Why is $\{x/2\}$ always either $\{x\}/2$ or $\{x\}/2 + 1/2$?

Question

I was working on a problem involving floor functions and fractional parts. The question is to prove that $\lfloor x \rfloor - 2 \cdot \lfloor \frac{x}{2} \rfloor$ is either $0$ or $1$. In one of the solutions provided, the statement was made that 'the fractional part of $\frac{x}{2}$ is always either ${\frac{\{x\}}{2}}$ or ${\frac{\{x\}}{2}} + \frac{1}{2}$.' While I have verified this statement through examples, I am seeking a clear explanation, supported by examples, as to why ${\frac{\{x\}}{2}}$ is consistently 'always' either ${\frac{\{x\}}{2}}$ or ${\frac{\{x\}}{2}} + \frac{1}{2}$.

Answer 1

I find the original problem $\lfloor x\rfloor-2\lfloor x/2\rfloor\in\{0, 1\}$ more elementary than what you are asking. In fact, only by using inequalities:

$$\lfloor x\rfloor\le x< \lfloor x\rfloor+1$$ $$\lfloor x/2\rfloor\le x/2< \lfloor x/2\rfloor+1$$

you conclude

$$x-1<\lfloor x\rfloor\le x$$ $$x/2-1<\lfloor x/2\rfloor\le x/2$$

The second inequality (when multiplied by $-2$ and reversed) gives you this:

$$-x\le-2\lfloor x/2\rfloor<-x+2$$

Add to the first:

$$-1<\lfloor x\rfloor-2\lfloor x/2\rfloor<2$$

so as $\lfloor x\rfloor-2\lfloor x/2\rfloor$ is an integer strictly between $-1$ and $2$, it can only be $0$ or $1$.

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Now, if you want to answer your question directly, I suggest you distinguish the cases $\lfloor x\rfloor$ odd and $\lfloor x\rfloor$ even.

If $\lfloor x\rfloor=2n$ is even, this means $x=2n+\{x\}$. Then,

$$\frac{x}{2}=\underbrace{n}_\text{integer}+\underbrace{\frac{\{x\}}{2}}_{\text{between }0\text{ and }1}$$

and, because $0\le\{x\}<1$ you have $0\le \frac{\{x\}}{2}<\frac{1}{2}<1$ and so $\frac{\{x\}}{2}$ must be the fractional part of $\frac{x}{2}$. (Why? Because we have just written $\frac{x}{2}$ as a sum of an integer $n$ and something in the interval $[0,1)$.) So in that case $\left\{\frac{x}{2}\right\}=\frac{\{x\}}{2}$.

If $\lfloor x\rfloor=2n+1$ is odd, this means $x=2n+1+\{x\}$. Then:

$$\frac{x}{2}=\underbrace{n}_\text{integer}+\underbrace{\frac{1}{2}+\frac{\{x\}}{2}}_{\text{between }0\text{ and }1}$$

and, because $0\le\frac{\{x\}}{2}<\frac{1}{2}$ we also have that $0\le\frac{1}{2}\le\frac{\{x\}}{2}+\frac{1}{2}<1$, so, similar to the first case, $\frac{\{x\}}{2}+\frac{1}{2}$ must be the fractional part of $\frac{x}{2}$. So in that case $\left\{\frac{x}{2}\right\}=\frac{\{x\}}{2}+\frac{1}{2}$.