Superabundant Numbers are Infinite in Number

Theorem

There are infinitely many superabundant numbers.

Proof

Aiming for a contradiction, suppose the set $S$ of superabundant numbers is finite.

Let $m$ be the greatest element of $S$.

By definition of superabundant, $m$ has the largest abundancy index of all the elements of $S$.

Consider the integer $2 m$.

From Abundancy Index of Product is greater than Abundancy Index of Proper Factors, $2 m$ has a higher abundancy index than $m$.

There are two possibilities:

$(1): \quad 2 m$ is the smallest integer greater that $n$ which has a higher abundancy index than $m$.

By definition, that would make $m$ superabundant.

$(2) \quad$ There exists a finite set $T := \set {n \in \Z: m < n < 2 m: \map A n > \map A m}$, where $\map A n$ denotes the abundancy index of $n$.

The smallest element $t$ of $T$ therefore has an abundancy index greater than all smaller positive integers.

Thus by definition $t$ is superabundant.

In either case, there exists a superabundant number not in $S$.

Thus $S$ cannot contain all superabundant numbers.

But this contradicts our initial assumption that the set $S$, containing all superabundant numbers is finite.

It follows by Proof by Contradiction that $S$ is infinite.

Hence the result.

$\blacksquare$

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $12$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $12$