Question - Möbius inversion formula

Question

I need your help in the next question:

Prove directly from the definition the Möbius inversion formula.

(Möbius function defined as follows:

• μ(n) = 1 if n is a square-free positive integer with an even number of prime factors.
• μ(n) = −1 if n is a square-free positive integer with an odd number of prime factors.
• μ(n) = 0 if n is not square-free.)

I'm not sure what I need to do and how

Thanks !

Answer 1

Well, since you already know about the abelian group structure, things are way easier:

Moebius Inversion Formula: For arithmetic functions $\,f\,,\,g\,$ we have

$$f(n)=\sum_{d\mid n}g(d)\iff g(n)=\sum_{d\mid n}f(d)\mu\left(\frac nd\right)$$

Proof: With the functions

$$u(n)=1\;\;\forall n\in\Bbb N\;,\;\;I(n)=\begin{cases}1&\,\;\;n=1\\0&,\,\,n>1\end{cases}$$ we get that

$$f=g*u\stackrel{\text{mult. by $\,\mu\,$}\;}\implies\,f*\mu=(g*u)*\mu=g*(u*\mu)=g*I=g\;\;\;\;\;\square$$