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So let's say that you have the number $n=12$. The factors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. I'm wondering if there's some definition that when you plug in a natural number $n$, it will give you the set of all of it's factors. So, if I plugged in $12$ for $n$, I would get the set ${1,2,3,4,6,12}$, in no particular order of course. Is there such a thing? If so, what is it?

Edit: Since $1$ and $n$ are always factors of $n$, I can get as far as ${1,n} \subseteq F $ where $F$ is the set of factors of $n$.

Andrés E. Caicedo
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Sam
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    You seem to have just defined it. I have no idea what else you are looking for. – Tobias Kildetoft Mar 04 '16 at 14:04
  • @TobiasKildetoft ??? So you're saying I can only divide 12 by 1 and 12, and not 2, 3, 4, or 6? That's just absurd. Reread the question and tell me what I didn't make clear, I'll fix it. – Sam Mar 04 '16 at 14:05
  • When did he say that? – 5xum Mar 04 '16 at 14:05
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    No, I am saying that "all the factors" is already a definition. – Tobias Kildetoft Mar 04 '16 at 14:05
  • @TobiasKildetoft No I mean like a mathematically written definition, not words. Maybe I should've said for formula? – Sam Mar 04 '16 at 14:06
  • @5xum I assumed he was mentioning my edit, in which I said that 1 and n are a subset of the factors of n. My mistake, he wasn't. – Sam Mar 04 '16 at 14:08
  • @Sam: What's wrong with words? – Ben Millwood Mar 04 '16 at 14:15
  • The set of all the numbers which are algebraic over $\mathbb Q$, it is known as the algebraic closure of $\mathbb Q$. This is something like this what it is asked by @Sam. – Piquito Mar 04 '16 at 14:16
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    The set you are looking for is $d(n)={m\in \mathbb{N} : m \text{ divides } n}$, but the question does not make much sense, since you can define whatever you want. If you are asking for some kind of map that given a natural number $n$ will return the set of all of its divisors, probably you can't do that via some formula. Otherwise cryptography would have no point at all. – Alberto Debernardi Mar 04 '16 at 14:18
  • You mean $1, n \in F$ and not $1, n \subseteq F$ which is something different. – Frunobulax Mar 04 '16 at 14:28
  • @Frunobulax the first symbol you mentioned is for elements. $1, n$ isn't an element, it's a set containing elements. That's what the subset symbol is for. – Sam Mar 04 '16 at 15:00
  • @Frunobulax so using the "element of" symbol would be incorrect... I think?? – Sam Mar 04 '16 at 15:06
  • If you want to combine objects into a set, then you need to add curly braces around them, so you'd need to write ${1,n}\subseteq F$. If you write something like $1,n\in F$ without braces, then the usual convention is to understand this as an abbreviation for $1 \in F \land n \in F$. – Frunobulax Mar 05 '16 at 14:59

2 Answers2

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Your question indicates that you do know the definition: the set of factors of the positive integer $n$ contains just the positive integers $m$ that divide $n$. If you'd rather not use the word "divide" in the definition, then it's the integers $m$ for which there is an integer $k$ such that $km=n$.

That definition does use words. Writing it more compactly this way (as in @AlbertoDebernardi 's comment):

$$ d(n)=\{m∈N:m | n\} $$

(where "$|$" is the symbol for "divides") doesn't really make it more "mathematical". Those symbols are all defined with words.

When you talk about "plugging in" an integer and getting the set of factors you may be thinking about a computer program. At http://rosettacode.org/wiki/Factors_of_an_integer you will find programs (in many languages) that accept an $n$ as input and return the list of its factors.

The python example:

factors(45)
[1, 3, 5, 9, 15, 45]

Ethan Bolker
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You need to define such a set for $n \in \mathbb{N}-\{0\}$ in the following way:

$$\{x\large|n=\prod_i^mp_i^{a_i} \land x=\prod_i^mp_i^{k_i} \,\, \text{where} \,\, 0\le k_i\le a_i\}$$

SchrodingersCat
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  • A Downvote without an explanation...? – SchrodingersCat Mar 04 '16 at 14:18
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    This answer is needlessly complicated, and fails to include that the $p_i$ should be prime numbers. – Ben Millwood Mar 04 '16 at 14:18
  • @BenMillwood This is a standard notation for any natural number from the Fundamental Theorem of Arithmetic. – SchrodingersCat Mar 04 '16 at 14:19
  • Why an answer that depends on unique factorization, while the concept makes sense in much more generality? – Ethan Bolker Mar 04 '16 at 14:23
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    The question asked for something where one could "plug in a natural number $n$". So we're in $\mathbb N$ and have unique factorization. The fact that the factorization can generate the factors in this way is a useful observation. I would possibly prefer a few more words and shorter formulas, but OP seems to want everything in a formula, so ... – David K Mar 04 '16 at 14:32