8
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I'm really confused about these two. For example if $n = 6$, then:

Divisors: $2, 3$
Proper Divisors: $1, 2, 3, 6$

Is it right?

Update
From Elementary Number Theory and Its Application by Kenneth H. Rosen 6th edition, page 256:

Because of certain mystical beliefs, the ancient Greeks were interested in those integers that are equal to the sum of all their proper positive divisors. Such integers are called perfect numbers.

Example:
$\sigma(6) = 1 + 2 + 3 + 6 = 12$, we see that $6$ is perfect.

Thanks,

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    I think most of the time the convention would be: divisors = $\{1,2,3,6\}$ and proper divisors =$\{1,2,3\}$. For instance 6 is perfect because $\sigma(n)=2n$ where $\sigma$ is the sum of it's _divisors_ or $\sigma_p(n)=n$ where $\sigma_p$ is the sum of it's _proper divisors_ . (Note that the notation $\sigma_p$ isn't standard.)2011-03-09
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    @MYself: Why don't you add an answer? Please also read this: http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments2011-03-09
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    @Myself: Thanks. But according to my the definition of perfect number in my text book, proper divisors are $1, 2, 3, 6$ :(.2011-03-09
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    @Moron: mostly because I'm confused about conventions about whether I should comment something that's more of a tiny answer or not. Thanks for the link, I'll try to figure out what's more appropriate.2011-03-09
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    @Chan: Then either your book is using a very odd definition or either you are somehow misinterpreting what the book says. If you're really in doubt you can reproduce the exact definition in an edit to your question.2011-03-09
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    @Myself: Just posted an update. The text interpretation really confused me.2011-03-10
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    @Chan: It's confusing indeed. The text first says a perfect number is the sum of its _proper divisors_ . But in the example for 6, the sum of all divisors turns out to be 12 and not 6. So one might think 6 isn't perfect after all. That's because the sum runs over all _divisors_ , not over all proper divisors. So the text and example are both correct but they don't really match up, hence the confusion.2011-03-10
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    @Myself: So I think I should use your definition from now on. Thank you.2011-03-10
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    @Chan: Well, it's not really _mine_ I guess :-) The thing is just that your book is correct but stating it in a confusing way, talking about proper divisors first and then summing up the actual divisors. So don't worry about it too much.2011-03-10

2 Answers 2

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I think most of the time the convention would be: divisors = $\{1,2,3,6\}$ and proper divisors= $ \{1,2,3\} $.

For instance 6 is perfect because $\sigma(n)=2n$ where σ is the sum of its divisors or $s(n)=n$ where $s$ is the sum of its proper divisors . (Note that the notation $s$ may not be completely standard.)

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    Indeed, nonstandard, since $\sigma_k(n)$ is usually the sum of the $k$-th powers of the divisors of $n$. Of course, here "p" is meant to the be the letter p for "proper", rather than the number $p$... perhaps you should use $\mathfrak{p}$ (`\mathfrak{p}`) instead? (-:2011-03-09
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    @Arturo Magidin: Even though I love the fraktur alphabet, I changed $\sigma_p$ into $s$. Apparently that notation is used at least by wikipedia and wolfram. (As the "aliquot sum", for instance on http://en.wikipedia.org/wiki/Divisor_function)2011-03-09
7

Generally for order relations, the adjective "proper" is usually used to denote a strict ordering, i.e. $\rm\ a \preceq b\ $ properly means $\rm\ a \preceq b\ $ but not $\rm\ b \preceq a\:.\:$ Thus we have proper divisors, proper subsets, etc.

Hence $\rm\:a\:$ is a proper divisor of $\rm\:b\:,\:$ or $\rm\ a\ |\ b\ $ properly, $\:$ simply means that $\rm\ a\ |\ b\ $ but not $\rm\ b\ |\ a\:.$