# 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:

$\dfrac {\map \sigma n} n > \dfrac {\map \sigma r} r$

and consequently also:

$\dfrac {\map \sigma n} n > \dfrac {\map \sigma s} s$

where $\map \sigma n$ 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 $d \divides r$, where $\divides$ indicates divisibility.

We have that:

$d \divides n$

and also that:

$d s \divides n$

Thus:

 $\displaystyle \map \sigma r$ $=$ $\displaystyle \sum_{d \mathop \divides r} d$ $\displaystyle \leadsto \ \$ $\displaystyle s \map \sigma r$ $=$ $\displaystyle \sum_{d s \mathop \divides n} d s$ $\displaystyle \leadsto \ \$ $\displaystyle \map \sigma n$ $>$ $\displaystyle \sum_{d s \mathop \divides n} d s$ as numbers of the form $d s$ do not exhaust divisors of $n$: note $1 \divides n$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {\map \sigma n} n$ $>$ $\displaystyle \dfrac {s \map \sigma r} n$ $\displaystyle$ $=$ $\displaystyle \dfrac {\map \sigma r} {n / s}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\map \sigma r} r$

Similarly for $s$.

$\blacksquare$