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:

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

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 $s$.

$\blacksquare$