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
, consider $r$.

We have by definition of $\sigma$ function and perfect number that:
 * $\dfrac {\sigma \left({r s}\right)} {r s} = 2$

But from Abundancy of Product is greater than Abundancy of Proper Factors:
 * $\dfrac {\sigma \left({r s}\right)} {r s} > \dfrac {\sigma \left({r}\right)} r$

That is:
 * $\dfrac {\sigma \left({r}\right)} r < 2$

Mutatis mutandis:
 * $\dfrac {\sigma \left({r}\right)} s < 2$

Hence the result by definition of deficient.