Elaboration of Möbius inversion formula.

Question

The Möbius Inversion Formula states:

Let $F$ and $f$ be two number-theoretic functions related by the formula, $$F(n) = \sum_{d\mid n} f(d),$$ Then, $$f(n) = \sum_{d\mid n} \mu(d) F(n/d) = \sum_{d\mid n} \mu(n/d)F(d).$$

But I do not know how the 2 sums mentioned in the conclusion of the theorem are the same, could anyone explain this for me please?

Answer 1

If $d$ is a divisor of $n$, so is $n/d$. Since $d$ runs through all divisors of $n$, both expressions are one and the same.

Answer 2

Example:

For $n=9$ the first sum is $$\mu(1)F(9)+\mu(3)F(3)+\mu(9)F(1)$$ and the second $$\mu(9)F(1)+\mu(3)F(3)+\mu(1)F(9)$$

Can you see the pattern?