Prove that $\lfloor x\rfloor \geq y$ if, and only if, $x\geq\lceil y\rceil$ I have some trouble proving that if $x,y\in\mathbb{R}$ then $\lfloor x\rfloor \geq y$ if, and only if, $x\geq\lceil y\rceil$.
I have tried some different approaches, the most recent being a proof by contradiction: Assume (for contradiction) that $x<\lceil y\rceil$, then $\lfloor x\rfloor < y +1$. However, I can not (of course) get rid of the $+1$ term because in general $y \leq\lceil y\rceil$.
One can perhaps prove that both the floor- and ceiling function preserve inequalities? If so, the result become trivial.
Any ideas are highly appreciated.
 A: Note that $ \lfloor x \rfloor \ge y$ implies $ \lfloor x \rfloor \ge \lceil y\rceil $, so $x\ge \lceil y\rceil$.
The other direction is similar.

If the above is not clear, recall that $\lceil y \rceil $ is defined as the least integer not smaller than $y$, i.e. if $C(y)$ is the set
$ C(y):= \{n\in\mathbb Z:n\ge y\}$,
then 
$$ \lceil y\rceil = \inf C(y).$$
The assumption $\lfloor x\rfloor \ge y $ is exactly that $\lfloor x\rfloor \in C(y)$. The only thing left is to use the defining property of an infimum,
$$c \in C(y) \implies \inf C(y) \le c.$$
A: Hint:
$$
\lfloor x\rfloor\geq y\iff \lfloor x\rfloor\geq\lceil y \rceil\iff x\geq \lceil y\rceil.
$$
A: We need to prove both direction, that is


*

*$\lfloor x\rfloor \geq y \implies x\geq\lceil y\rceil$
and


*

*$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq y $
For the first one we have that
$$\lfloor x\rfloor \geq y \implies \lfloor x\rfloor \geq \lceil y\rceil $$
and therefore
$$x\ge \lfloor x\rfloor \geq \lceil y\rceil  \implies x\geq\lceil y\rceil$$
For the second one we have that
$$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq \lceil y\rceil $$
and therefore
$$\lfloor x\rfloor \geq \lceil y\rceil \ge y  \implies \lfloor x\rfloor \ge y$$
A: We need the following elementary properties of the floor and ceiling functions: $a\ge\lfloor a\rfloor$, $\lceil\lfloor a\rfloor\rceil=\lfloor a\rfloor$, $a\ge b\implies\lceil a\rceil\ge\lceil b\rceil$, and $\lceil a\rceil=-\lfloor-a\rfloor$.
From the first three elementary propeties, we have, for any real $u$ and $v$,
$$\lfloor u\rfloor\ge v
\implies u\ge\lfloor u\rfloor=\lceil\lfloor u\rfloor\rceil\ge\lceil v\rceil$$
so letting $u=x$ and $v=y$ gives 
$$\lfloor x\rfloor\ge y\implies x\ge\lceil y\rceil\qquad(*)$$
while letting $u=-y$ and $v=-x$ gives
$$\lfloor-y\rfloor\ge-x\implies-y\ge\lceil-x\rceil$$
Invoking now the fourth elementary property turns that last implication into
$$-\lceil y\rceil\ge-x\implies-y\ge-\lfloor x\rfloor$$
which can be re-expressed as
$$x\ge\lceil y\rceil\implies\lfloor x\rfloor\ge y\qquad(**)$$
Putting $(*)$ and $(**)$ together, we have
$$\lfloor x\rfloor\ge y\iff x\ge\lceil y\rceil$$
