Divisor of Deficient Number is Deficient

Theorem
Let $n$ be a perfect number.

Let $n = k d$ where $r$ is a positive integer.

Then $k$ is deficient.

Proof
We have by definition of $\sigma$ function and perfect number that:
 * $\dfrac {\sigma \left({k d}\right)} {k d} < 2$

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

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

Hence the result by definition of deficient.