Help with equation $2\lfloor x/2\rfloor =\lfloor x\rfloor $ 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 –
 A: 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.
A: 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.
