# Math Help - Deduce from Mobius inversion formula

1. ## Deduce from Mobius inversion formula

I saw this question in few book but no detail, hope someone can show the working. Use the Mobius inversion formula to prove

1. $\sum_{d\,\mid \,n}\mu(d)\tau(\displaystyle{\frac{n}{d}})=1$

2. $\sum_{d\,\mid \,n}\mu(d)\sigma(\displaystyle{\frac{n}{d}})=n$

2. Both follow directly from Möbius inversion formula.

Möbius inversion formula

Given an arithmetic fuction $f(n)$

Define: $F(n)=\sum_{d|n}{f(d)}$

Then we have $f(n)=\sum_{d|n}{\mu(d)\cdot{F\left(\frac{n}{d}\rig ht)}}$

$\tau(n)=\sum_{d|n}{1}$ (number of positive divisors) and $\sigma(n)=\sum_{d|n}{d}$