Intuitively, the proper divisors of an integer $n$ don't include $n$ because trivially, any number divides itself; but $1$ divides any integer as well.

What is the rationale for including $1$?

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    $\begingroup$ Proper divisor of a number $n$ means a divisor $d$ smaller than the original number $n$. This definition is diferent from being a trivial divisor. Every number $n$ has two trivial divisors: $n$ itself and $1$. $\endgroup$ – Xam Oct 28 '16 at 1:57

The word "proper" commonly connotes "but different than" - for example, if A is a proper subset of B then A is a subset of B but different than B.

So 1 is a divisor of 6 but different than 6, making 1 a proper divisor of 6.

  • $\begingroup$ Is the empty set then considered a proper subset of any non-empty set? $\endgroup$ – Mark Fischler Oct 28 '16 at 1:58
  • $\begingroup$ If the adjective 'proper' in mathematics has the meaning you are saying it does, then everything makes sense. $\endgroup$ – user376034 Oct 28 '16 at 1:58
  • $\begingroup$ Yes, the empty set is a proper subset of any non-empty set. It is often referred to as the trivial subset. $\endgroup$ – Browning Oct 28 '16 at 2:08

One decent reason for this definition is that when studying perfect numbers (which is an interesting field of study) it makes it easy to say that a perfect number is equal to the sum of its proper divisors.

Similarly, you can say that we count $1$ as relatively prime to all integers so that the totient function (number of integers less than $n$ which are relatively prime to $n$) will be multiplicative. That is, if $m$ and $n$ are relatively prime, then $$ \varphi(mn) = \varphi(m) \varphi(n) $$


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