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Can I use

$k \in \mathbb{Q}, M_{k} = \{c~|~c \in \mathbb{Q} \land k|c \}$

to mean the set of all multiples of $k$? Or in other words, is the $|$ operator defined for rational numbers?

I was confused as the German version of the wikipedia entry for the vertical bar seems to limit its applicability to integers:

https://de.wikipedia.org/wiki/Senkrechter_Strich

(for those who can read it)

Edit:

I guess I confused $x|y$ to evaluate to true for rational numbers if $\frac{x}{y} \in \mathbb{Z}$. I guess @G Tony Jacobs anticipated my mistake and recommend I use $\{ nk | n \in \mathbb{Z} \}$.

Edit:

In fact, I wanted to use the vertical bar notation as building block to express the following:

$f(x,y) = \text{the smallest number $z$ so that $z = xa, a \in \mathbb{Z} \land z = xb, b \in \mathbb{Z}$}$

Maybe there is an idiomatic mathematical notation for this?

I posted somewhat of a follow-up here: Does the following hold true and how to learn how to solve this?

Bill Dubuque
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apriori
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  • I would edit your question to ask if “divisibility” is defined between two rational numbers. – Joe Aug 15 '19 at 18:28
  • In the integers, $a \mid b$ means that there exists some integer $k$ such that $b = ka$. Presumably, the same definition would apply to the rationals, but then $a \mid b$ for any rational numbers $a$ and $b$, as we can always find some rational number $k$ such that $b = ka$ (indeed, this number is $\tfrac{b}{a}$). This doesn't seem like a meaningful definition. – Xander Henderson Aug 15 '19 at 18:40
  • @XanderHenderson It is a meaningful definition, as it is defined in all algebra books for all integral domains, so even for fields. Of course, it is somehow like the empty set in set theory. But this is important, too. – Dietrich Burde Aug 15 '19 at 18:44
  • @DietrichBurde I should rephrase: this doesn't seem like a useful (rather than meaningful) definition in a field, such as $\mathbb{Q}$. Obviously, the definition makes sense, I just don't see the utility of it in this setting. – Xander Henderson Aug 15 '19 at 18:47
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    Does this answer your question? Divisibility, LCM and GCD for fractions (rationals) – Bill Dubuque Jan 31 '21 at 20:49

2 Answers2

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If you want the set of all integer multiples of $k$, you should write $$\{nk:n\in\Bbb{Z}\},$$ or more compactly: $$k\Bbb{Z}.$$

If you want the set of all rational multiples of $k$, that's just $\Bbb{Q}$ again anyway, as long as $k\ne 0$.

The divisibility relation can be defined in any commutative ring $R$, by the rule: For $a,b\in R$, we say $a|b$ if there is a $c\in R$ such that $ac=b$.

The rational numbers do form a ring, but it is also a field, and we usually don't bother talking about divisibility in a field (although it is well-defined), because every non-zero element divides every element, so there's nothing to talk about. Divisibility in interesting in $\Bbb{Z}$ and other rings that aren't fields, because some integers divide each other, and some do not!

If you're using the symbol to mean "divisibility with an integer quotient".... What you've written isn't a good notation for that, because the symbol is well-defined in any ring. Here, $\Bbb{Q}$ is the only ring mentioned in your OP, so readers would be liable to interpret the symbol as divisibility in $\Bbb{Q}$. If that is what you intend, for whatever reason, then it might still be worth specifying, for those unaccustomed to talking about divisibility in fields.

