# Functional Equation involving Fractional Part

I'm stuck on the following problem:

Find (or classify) all functions $f$ with the property $$f(\{a+b\})=\{f(a)f(b)\}$$ Where $\{\}$ is the "fractional part" function.

So far, I've determined that $$f(0)=0$$ and that $$f(a)f(-a) \in \mathbb Z, \forall a \in \mathbb R$$

Can anyone help?

If it helps, I solved a similar functional equation $$f(a+\{b\})=f(b+\{a\})$$ and found that the solution set of functions was the set of functions with period $1$, or a period that goes evenly into one, like $0.5$ or $0.2$.

Any help is appreciated!

• the statement is equivalent to find functions $f$ such that $$f(\{x\})=\{f(x-z)f(z)\},\quad\forall z\in\Bbb R$$ Are you sure that the exercise says $\{a+b\}$? By curiosity, from what book is this exercise? – Masacroso Jul 6 '17 at 19:27
• @Masacroso Unfortunately, I'm not sure where it is from. I remember seeing it somewhere on the internet and writing it down, but when I got stuck and went back to look for the source to hopefully find an explanation, I could not find the place where I first found it. :( – Franklin Pezzuti Dyer Jul 6 '17 at 19:29
• @Masacroso Also, how did you get to that statement? – Franklin Pezzuti Dyer Jul 6 '17 at 19:45
• @Masacroso Oh, duh, sorry. :P Thanks for the help! If you post it as an answer, I will up vote it. – Franklin Pezzuti Dyer Jul 6 '17 at 19:50

This is just a partial solution to the question

We can rewrite the functional equation as

$$f(\{x\})=\{f(x-z)f(z)\},\quad \forall \{x\},z,x-z\in\operatorname{dom}(f)\tag1$$

Suppose that $0\in\operatorname{dom}(f)$, then we found that

$$f(0)=\{f(0)^2\}\implies |f(0)|<1\implies f(0)=f(0)^2\implies f(0)=0\tag2$$

By $(1)$ we have that $f(\{x\})=\{f(x)f(0)\}=0$, what imply that $f$ is zero if $|x|<1$. In general we have that

$$f(x)f(y)\in\Bbb Z,\quad \forall \{x+y\},x,y\in\operatorname{dom}(f)$$

There are infinite solutions from the assumption on the domain of $f$. Just we need that

• $f(x)=0$ when $|x|<1$ and

• $f(x)\in\Bbb Z$ when $|x|\ge 1$

• Awesome, thanks! – Franklin Pezzuti Dyer Jul 6 '17 at 20:43
• @Nilk I had a mistake (that I already fixed): ceil, floor and nearest integer are not, in general, solutions, because (by example) we have that $\lfloor -0.7\rfloor=-1\neq 0$. – Masacroso Jul 6 '17 at 20:47
• Oh, that's okay. :P – Franklin Pezzuti Dyer Jul 6 '17 at 21:08

HINT

Consider the properties of $$f(x) = \alpha^x$$ for $0<\alpha < \sqrt{2}$.

What happens to ruing things when $\alpha >= \sqrt{2}$?