# Abundancy Index of Product is greater than Abundancy Index of Proper Factors

## Theorem

Let $n \in \Z_{>0}$ be a composite number such that $n = r s$, where $r, s \in \Z_{>1}$.

Then:

$\dfrac {\sigma \left({n}\right)} n > \dfrac {\sigma \left({r}\right)} r$

and consequently also:

$\dfrac {\sigma \left({n}\right)} n > \dfrac {\sigma \left({s}\right)} s$

where $\sigma \left({n}\right)$ denotes the $\sigma$ function of $n$.

That is, the abundancy index of a composite number is strictly greater than the abundancy index of its divisors.

## Proof

Consider the divisors of $r$.

Let $d \mathrel \backslash r$, where $\backslash$ indicates divisibility.

We have that:

$d \mathrel \backslash n$

and also that:

$d s \mathrel \backslash n$

Thus:

 $\displaystyle \sigma \left({r}\right)$ $=$ $\displaystyle \sum_{d \mathop \backslash r} d$ $\displaystyle \leadsto \ \$ $\displaystyle s \sigma \left({r}\right)$ $=$ $\displaystyle \sum_{d s \mathop \backslash n} d s$ $\displaystyle \leadsto \ \$ $\displaystyle \sigma \left({n}\right)$ $>$ $\displaystyle \sum_{d s \mathop \backslash n} d s$ as numbers of the form $d s$ do not exhaust divisors of $n$: note $1 \mathrel \backslash n$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\sigma \left({n}\right)} n$ $>$ $\displaystyle \dfrac {s \sigma \left({r}\right)} n$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sigma \left({r}\right)} {n / s}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sigma \left({r}\right)} r$

Similarly for $s$.

$\blacksquare$