Lets assume that we have some interval:
[3.5,5.5]
Now, lets suppose that we x = 4.5
Now, let
y = x + 1 = 5.5
y = x + 2 = 3.5
y = x + 3 = 4.5
y = x + n = ?
I am searchin for formula for arbitrary $n$. Can you help me ? I can only for small numbers do it. Keep in mind that our interval is in real - it means that number can be non-integer and not positive. Of course, sometimes I would like to count x-n=?
The question seems to be for: $a,b,x,z \in \mathbb{R} \land x \in [a,b]$, provide a function $f(x+z) \in [a,b]$.
Even though $x \in [a,b]$, because $z$ can be any real number, there's no essential difference between $z$ and $x+z$: they both cover the same range. Perhaps $z \geq 0$ is what was intended, in which case $z+x \geq a$.
Of course, multiple definitions of $f$ are possible. $f$ needs to map a domain of infinite length into a range of finite length. The reciprocal function does that. Using the interval length as a modulus is another way. arccos and arcsin also map an infinite domain to a finite range. Most any function that maps an infinite domain to a finite range could be adopted by appropriate scaling and translation of both the argument and the result.
Using the interval length as a modulus could be expressed as $f(x)= ((x+z) \mod (b-a)) +a$. Since $(b-a)$ is real $(v\mod(b-a))$ can be understood as $0 \leq v - k(b-a) < (b-a), v \in \mathbb{R}$ for some $k \in \mathbb{Z}$. A small tweak would be required if $f$ must be able to yield $b$ (which it doesn't with this definition).
Using the reciprocal, and assuming $z \geq 0$ then $f(x) = a + \frac{a(b-a)}{x+z}$. Note that $f$ goes from $b$ down to $a$ as $x+z$ increases from $a$ to infinity. A tweak would be required if $f$ must be able to yield $a$ (which it doesn't with this definition).