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I need to find the answer to this equation: $2\lfloor x/2\rfloor =\lfloor x\rfloor $ Where $\lfloor\; \rfloor$ is the floor function.

I found that it's true for every $x$ who's $\lfloor x\rfloor$ is even, but I'm not sure how to write the answer –

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    $\begingroup$ What have you tried? Can you find some x for which it is true? Can you find some x for which it is false? $\endgroup$ – lulu Jul 17 '15 at 12:49
  • $\begingroup$ Help? Substitute in some typical numbers for $x$, experiment a little, see what happens. $\endgroup$ – Gerry Myerson Jul 17 '15 at 12:49
  • $\begingroup$ I found that it's true for every x who's [x] is even, but I'm not sure how to write the answer $\endgroup$ – Noam Dolovich Jul 17 '15 at 12:53
  • $\begingroup$ @NoamDolovich: What happens when $x$ is an odd integer? Are you only interested in integer solutions? $\endgroup$ – Thomas Jul 17 '15 at 12:55
  • $\begingroup$ No I'm interested in all real number solution, that's just something I observed and frankly I have no idea where to start. $\endgroup$ – Noam Dolovich Jul 17 '15 at 12:58
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You said in the comments:

I found that it's true for every x who's [x] is even, but I'm not sure how to write the answer

Well, $\lfloor x\rfloor$ is odd (let's get rid of this) means that $\exists k\in\mathbb{Z},\,\lfloor x\rfloor=2k+1$. But $2\left\lfloor \dfrac{x}{2}\right\rfloor$ is even since $\left\lfloor \dfrac{x}{2}\right\rfloor\in\mathbb{Z}$ so what can you conclude?

Now for the case where $\lfloor x\rfloor$ is even. So $\exists k\in\mathbb{Z},\,\lfloor x\rfloor=2k$. What interval does $x$ belongs to? (You already know some inequalities that compares $\lfloor x\rfloor$ and $x$). If you do that and see then how is $\left\lfloor \dfrac{x}{2}\right\rfloor$ in this case, you're going to be able to solve your problem.

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You're right, $\lfloor x \rfloor$ is even, since $\lfloor x \rfloor = 2 \left\lfloor \frac x2 \right\rfloor$ and $\left\lfloor \frac x2 \right\rfloor$ is an integer.

If you write the solution to $\lfloor x \rfloor = k$ as $x \in [k, k+1),$ then you might write the solution of $2 \left\lfloor \frac x2 \right\rfloor = \lfloor x \rfloor$ like this:

$$ x \in [2n, 2n+1) \quad \text{for some $n\in \mathbb Z$}$$

or even

$$ x \in \bigcup_{n\in \mathbb Z} \,[2n, 2n+1)$$ or $$ x \in \bigcup_{n\in \mathbb Z} \{x \mid 2n \leq x < 2n + 1\}.$$

The notation $\displaystyle\bigcup_{n\in \mathbb Z} S(n)$ means you have an infinite collection of sets $S(n)$, one such set for each integer $n$, and the result is the union of all those sets. Note that for any given value of $n$, $[2n, 2n+1)$ and $\{x \mid 2n \leq x < 2n + 1\}$ are just two ways of writing the same set.

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