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
 A: 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.
A: As noted in the comments, the assertion is false.
What is true is that if $x\le y$, then $\lceil x\rceil\le\lceil y\rceil$. Suppose, on the contrary, that $\lceil x\rceil>\lceil y\rceil$. For convenience let $n=\lceil x\rceil$ and $m=\lceil y\rceil$. Then $n-1<x\le n$ and $m-1\le y<m$, by the definition of the ceiling function. But we also know that $n>m$, where $n$ and $m$ are integers, so $n-1\ge m$, and therefore $x>n-1\ge m>y$, contradicting the hypothesis that $x\le y$.
