I want to proof the following, with elementary properties of the integers and reals:
Let $x\in \mathbb{R}$. Then there are unique $p,q\in \mathbb{Z}$, such that:
$$p\leq x < p+1\text{ and }q-1<x\leq q$$
In that case, we have:
$$p = \max \{z \in \mathbb{Z} : z \leq x\}\text{ and } q = \min \{z\in \mathbb{Z} : x\leq z\}$$
This implies the existence of the floor and ceiling functions.
Finding some proof is not so hard (I suppose):
Let $x\geq 0$. Then by the archimedian property, the set $A:=\{z\in \mathbb{N}_0 : x\leq z\}$ is nonempty
and, by the well-ordering principle of the nonnegative integers, has a minimum $q\in \mathbb{N}_0$.
Then $x\leq q$. If $q-1\geq x$, then $q-1\in A$, but $q-1<q=\min A$ (Contradiction). Hence: $q-1<x$.Suppose $q'\in \mathbb{Z}$, such that $q'-1<x\leq q'$. Because $q'\in A$, we have $q\leq q'$. If $q<q'$,
then $q\leq q'-1$ and consequently $q'-1<x\leq q'-1$ (Contradiction). Therefore: $q=q'$.Let $x<0$. Then by the archimedian property, there is a $k\in \mathbb{N}_0$ with $-x\leq k$, that is $0\leq x+k$.
Then there is a unique $q'\in \mathbb{N}_0$ with $q'-1<x+k\leq q'$. We let $q:=q'-k$, then $q$ is the unique integer
with $q-1<x\leq q$.Now, the set $B:= \{z \in \mathbb{Z} : z \leq x\}$, by the previous argument, is nonempty and bounded above, therefore
it has a maximum $p$. …
However, I think my proof looks like complete overkill, since the only way I see, to prove the omitted second part (concerning the floor of $x$), is to basically repeat the same argument as above, with slight changes.
Is there an easier way? (Is my proof correct so far, btw.?)
Best Answer
For the floor function alone, you can refer to this question and its accepted answer. I regard your question as a different question, namely, once you have proved the existence of the floor, must you expend equal effort to prove the existence of the ceiling?
Instead, given any $x \in \mathbb R,$ use the proof of the floor function to show that there is a unique integer $-q$ such that $-q \leq -x < -q + 1.$ From this you can easily show that $\lceil x \rceil$ is uniquely defined for all $x.$ As a bonus, you will then have established that $\lceil x \rceil = -\lfloor -x \rfloor.$