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

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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 $s$.

$\blacksquare$