathbb{N}$ (which means that forall $n$ that's a natural number)
$\endgroup$ 2 $\begingroup$You're right $2|5$ is a false statement. Here are some true statements using |, see if you can see why they are true:
$$a|b\wedge b|c\Longrightarrow a|c\\\{n|ab\Longrightarrow n|a \vee n|b\}\Longleftrightarrow \text{$n$ is prime}$$
$\endgroup$ 2 $\begingroup$The rule says that if $a\mid b$, then $ b = k \cdot a$ for some integer $k$. Since $2.5\notin\mathbb{Z}$, $2\nmid5$.
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