# positive multiple of an abundant number is abundant

Theorem. A positive multiple of an abundant number is abundant.
Proof. Let $n$ be abundant and $m>0$ be an integer. We have to show that $\sigma(mn)>2mn$, where $\sigma(n)$ is the sum of the positive divisors of $n$. Let $d_{1},\ldots,d_{k}$ be the positive divisors of $n$. Then certainly $md_{1},\ldots,md_{k}$ are distinct divisors of $mn$. The result is clear if $m=1$, so assume $m>1$. Then

 $\displaystyle\sigma(mn)$ $\displaystyle>$ $\displaystyle 1+\sum_{i=1}^{k}md_{i}$ $\displaystyle>$ $\displaystyle m\sum_{i=1}^{k}d_{i}$ $\displaystyle>$ $\displaystyle m(2n)$ $\displaystyle=$ $\displaystyle 2mn.$

As a corollary, the positive abundant numbers form a semigroup.

Title positive multiple of an abundant number is abundant PositiveMultipleOfAnAbundantNumberIsAbundant 2013-03-22 16:17:07 2013-03-22 16:17:07 Mathprof (13753) Mathprof (13753) 13 Mathprof (13753) Theorem msc 11A05 TheoremOnMultiplesOfAbundantNumbers