Intuitively, why can $a, b$ cycle in ${\color{red}{b}} = \frac c{\color{red}{a}} \iff {\color{red}{a}} = \frac c{\color{red}{b}}$?
I'm NOT asking about algebra behind $ab = c \iff {\color{red}{b}} = \frac c{\color{red}{a}} \iff {\color{red}{a}} = \frac c{\color{red}{b}}.$

Rather, what's the intuition why $\color{red}{a, b}$ can swap places, whilst c remains in the numerator?

What's this phenomenon or behavior called? A cyclic permutation?
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User  Comment  Date 

whybecause  (no comment)  Dec 30, 2021 at 22:42 
 Naming
"What's this phenomenon or behavior called?"
Sometimes we feeding one initial value into a function to get a first result, and then feeding that result into the same function to get a second result.
If the second result is always the same as the initial value, no matter what the initial value is, the behavior is called an "involution".
In cryptography, this behavior is called a "reciprocal cipher".
 Intuition
"what's the intuition ...?"
x columns
o o o o o o o o o
o o o o o o o o o y rows
...
o o o o o o o o o
o o o o o o o o o
Say we have some total number of rocks C and we line them up in rows and columns.
After we start placing C in x columns, there's going to be some number y=C/x of rows of rocks.
If we take those same C rocks and start placing them in y columns, there's going to be some number z=C/y of rocks per column.
It's relatively intuitive to me that we can rotate how we look at that rectangle without changing C, so we can see it as either dividing those C rocks into x columns, or dividing those C rocks into y columns, so z must equal x. (At least, it's clear to me when x and y are both integers; perhaps there's a more intuitive approach when x or y or both are not integers).
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DNB  (no comment)  Dec 31, 2021 at 06:54 
A possibly helpful further note beyond the answers given elsewhere here: Do you intuitively understand why $ab=c$ is equivalent to $a=c/b$ for all $b\ne 0$? If so, and if you intuitively understand the commutativity of multiplication, then these two intuitions put together give you everything that you need.
The intuition for the commutativity of multiplication is really easy to come up with so I won't argue that case.
But the equivalence of $ab=c$ and $a=c/b$ is basically just down to what we mean by division.
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