Sum of Möbius Function over Divisors/Lemma

Lemma to Sum of Möbius Function over Divisors
Let $n \in \Z_{>0}$, i.e. let $n$ be a strictly positive integer.

Let $\displaystyle \sum_{d \mathop \backslash n}$ denote the sum over all of the divisors of $n$.

Let $\mu \left({d}\right)$ be the Möbius function.

Then:
 * $\displaystyle \sum_{d \mathop \backslash n} \mu \left({d}\right) = \left \lfloor {\frac 1 n} \right \rfloor$

That is:


 * $\mu * u = \iota$

where $u$ and $\iota$ are the unit arithmetic function and identity arithmetic function respectively.

Proof
The lemma is clearly true if $n=1$.

Assume, then, that $n > 1$ and write, by the Fundamental Theorem of Arithmetic:
 * $n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$

In the sum $\displaystyle \sum_{d \mathop \backslash n} \mu \left({d}\right)$ the only non-zero terms come from $d=1$ and the divisors of n which are products of distinct primes.

Thus from Alternating Sum and Difference of Binomial Coefficients for Given n:

Hence, the sum is $1$ for $n=1$, and $0$ for $n>1$, which are precisely the values of $\left \lfloor {\dfrac 1 n} \right \rfloor$.