What is the clearest way of proving of $[x]+[x+1/2]=[2x]$?

Question

What is the clearest and simplest way of proving that $[x]+[x+1/2]=[2x]$? (Where $[x]$ is the greatest integer function)

According to Bartleby, if $x=m$ for $m\in \mathbb{Z}$, then $[x]=m$ and $[x+1/2]=m$. Hence, $[x]+[x+1/2]=2m=[2x]$. Hence when $m\in\mathbb{Z}$, $[x]+[x+1/2]=[2x]$.

When $m<x<m+1$, $[x]=[m+\{x\}]=m+[\{x\}]$ and $[x+1/2]=[m+\{x\}+1/2]=m+[\{x\}+1/2]$. Hence $[x]+[x+1/2]=2m+[\{x\}]+[\{x\}+1/2]$. Since $0 \le \{x\} < 1$ and $1/2 \le \{x\}+1/2 < 3/2$; we have $0 \le [\{x\}] < 1$ and $0\le[\{x\}+1/2]< 1$. Hence $2m < 2m+[\{x\}]+[\{x\}+1/2]\le 2m+1$ which implies $[2x]$. Hence $[x]+[x+1/2]=[2x]$.

Hence $[x]+[x+1/2]=[2x]$

Is my solution clear or is there a better one?

Answer 1

My favourite, and the cleanest way in my oppinion is the following classical solution:

Let $f(x)= \lfloor x\rfloor+\lfloor x+\frac{1}{2}\rfloor-\lfloor 2x \rfloor$.

Then, it is trivial to see that

• $f(x+\frac{1}{2})=f(x)$.
• $f(x)=0 \forall x \in [0, \frac{1}{2})$.

From here it follows immediatelly that $f \equiv 0$.

P.S. By using $f(x)= \lfloor x\rfloor+\lfloor x+\frac{1}{n}\rfloor+ \ldots +\lfloor x+\frac{n-1}{n}\rfloor-\lfloor nx \rfloor$ you can deduce the same way that $$\lfloor x\rfloor+\lfloor x+\frac{1}{n}\rfloor+ \ldots +\lfloor x+\frac{n-1}{n}\rfloor=\lfloor nx \rfloor$$

Answer 2

For $x > 0$:

$[2x]$ is the number of positive integers $\leq 2x$. Count them by separating the even and odd numbers.

For general $x$: add a sufficiently large integer to it.