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

Let $\map \mu d$ be the Möbius function.

Then:
 * $\displaystyle \sum_{d \mathop \divides n} \map \mu d = \floor {\frac 1 n}$

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 $\displaystyle \sum_{d \mathop \divides n} \map \mu d$ the only non-zero terms come from $d = 1$ and the divisors of $n$ which are products of distinct primes.

Thus:

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