Clarification on Mobius inversion formula proof . I was looking at a proof for the Mobius inversion formula and the first line states that 
$\sum_{d|n}\mu(d)F(n/d)=\sum_{d|n}\mu(d)\sum_{d|n}f(n/d)$
But this confuses me as it seems to imply that $F(n/d)=\sum_{d|n}f(n/d)$
But a requisite to the theorem is that $F(n)=\sum_{d|n}f(d)$
So then shouldn't we have that $F(n/d)=\sum_{d|n}f(1)$ ?
(And by that notation I mean add one f(1) to the sum for every d that divides n)
Edit: 
Mobius inversion formula states that if $f$ is an arithmetic function and $F(n)=\sum_{d|n}f(d)$ then $f(n)=\sum_{d|n}\mu(d)F(n/d)$
The full proof I'm looking at is found here http://mathonline.wikidot.com/the-moebius-inversion-formula
 A: For every $d \mid n$, there is a unique corresponding $\frac{n}{d} \mid n$ since, obviously, $d \times \frac{n}{d} = n$. This means there is a $1-1$ correspondence between the elements of $d$ where $d\mid n$, and those of $\frac{n}{d}$ where $d \mid n$. Thus,
$$F(n)=\sum_{d|n}f(d)=\sum_{d|n}f(n/d) \tag{1}\label{eq1}$$
As such, with the right side of what you're asking, you get
$$\sum_{d\mid n}\mu(d)\sum_{d|n}f(n/d) = \sum_{d\mid n}\mu(d)F(n) = \sum_{d\mid n}\mu(n/d)F(n) \tag{2}\label{eq2}$$
Your left side, along with the equivalent of your right side in \eqref{eq2}, becomes
$$\sum_{d\mid n}\mu(d)F(n/d) = \sum_{d\mid n}\mu(n/d)F(n) \tag{3}\label{eq3}$$
This is specifically asked to be proven in Mobius Inversion Formula, and it has an answer. Specifically, the solution involves changing the order of summations, along with making the appropriate indices changes. Also, Understanding the proof of Möbius inversion formula asks a quite similar question, and it has $2$ answers, each of which also provide more details about changing the summation orders.
I'm somewhat surprised, and disappointed, that the Mathonline proof makes the
statement of $\sum_{d|n}\mu(d)F(n/d)=\sum_{d|n}\mu(d)\sum_{d|n}f(n/d)$ without any additional explanation. I consider it a bit questionable to do this even if the target audience were number theorists, but it's especially so for a more general audience.
