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

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

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

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

From the definition of the Möbius function, the Möbius function of a product of $m$ distinct primes is $\paren {-1}^m$.

There are $\dbinom k m$ ways to choose $m$ primes from $p_1, p_2, \dots, p_k$ to multiply together.

So among the divisors of $p_1 p_2 \dotsm p_k$, there are exactly $\dbinom k m$ numbers that are the product of $m$ distinct primes.

Thus:

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

Now we show that $\mu * u = \iota$.

Pick any $n \in \mathbb N$.

Then:

Hence $\mu * u = \iota$.