G Tony Jacobs
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  • Only because some divisibility isn't really something we talk about, it doesn't mean it is not good. For example, in the integers, we don't talk too much about $1\mid b$, but it is not bad. So, divisibility in $\Bbb Q$ is not bad, just not too interesting. – Dietrich Burde Aug 15 '19 at 18:39
  • I don't think anyone's claiming that divisibility in $\Bbb{Q}$ is bad :) I'm saying it's a bad notation if we're trying to use it to mean divisibility by an integer only. – G Tony Jacobs Aug 15 '19 at 18:41
  • You wrote:"Divisibility in $\Bbb Q$..., so it's not a good notation". I think, the notation is perfectly correct and good - just not too exciting. The OP asked in his title only, if he can use the symbol for rational numbers. Answer: yes. – Dietrich Burde Aug 15 '19 at 18:42
  • Ok, thanks. I wrote that in the context of the paragraph, which made the intent clear to me, but if you want to disagree with how I worded it, that's fine :) Maybe I'll reword it, if I think I can improve it. Thanks again. – G Tony Jacobs Aug 15 '19 at 18:43
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    I think your answer is fine. I just disagree that the notations should not be used for fields. It is defined for all integral domains. – Dietrich Burde Aug 15 '19 at 18:46
  • Ok, I haven't claimed that the notation shouldn't be used for fields. I think you're right :) – G Tony Jacobs Aug 15 '19 at 18:47
  • I have edited my answer to clarify – G Tony Jacobs Aug 15 '19 at 18:49
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    or even just $k\mathbb Z$ – gen-ℤ ready to perish Aug 15 '19 at 19:02
  • I don't know the definitions but it seems to me that $3|\frac{1}{3}$ qualifies. I don't know if $3|\frac{3}{16}$ qualifies. Food for thought. – poetasis Aug 15 '19 at 19:05
  • @poetasis, if 1/3 is as good as 9, then why isn't 3/16 as good as 24? – G Tony Jacobs Aug 15 '19 at 21:26
  • @G Tony Jacobs Sorry, I mean, when When $\frac{1}{3}$ divides $\frac{1}{3}$ you get $1$, an integer. When $3$ divides $\frac{3}{16}$. you get $\frac{1}{16}$ – poetasis Aug 16 '19 at 11:47
  • @poetasis Oh, I see. Then you're saying what I thought OP was saying. When you wrote $3|\frac13$, I got confused. – G Tony Jacobs Aug 16 '19 at 13:01
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No, divisibility is not restricted to integers. We can define $a\mid b$ in any integral domain $R$ by the property that there exists an element $c\in R$ such that $b=ac$. This can be translated to principal ideals, i.e., to contain is to divide: we have $a\mid b$ if and only if $(a)\supseteq (b)$. Of course, if $R$ is a field, this is not very useful, as all nonzero elements are units.

Dietrich Burde
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  • But $\mathbb{Q}$ as an integral domain is really boring as every element is a unit. Hence, $a|b$ is always true for $a\neq 0$. – nowhere dense Aug 15 '19 at 18:32
  • @yametekudasai Yes, but this was not the question (see the title). Nevertheless $a\mid b$ is defined and makes sense. – Dietrich Burde Aug 15 '19 at 18:33
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    I think the correct generalization uses that $\mathbb{Q}$ is the fraction field of $\mathbb{Z}$. So $s|r$ happens iff $\frac{r}{s}\in \mathbb{Z}$. This translate to $(s)\supset (r)$ as fractional ideals of $\mathbb{Z}$ (i.e, here $(s)=s\mathbb{Z}\subset \mathbb{Q}$). – nowhere dense Aug 15 '19 at 18:37
  • Of course you are right with that. The thing is that $\mathbb{Q}$ has much more arithmetic structure than an arbitrary field, so forgetting that extra structure is just too naive. For example $\mathbb{Q}$ is also a UFD, every element has a unique factorization equal to just itself, but that is not interesting at all. Much more interesting is the fact that every element can be written uniquely as $$\pm p_1^{\alpha_1}\dots p_k^{\alpha_k}$$ for some prime numbers $p_i$ and $\alpha_i\in \mathbb{Z}$. This comes from being the fraction field of a non trivial UFD. – nowhere dense Aug 15 '19 at 18:57
  • Yes, you are also right with that. I think, I focused more (too much) on the title question rather than on arithmetic. And for the title question the answer is (quite trivially) "yes". – Dietrich Burde Aug 15 '19 at 19:00