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:

$\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} r} r$

and consequently also:

$\dfrac {\map {\sigma_1} n} n > \dfrac {\map {\sigma_1} s} s$

where $\sigma_1$ denotes the divisor sum function.

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:

$\ds \map {\sigma_1} r$

$=$

$\ds \sum_{d \mathop \divides r} d$

$\ds \leadsto \ \ $

$\ds s \map {\sigma_1} r$

$=$

$\ds \sum_{d s \mathop \divides n} d s$

$\ds \leadsto \ \ $

$\ds \map {\sigma_1} n$

$>$

$\ds \sum_{d s \mathop \divides n} d s$

as numbers of the form $d s$ do not exhaust divisors of $n$: note $1 \divides n$

$\ds \leadsto \ \ $

$\ds \dfrac {\map {\sigma_1} n} n$

$>$

$\ds \dfrac {s \map {\sigma_1} r} n$

$\ds $

$=$

$\ds \dfrac {\map {\sigma_1} r} {n / s}$

$\ds $

$=$

$\ds \dfrac {\map {\sigma_1} r} r$

Similarly for $s$.

$\blacksquare$