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 deficient number that:
 * $\sigma \left({n}\right) < 2 n$

Each of the divisors of $k$ can be multiplied by $d$, and these numbers will all be divisors of $n$.

Thus all the divisors of $k$ can be expressed in the form $\dfrac r d$ whose sum will be less than $\dfrac {2 n} d$.

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

Hence the result by definition of deficient.