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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.) –  Myself Mar 9 '11 at 17:41
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@MYself: Why don't you add an answer? Please also read this: meta.math.stackexchange.com/questions/1559/… –  Aryabhata Mar 9 '11 at 17:42
    
@Myself: Thanks. But according to my the definition of perfect number in my text book, proper divisors are $1, 2, 3, 6$ :(. –  Chan Mar 9 '11 at 17:43
    
@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. –  Myself Mar 9 '11 at 17:45
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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. –  Myself Mar 9 '11 at 17:51
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2 Answers

up vote 12 down vote accepted

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? (-: –  Arturo Magidin Mar 9 '11 at 19:12
    
@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 en.wikipedia.org/wiki/Divisor_function) –  Myself Mar 9 '11 at 19:51
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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 $\rm\ b\ |\ a\:.$

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