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 $m n$ . 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
Canonical name	PositiveMultipleOfAnAbundantNumberIsAbundant
Date of creation	2013-03-22 16:17:07
Last modified on	2013-03-22 16:17:07
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	13
Author	Mathprof (13753)
Entry type	Theorem
Classification	msc 11A05
Related topic	TheoremOnMultiplesOfAbundantNumbers