Divisor of Perfect Number is Deficient

Theorem

Let $n$ be a perfect number.

Let $n = r s$ where $r$ and $s$ are positive integers such that $r > 1$ and $s > 1$.

Then $r$ and $s$ are both deficient.

Proof

Without loss of generality, consider $r$.

We have by definition of $\sigma$ function and perfect number that:

$\dfrac {\map \sigma {r s} } {r s} = 2$
$\dfrac {\map \sigma {r s} } {r s} > \dfrac {\map \sigma r} r$

That is:

$\dfrac {\map \sigma r} r < 2$
$\dfrac {\map \sigma r} s < 2$

Hence the result by definition of deficient.

$\blacksquare$