Sum of Möbius Function over Divisors

Theorem
Let $$n \in \mathbb{Z}^*_+$$, i.e. let $$n$$ be a positive integer.

Then $$\sum_{d \backslash n} \mu \left({d}\right) \frac {n} {d} = \phi \left({d}\right)$$

where:
 * $$\sum_{d \backslash n}$$ denotes the sum over all of the divisors of $$n$$;
 * $$\phi \left({d}\right)$$ is the Euler $\phi$ function, the number of integers less than $$d$$ that are prime to $$d$$;
 * $$\mu \left({d}\right)$$ is the Moebius function.