Prove or disprove $[[x] + x] = [2x]$ for all real numbers $x$ and $y$

I tried to prove that through giving a value which is floating-point but I have to solve this using letters like the proving that if $n^2$ is odd , then $n$ is odd when I say $n=2k+1$.

The question is, prove or disprove: For all real numbers $x$ and $y$, $[[x] + x] = [2x]$ where $[x]$ is the greatest integer less than or equal to $x$.

Thank you for response.

• Take $x=0.5$, then $[x]=0$, $[[x]+x]=[x]=0$, $[2x]=1$. – m-agag2016 Mar 9 '17 at 10:04
• You can never prove by example! Only disprove! What is the meaning of $[x]$? Rounding? What kind of rounding? – M. Winter Mar 9 '17 at 10:04
• What does $y$ have to do with it? You say "for all real numbers $x$ and $y$ but then $yR is never mentioned again. – bof Mar 9 '17 at 10:08 • Thanks your for suggestions but I want to disprove it without using numebr values , how can I do it in that way ? – invictum Mar 9 '17 at 10:24 2 Answers You can see how to construct a counter-example as follows. Set$x = n + \epsilon$,$n \in \mathbb{Z}$,$\epsilon \in [0, 1)$. Then we have$[x] = [n + \epsilon] = n$. Looking at the L.H.S. we have $$[[x] + x] = [2n + \epsilon] = 2n\ .$$ On the R.H.S. we have $$[2x] = [2n + 2\epsilon] = \begin{cases} 2n\ &\quad 0 < \epsilon < 0.5 \\ 2n + 1\ &\quad 0.5 \geq \epsilon < 1 \end{cases}$$ So the original equality only holds if$\epsilon < 0.5$, so choosing any$\epsilon \geq 0.5$will break the equality. If we try$\epsilon = 0.5$, we see immediately that$[[0.5] + 0.5] = [0 + 0.5] = 0 \neq [2(0.5)] = 1$. This statement is false. You can thus not prove it, only disprove it. You can disprove statements by example. I suggest trying$x=0.6$. • Why would you assume$[x]$means rounded to the nearest integer when it says in the question$[x]$is the greatest integer less than or equal to$x$? In other words,$[x] = \mathrm{floor}(x)$. – quantumkid Mar 9 '17 at 10:41 • If you take$[x] = \mathrm{floor}(x)$then$x = 0.6$still works as a counterexample, since we have$[[0.6] + 0.6] = [0 + 0.6] = [0.6] = 0 \neq [2(0.6)] = [1.2] = 1\$. – quantumkid Mar 9 '17 at 10:44
• Because I somehow missed that part. Thanks for pointing it out, I fixed the reply. – Shinja Mar 9 '17 at 10:57