Number Theory: Möbius Function Background



Definition:
If $S$ with partial order ≤ is locally finite with minimal element, then the generalized Möbius function $\mu$: S $\times$ S $\rightarrow$ $\mathbb{N}$ by
(a) $\forall$ s $\in$S, $\mu$ (s, s)=1
(b) for r< s, $\mu$ (r, s) = - $\sum_{t \in [r, s) } μ(r, t)$\
Theorem to be proved:
Let S $\neq$ ∅ and define if and only if a | b, then a ≤ b. If {d, n}$\mathbb{Z}$, μ(d, n)= $\frac{n}{d}$
Source:
Number Theory with Applications (James A. Anderson, James M. Bell)
My Question
Sorry that I have tried to prove this theorem for many times but I fail. Could somebody help me?
 A: To avoid any possible confusion, I’ll use $\mu'$ for the number theoretic function.
You can do it by induction on $n$. Assume that
$$\mu(d,m)=\mu'\left(\frac{m}d\right)$$
whenever $1\le d\mid m<n$, and let $1\le d\mid n$. If $d=n$, we have
$$\mu(d,n)=\mu(n,n)=1=\mu'(1)=\mu'\left(\frac{n}d\right)\,,$$
so assume that $d<n$.
Let $\prod_{k=1}^\ell p_k^{\alpha_k}$ be the prime factorization of $\frac{n}d$, where each $\alpha_k\ge 1$. then $m\in[d,n)$ iff $\frac{m}d=\prod_{k\in S}p_k^{\beta_k}$, where $S\subseteq\{1,\ldots,\ell\}$, $1\le\beta_k\le\alpha_k$ for $k\in S$, and either $S\ne\{1,\ldots,\ell\}$, or there is a $k\in S$ such that $\beta_k<\alpha_k$.
Suppose first that $\alpha_k=1$ for $k=1,\ldots,\ell$, so that $\mu'\left(\frac{n}d\right)=(-1)^\ell$. Then $m\in[d,n)$ iff there is an $S\subsetneqq\{1,\ldots,\ell\}$ such that $\frac{m}d=\prod_{k\in S}p_k$, and therefore
$$\begin{align*}
\mu(d,n)&=-\sum_{m\in[d,n)}\mu(d,m)\\
&=-\sum_{m\in[d,n)}\mu'\left(\frac{m}d\right)\\
&=-\sum_{S\subsetneqq\{1,\ldots,\ell\}}(-1)^{|S|}\\
&\overset{(1)}=-\sum_{k=0}^{\ell-1}(-1)^k\binom{\ell}k\\
&\overset{(2)}=-\left((1-1)^\ell-(-1)^\ell\right)\\
&\overset{(3)}=(-1)^\ell\\
&=\mu'\left(\frac{n}d\right)\,.
\end{align*}$$
Here $(1)$ is because $\{1,\ldots,\ell\}$ has $\binom{\ell}k$ subsets of cardinality $k$, $(2)$ follows from the binomial theorem, and $(3)$ holds because $0^\ell=0$, since $\ell\ge 1$.
The other possibility is that $\alpha_k>1$ for some $k\in\{1,\ldots,\ell\}$, so that $\mu'\left(\frac{n}d\right)=0$. See if you can carry out the induction step in this case: the computation is actually very similar to the one that I just did, though for some $m\in[d,n)$ you’ll have $\mu'\left(\frac{m}d\right)=0$.
Added: If $\alpha_k>1$ for some $k\in\{1,\ldots,\ell\}$, we have to split the $m\in[d,n)$ into those that will actually contribute to $\mu(d,n)$ and those that contribute $0$. For each $m\in[d,n)$ let $\frac{m}d=\prod_{k=1}^\ell p_k^{\beta_k(m)}$, where $0\le\beta_k(m)\le\alpha_k$ for each $k\in\{1,\ldots,\ell\}$. Let
$$S_m=\{k\in\{1,\ldots,\ell\}:\beta_k(m)=1\}$$
and
$$T_m=\{k\in\{1,\ldots,\ell\}:\beta_k(m)>1\}\,.$$
If $T_m\ne\varnothing$, then $\mu(d,m)=\mu'\left(\frac{m}d\right)=0$. Moreover, for each $S\subseteq\{1,\ldots,\ell\}$ there is exactly one $m\in[d,n)$ such that $S_m=S$ and $T_m=\varnothing$, and each of these is a proper divisor of $n$, so
$$\begin{align*}
\mu(d,n)&=-\sum_{m\in[d,n)}\mu(d,m)\\
&=-\sum_{m\in[d,n)}\mu'\left(\frac{m}d\right)\\
&=-\sum_{\substack{m\in[d,n)\\T_m=\varnothing}}\mu'\left(\frac{m}d\right)\\
&=-\sum_{S\subseteq\{1,\ldots,\ell\}}(-1)^{|S|}\\
&=-\sum_{k=0}^\ell(-1)^k\binom{\ell}k\\
&=-(1-1)^\ell\\
&=0\\
&=\mu'\left(\frac{n}d\right)\,.
\end{align*}$$
