# alternate proof of Möbius inversion formula

The Möbius inversion theorem can also be proved elegantly using the fact that arithmetic functions^{} form a ring under $+$ and $*$.

Let $I$ be the arithmetic function that is everywhere $1$. Then obviously if $\mu $ is the Möbius function^{},

$$(\mu *I)(n)=\sum _{d|n}\mu (d)I\left(\frac{n}{d}\right)=\sum _{d|n}\mu (d)=\{\begin{array}{cc}1\hfill & \text{n=1}\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}$$ |

and thus $I*\mu =e$, where $e$ is the identity of the ring.

But then

$$f(n)=\sum _{d|n}g(d)=\sum _{d|n}g(d)I\left(\frac{n}{d}\right)$$ |

and so $f=g*I$. Thus $f*\mu =g*I*\mu =g$. But $g=f*\mu $ means precisely that

$$g(n)=\sum _{d|n}\mu (d)f\left(\frac{n}{d}\right)$$ |

and we are done.

The reverse equivalence is similar ($f*\mu =g\Rightarrow f*\mu *I=g*I\Rightarrow f=g*I$).

Title | alternate proof of Möbius inversion formula^{} |
---|---|

Canonical name | AlternateProofOfMobiusInversionFormula |

Date of creation | 2013-03-22 16:30:31 |

Last modified on | 2013-03-22 16:30:31 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 4 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 11A25 |