# divisor function is multiplicative, the

###### Theorem.

The divisor function^{} (http://planetmath.org/TauFunction) is multiplicative.

Proof.
Let $t=mn$ with $m,n$ coprime^{}.
Applying the fundamental theorem of arithmetic^{}, we can write

$$m={p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\mathrm{\cdots}{p}_{r}^{{a}_{r}},n={q}_{1}^{{b}_{1}}{q}_{2}^{{b}_{2}}\mathrm{\cdots}{q}_{s}^{{b}_{s}},$$ |

where each ${p}_{j}$ and ${q}_{i}$ are prime. Moreover, since $m$ and $n$ are coprime, we conclude that

$$t={p}_{1}^{{a}_{1}}{p}_{2}^{{a}_{2}}\mathrm{\cdots}{p}_{r}^{{a}_{r}}{q}_{1}^{{b}_{1}}{q}_{2}^{{b}_{2}}\mathrm{\cdots}{q}_{s}^{{b}_{s}}.$$ |

Now, each divisor^{} of $t$ is of the form

$$t={p}_{1}^{{k}_{1}}{p}_{2}^{{k}_{2}}\mathrm{\cdots}{p}_{r}^{{k}_{r}}{q}_{1}^{{h}_{1}}{q}_{2}^{{h}_{2}}\mathrm{\cdots}{q}_{s}^{{h}_{s}}.$$ |

with $0\le {k}_{j}\le {a}_{j}$ and $0\le {h}_{i}\le {b}_{i}$, and for each such divisor we get a divisor of $m$ and a divisor of $n$, given respectively by

$$u={p}_{1}^{{k}_{1}}{p}_{2}^{{k}_{2}}\mathrm{\cdots}{p}_{r}^{{k}_{r}},v={q}_{1}^{{h}_{1}}{q}_{2}^{{h}_{2}}\mathrm{\cdots}{q}_{s}^{{h}_{s}}.$$ |

Now, each respective divisor of $m$, $n$ is of the form above, and for each such pair their product is also a divisor of $t$. Therefore we get a bijection between the set of positive divisors of $t$ and the set of pairs of divisors of $m$, $n$ respectively. Such bijection implies that the cardinalities of both sets are the same, and thus

$$d(mn)=d(m)d(n).$$ |

Title | divisor function is multiplicative, the |
---|---|

Canonical name | DivisorFunctionIsMultiplicativeThe |

Date of creation | 2013-03-22 15:03:47 |

Last modified on | 2013-03-22 15:03:47 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 9 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 11A25 |