Proof ceiling function monotonocity? Is there any indication about the proof of the following statement:
$\forall x,y. x \le y \Rightarrow \left \lceil x \right \rceil \le \left \lceil y \right \rceil$
Yes it's $\le$. Sorry.
Thanks

We need only one counterexample to show that the conjecture is false. 
Take $x = 1.5$ and $y = 2$: $$1.5 \lt 2,\;\;\text{but} \;\; \lceil 1.5 \rceil = \lceil 2\rceil = 2$$
However, if you replace the second $\lt$-sign to $\leq$, then it is true that$$\forall x,y. x < y \Rightarrow \left \lceil x \right \rceil \leq \left \lceil y \right \rceil$$ This conjecture $(1)$ can easily be proven using a proof by contradiction: assume $x \lt y$ AND that $\left \lceil x \right \rceil \gt \left \lceil y \right \rceil,$ and you'll soon find a contradiction!.
If you are interested in whether the ceiling function  is monotonic: Monotonicity is still preserved by the ceiling function, given that it is non-decreasing.
