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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.

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    $\begingroup$ Take $x=0.5$, then $[x]=0$, $[[x]+x]=[x]=0$, $[2x]=1$. $\endgroup$ – m-agag2016 Mar 9 '17 at 10:04
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    $\begingroup$ You can never prove by example! Only disprove! What is the meaning of $[x]$? Rounding? What kind of rounding? $\endgroup$ – M. Winter Mar 9 '17 at 10:04
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    $\begingroup$ What does $y$ have to do with it? You say "for all real numbers $x$ and $y$ but then $yR is never mentioned again. $\endgroup$ – bof Mar 9 '17 at 10:08
  • $\begingroup$ Thanks your for suggestions but I want to disprove it without using numebr values , how can I do it in that way ? $\endgroup$ – invictum Mar 9 '17 at 10:24
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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 $[[0.5] + 0.5] = [0 + 0.5] = 0 \neq [2(0.5)] = 1$.

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This statement is false. You can thus not prove it, only disprove it. You can disprove statements by example. I suggest trying $x=0.6$.

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  • $\begingroup$ 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, $[x] = \mathrm{floor}(x)$. $\endgroup$ – quantumkid Mar 9 '17 at 10:41
  • $\begingroup$ If you take $[x] = \mathrm{floor}(x)$ then $x = 0.6$ still works as a counterexample, since we have $[[0.6] + 0.6] = [0 + 0.6] = [0.6] = 0 \neq [2(0.6)] = [1.2] = 1$. $\endgroup$ – quantumkid Mar 9 '17 at 10:44
  • $\begingroup$ Because I somehow missed that part. Thanks for pointing it out, I fixed the reply. $\endgroup$ – Shinja Mar 9 '17 at 10:57

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