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I am not a number theorist and I am learning about relations.

I encountered a relation that says

$a \leq b$ if $a$ divides $b$

Can someone clarify what it means to a number to divide another number?

Does it mean what I think? $a$ divides $b$ if $a | b \in \mathbb{Z}$?

So given a set $S = \{a, a^2, a^3, \ldots\}$, with relation $a | b \leftrightarrow a \leq b$, does the relation hold going from left to right or right to left? i.e. $a|a^2, a^2|a^3, \ldots$

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    $\begingroup$ Let $(G,\circ)$ be a group. We then say that $a\mid b$ (read as "$a$ divides $b$") if, $$\exists c\in G\mid b=a\circ c$$ $\endgroup$
    – user170039
    Jun 24, 2016 at 4:31
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    $\begingroup$ A common mistake is to confuse the statement $a \mid b$, which may be either true or false but is nonetheless a relationship between two numbers, with the quotient $a/b$ (i.e., a number). So "$a/b \in \Bbb Z$" makes sense, and may be true or false. But "$a \mid b \in \Bbb Z$" is not a well-formed statement. $\endgroup$
    – pjs36
    Jun 24, 2016 at 4:37
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    $\begingroup$ "a divides b" means a and b are integers and there is an integer n, such that n x a = b; or, if you prefer $b/a \in \mathbb Z$, or if you prefer "a divides into b evenly with no remainder". The notation $a | b$ doesn't mean what you think it does. "|" isn't an operation that give a third value. $a | b$ is shorthand for the sentence "a divides b". So it goes left to right $a|a^2$ and $a^k | a^{k+m}$ etc. $\endgroup$
    – fleablood
    Jun 24, 2016 at 4:49
  • $\begingroup$ @pjs36 Not that $a | b \iff b/a \in \mathbb Z$. and not the other way. $\endgroup$
    – fleablood
    Jun 24, 2016 at 4:51
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    $\begingroup$ If the terminology is more familiar, "$a$ divides $b$" is the same as "$b$ is divisible by $a$". $\endgroup$
    – Mike
    Jun 24, 2016 at 6:46

2 Answers 2

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Given two integers $a$ and $b$, we say $a$ divides $b$ if there is an integer $c$ such that $b=ac$.
Source.

This is what $a$ divides $b$ means. The shorthand notation is $$a|b$$.

In your example, $$a|a^2\iff a\leq a^2$$ since by definition there exists $c$ such that $a^2 = ac$, namely $a = c$.

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  • $\begingroup$ Yikes, shouldn't it be $b$ divides $a$, since "b divison operator a = b/a = c$ $\endgroup$
    – Olórin
    Jun 24, 2016 at 4:33
  • $\begingroup$ No, you are reading the definition incorrectly. $\endgroup$
    – Em.
    Jun 24, 2016 at 4:37
  • $\begingroup$ Wait, were you calling the pipe symbol $|$ the division operator? I'm not even sure what that means, but no $a\mid b$ is not the same as $a/b$. $a/b$ may not be an integer, like $3/4$, while $a\mid b$ guarantees all intgers $b = ac$, like $4\mid 8$ implies $8 = 4(2)$. $\endgroup$
    – Em.
    Jun 24, 2016 at 4:55
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    $\begingroup$ The pipe symbol "|" and the quotient symbol "/" are completely different things. That's the source of your confusion. "a|b" is shorthand for "a divides evenly into b with no remainder" whereas "a/b" is the result you get when you divide b into a. Different concepts. Note: $a | b \iff b/a \in \mathbb Z$. $\endgroup$
    – fleablood
    Jun 24, 2016 at 5:03
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    $\begingroup$ @Niing Yes, $\frac{b}{a}$ must be an integer. Hence, $a\neq 0$ for $\frac{b}{a}$. For example, take a look at this question regarding $0|0$ and why it can make sense. $\endgroup$
    – Em.
    Mar 18, 2018 at 9:23
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We say $a$ divides $b$, denoted by $a | b$, if $b$ is a multiple of $a$ (ie, $b$ is an integer multiple of $a$). Equivalently, $a |b$ iff $b=ka$ for some integer $k$.

To remember what "$2$ divides $6$" means, perhaps you can remember the phrase "$2$ divides $6$ into $3$ parts". Hence, $2 | 6$.

Note that $2 | 0$ because $0$ is an integer multiple of $2$: $0 = k2$ for some integer $k$. Just take $k=0$.

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