How do I prove that $f(x) = x - {\lfloor}x{\rfloor}$ is periodic and find its minimal period?

I've taken the following steps:

Let $x = x_0 + \Delta{x}$ where $x_0 \in \mathbb Z$ and $\Delta{x} \in [0;1)$. I need to prove that for given $x$ and $T$: $f(x) = f(x+T)$ where $T$ is some period to be defined. Let $T = n + \Delta{T}$ where $n \in \mathbb N$ and $\Delta{T} \in [0, 1)$

$$ f(x) = f(x+T) \\ x- {\lfloor}x{\rfloor} = x+T - {\lfloor}x+T{\rfloor} \\ x_0+\Delta{x} - {\lfloor}x_0+\Delta{x}{\rfloor} = x_0 + \Delta{x} + n +\Delta{T} - {\lfloor}x_0 + \Delta{x} + n + \Delta{T}{\rfloor} $$

So since $\Delta{x} \in [0;1)$ and $x_0 \in \mathbb Z$ then ${\lfloor}x_0+\Delta{x}{\rfloor} = x_0 +{\lfloor}\Delta{x}{\rfloor}$. Based on that LHS may be rewritten as:

$$ x_0+\Delta{x} - {\lfloor}x_0+\Delta{x}{\rfloor} = \Delta{x} - {\lfloor}\Delta{x}{\rfloor} = \Delta{x} $$

At the same time:

$$ x_0 + \Delta{x} + n +\Delta{T} - {\lfloor}x_0 + \Delta{x} + n + \Delta{T}{\rfloor} = \\ = (x_0 + n) + \Delta{x} + \Delta{T} - (x_0 + n) - {\lfloor}\Delta{x} + \Delta{T}{\rfloor} = \\ = \Delta{x} + \Delta{T} - {\lfloor}\Delta{x} + \Delta{T}{\rfloor} $$ That means that for LHS and RHS to be equal $\Delta{T}$ must be equal to $0$, hence $T=n+\Delta{T} \in \mathbb N$. And the smallest natural number is $1$, which gives that the function is indeed periodic and its minimal period is $1$.

Is the proof above valid?

  • 2
    $\begingroup$ It's valid, but overkill. $f(x)=f(y)$ iff they have the same fractional part, which happens iff they differ by an integer. Since the integer with smallest nonzero absolute value is 1, it follows that 1 is the minimal period. $\endgroup$ Jun 27, 2018 at 18:03
  • $\begingroup$ Before trying to prove it's periodic, just ask how in general this function behaves. After you answer that it will be clear what the period is. $\endgroup$ Jun 27, 2018 at 19:20
  • $\begingroup$ Too long. I didn't read it. $[x+n]=[x]+n;n\in\mathbb Z$ is easy to prove so $f(x+1)=(x+1)-[x+1]=(x+1)-[x]-1=x-[x]=f(x)$ so $f(x)$ has a period of $1$. If $k<1$; $n-1< x <n-k$ for some integer $n$ then $f(x)=x-n+1$ but $f(x+k)=x+k-n+1$. So $f$ does not have period $k$. So $1$ is minimum period. $\endgroup$
    – fleablood
    Jun 27, 2018 at 21:26

2 Answers 2


Note that $$f(x+T) = f(x) \iff x+T - \lfloor x+T\rfloor = x -\lfloor x\rfloor \iff T=\lfloor x+T\rfloor-\lfloor x\rfloor.$$ Since the RHS is the difference of two integers, $T$ must be an integer. Also, $T=\lfloor x+T\rfloor-\lfloor x\rfloor$ for all integers. As such, the minimal period is $T=1$.

  • $\begingroup$ This one is much more elegant than mine, thanks $\endgroup$
    – roman
    Jun 27, 2018 at 19:13
  • $\begingroup$ I do have to admire the eloquence. $\endgroup$
    – fleablood
    Jun 27, 2018 at 21:28


$x= \text{floor}(x) + \{x\}$ where ${x}$ denotes the fractional part function.

$\implies f(x)= \{x\}$

$\implies f(x+1)= \{x+1\}$

$\implies f(x+1)=\{x\} $

$\implies f(x)=f(x+1)$.

Thus, $f(x)$ is periodic with period $1$


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