# 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 ${d}_{1},\mathrm{\dots},{d}_{k}$ be the positive divisors of $n$. Then
certainly $m{d}_{1},\mathrm{\dots},m{d}_{k}$ are distinct divisors of $mn$. The result is
clear if $m=1$, so assume $m>1$. Then

$\sigma (mn)$ | $>$ | $1+{\displaystyle \sum _{i=1}^{k}}m{d}_{i}$ | ||

$>$ | $m{\displaystyle \sum _{i=1}^{k}}{d}_{i}$ | |||

$>$ | $m(2n)$ | |||

$=$ | $2mn.$ |

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

Title | positive multiple of an abundant number is abundant |
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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 |