# 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,

 $(\mu*I)(n)=\sum_{d|n}\mu(d)I\left(\frac{n}{d}\right)=\sum_{d|n}\mu(d)=\begin{% cases}1&n=1\\ 0&\text{otherwise}\end{cases}$

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 AlternateProofOfMobiusInversionFormula 2013-03-22 16:30:31 2013-03-22 16:30:31 rm50 (10146) rm50 (10146) 4 rm50 (10146) Proof msc 11A25