For all $x$ which are real numbers, prove that $$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$
I know that
Let $\lfloor x\rfloor = n$
$n \leq x < n+1$
Let $ x = i + d $ where $ 0 \le d < 1 $
$ \lfloor 2x \rfloor = \lfloor 2i + 2d \rfloor = 2i + \lfloor 2d \rfloor $
$ \lfloor x \rfloor + \lfloor x + 0.5 \rfloor = 2i + \lfloor d \rfloor + \lfloor d + 0.5 \rfloor $
So we reduced the problem to just show for $ d $
Suppose $ 0 \le d < 0.5 $, then both sides are just $ 2d $
Suppose $ 0.5 \le d < 1 $, then both sides are just $ d + 1 $.
QED $ \blacksquare $
I am assuming $\lfloor x\rfloor = n$
Let's take two cases - one - the fractional part of $x$ is less than $0.5$
LHS will be $2n$ RHS will be $n+n$
case two - The fractional part of $x$ is greater than $0.5$
LHS will be $2n+1$ RHS will be $n+n+1$
Hence, proved.
I will try to show $\lfloor nx \rfloor = \sum_{k=0}^{n-1} \lfloor x+\frac{k}{n} \rfloor$.
Let $m = \lfloor x \rfloor$ and $d = x - m$, so $0 \le d < 1$.
Let $j =\lfloor nd \rfloor $, so $0 \le j \le n-1 $ and $\frac{j}{n} \le d < \frac{j+1}{n} $. If $0 \le k \le n-j-1$,
$\begin{array}\\ m &\le m+d+\frac{k}{n}\\ &< m+\frac{n-j}{n}+\frac{j}{n}\\ &= m+\frac{n}{n}\\ &= m+1\\ \end{array} $
so $\lfloor m+d+\frac{k}{n} \rfloor =m $.
If $n-j \le k \le n-1$,
$\begin{array}\\ m+d+\frac{k}{n} &\ge m+\frac{n-j}{n}+\frac{j}{n}\\ &= m+\frac{n}{n}\\ &= m+1\\ \end{array} $
and
$\begin{array}\\ m+d+\frac{k}{n} &\lt m+\frac{n-j+1}{n}+\frac{n-1}{n}\\ &= m+\frac{2n-j}{n}\\ &= m+2-\frac{j}{n}\\ \end{array} $
so $\lfloor m+d+\frac{k}{n} \rfloor =m+1 $.
Therefore
$\begin{array}\\ \sum_{k=0}^{n-1} \lfloor x+\frac{k}{n} \rfloor &=\sum_{k=0}^{n-1} \lfloor m+d+\frac{k}{n} \rfloor\\ &= \sum_{k=0}^{n-j-1} \lfloor m+d+\frac{k}{n} \rfloor +\sum_{k=n-j}^{n-1} \lfloor m+d+\frac{k}{n} \rfloor\\ &= (n-j)m+j(m+1)\\ &= nm+j\\ \end{array} $
and $\lfloor nx \rfloor =\lfloor n(m+d) \rfloor =nm+\lfloor nd \rfloor =nm+j $.
We are done.
$n\leq x< n+1$ , then $f(x)=n$ for all $x$
I) If $n<x< \frac{2n+1}{2}$
$f(x)+f(x+0.5)= n+n=2n=f(2x)$
since $x+0.5<\frac{2n+1}{2} + \frac{1}{2}=n+1$
II) If $n+1>x\geq \frac{2n+1}{2}$
$f(x)+f(x+0.5)= n+n+1=2n+1=f(2x)$
Since $x+0.5\geq\frac{2n+1}{2} + \frac{1}{2}=n+1$
$\lfloor x\rfloor$
to show $\lfloor x\rfloor$. Formatting tips here. $\endgroup$