Slightly confused about Möbius' inversion theorem Alright, I understand the proof of the theorem and everything, but I don't have much intuition about the definition. I think the theorem would work using any function $\psi : \mathbb{N} \rightarrow \mathbb{N}$ such that $\psi(1) = 1$ and $\displaystyle\sum_{d | n} \psi(d) = 0$ whenever $n > 1$. So is there some specific reason as to why $\mu$ is defined the way it is?
As another question, I am using the book "A classical introduction to modern number theory" by Ireland & Rosen and on chapter 2 there is the exercise 21 which I just cannot make sense of. It reads:
"Define $f(n) = p$ if $n$ is a power of p and zero otherwise. Prove that $\sum_{d | n} \mu(n/d) \log d = f(n)$. [Hint: First calculate $\sum_{d | n} f(d)$ and then apply the Möbius inversion formula]"
It doesn't make sense to me, because which is this $p$ that the question references? Is it some fixed constant? I guess this would make sense if $\sum_{d | n} f(d) = \log n$ because then the formula would work out, but why would that be true? Well I guess if we consider that $n = p^{\epsilon}$ then the sum would give $\epsilon \log p = \log p^{\epsilon} = \log n$ (if we consider 1 as not a power of $p$), but that's just not a safe assumption, is it? I'm rambling here, any ideas?
 A: The reason $\mu$ is defined the way it is is that this gives the only function $\psi$ with the properties you required. This is trivial to prove by induction.
As for your exercise, I think you must have not transcribed it correctly. Certainly $p$ is assumed to be a prime number, but from what you write it seems that $f(p^m)=\log p$ for $m>0$ and $f(1)=0$, since that is what the formula you are required to prove gives. Also $f(n)=0$ for non prime-powers, and one can check that $\sum_{d|n} f(d) = \log n$ for all $n>0$ by using the prime factorization of $n$ (if $p$ has multiplicity $m$ in $n$ then there are $m$ positive powers of $p$ that divide $n$), so your argument basically works.
A: Euler showed that the Dirichlet series of a multiplicative function $f$ can be written as a product (now called an Euler product):
$$
L(s, f) := \sum_{n \geq 1} \frac{f(n)}{n^s} = \prod_p \left(1 + f(p)p^{-s}+ f(p^2) p^{-2s} + \dots \right).
$$
In particular the $\zeta$-function corresponds to $f(n) \equiv 1$ and yields:
$$
\zeta(s) = L(s,1) = \sum_{n \geq 1} n^{-s} = \prod_{p} \left((p^{-s})^0 + (p^{-s})^1 + (p^{-s})^2 + \dots \right) = \prod_{p} \left(1 - p^{-s} \right)^{-1}.
$$
This implies that
$$
\frac{1}{\zeta(s)} = \prod_{p} \left( 1 - p^{-s}\right) = \sum_{n \geq 1} \frac{\mu(n)}{n^s}
$$
for some multiplicative function $\mu$ such that $\mu(p) = -1$ and $\mu(p^k) =0$ for any prime $p$ when $k \geq 2$.  This yields the standard definition of the Möbius function and is possibly how the function was discovered.  This should at least provide an intuitive understanding of the function $\mu$.
