Periodic function

Question

Considering the function

$F(x)=x-E(x)$ such as $E(x)$ is the integer part of $x$

So here just with observation we can see that :

$f(x+1)=(x+1)-E(x+1)=f(x)$.

But here mathematics is not based just of observation .

how can one prove that the function f(x) is periodic with 1 without starting by f(x+1) ?

Answer 1

One can prove periodicity of $f(x)$ by direct substitution.

First, note that the floor function $E(x)$ $\big($which is, by the way, also denoted as $ \lfloor x \rfloor$ and defined as $E(x) = \lfloor x \rfloor := \max\big\{n\in\mathbb{Z}\:\big| \ \ n<x \big\}$ $\big)$ have the following property for any integer $k$:

$$ \forall k \in\mathbb{Z}, \forall x \in \mathbb{R} \quad E(x + k) = E(x) + k \tag{1} $$

following from definition of $E(x)$.

Second, substitute the equality $\eqref{1}$ into expression for $f(x+1)$ we get

$$ f(x+1) \stackrel{\text{def}}{=} (x+1) - E(x+1) = (x+1) - \big(E(x) + 1\big) = x - E(x) = f(x), $$ i.e. $\ f\ $ is periodic with periond $1$ as $f(x+1) = f(x)\ \ \forall x\in\mathbb{R}$.

Q.E.D.

Answer 2

A function is said to be periodic if $\ f(x + P) = f(x), P > 0 $
http://en.wikipedia.org/wiki/Periodic_function

I think that "observation" is enough here to prove that f(x) is periodic.