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,

  • 0
    @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.)

  • 0
    @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 $\rm\ b\ |\ a\:.$