# 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 $s$.

$\blacksquare$