# 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:

$\dfrac {\map \sigma {k d} } {k d} < 2$
$\dfrac {\map \sigma {k d} } {k d} > \dfrac {\map \sigma k} k$

That is:

$\dfrac {\map \sigma k} k < 2$

Hence the result by definition of deficient.

$\blacksquare$