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# 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.

Related:

TheoremOnMultiplesOfAbundantNumbers

Major Section:

Reference

Type of Math Object:

Theorem

Parent:

## Mathematics Subject Classification

11A05*no label found*

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