What is the period of $f(x)=x-[x]$?

Question

What is the period of the function $f(x)=x-[x]$?

(Here, $[\,.]$ represents the greatest integer function)

On a tangential note: does that affect the periodicity of $e^{x-[x]}$?

Answer 1

The period is 1, because $f(x+1) = (x+1) - [x+1] = x+1-([x]+1) = x-[x] = f(x)$. Common sense suggest that it cannot be less than 1, but to prove this rigorously you only need to see that for $x \in [0,1)$, $f(x) = x$ so it's injective on an interval of length 1.

This also influences the other function: if $f(x)$ is periodic with a period $a$, $g(f(x))$ is also periodic with period $a$ (or smaller).

Answer 2

We know that, for all $x\in\mathbb{R}$ :

$$[x]\le x<[x]+1$$

Hence :

$$[x]+1\le x+1<([x]+1)+1$$

Which proves that $[x]+1$ is the greatest integer less that $x+1$. In other words :

$$[x+1]=[x]+1$$

which can be written :

$$x+1-[x+1]=x-[x]$$

As a conclusion, $f:x\mapsto x-[x]$ is $1$-periodic.