## Inversion of an Equation with Floor

Whilst working with the equations for the previous post on the fusc function, I encountered one I wished to invert. That equation was essentially

$y = x - 2 \lfloor x \rfloor$

but I had not noticed that this equation is special, and that the ‘2’ is significant.

It turns out that the inverse is identical:

$x = y - 2 \lfloor y \rfloor$

Similarly, the inverse of

$y_2 = 2 \lfloor x_2 \rfloor - x_2$

is given by

$x_2 = 2 \lceil y_2 \rceil - y_2$

but note the switch from floor to ceiling in this case.

I wondered if equations slightly more general than this had a solution, and it seems that the answer is essentially ‘no’. To be more specific, consider the equation

$y = f(x) = x + c \lfloor x \rfloor$

Following is a table of some properties of this function of x, for different values of c.

c inj surj cts mono bound
(−∞,−2)  ✔  ✗  ✗  ✗  ✗
−2  ✔  ✔  ✗  ✗  ✗
(−2,−1)  ✗  ✔  ✗  ✗  ✗
−1  ✗  ✗  ✗  ✗  ✔ to [0,1)
(−1,0)  ✗  ✔  ✗  ✗  ✗
0  ✔  ✔  ✔  ✔ (<)  ✗
(0,+∞)  ✔  ✗  ✗  ✔ (<)  ✗

where

inj = injective or 1-to-1 ($f(y_1) = f(y_2) \implies y_1 = y_2$)
surj = surjective or onto ($\forall y \exists x \bullet f(x) = y$)
cts = continuous
mono = monotonic:
(<) for strictly increasing ($x_1 < x_2 \implies f(x_1) < f(x_2)$)
bound = bounded range

For f to be invertible (i.e. for f to have an inverse), f must be bijective, that is, both injective and surjective. There are only two values for which that is the case: $c \in \{-2,0\}$. When c=0, f is just the identity function, and thus the case c=−2 is the only interesting case.