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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.?)

Stefan Perko
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  • This looks very complicated. Can you not just define $q$ to be the minimum of ${z\in\mathbb{Z}:,x\leq z}$? A set of integers bounded from below always has the minimum (that follows easily from the property that any set of natural numbers has a minimum) and then the property $q-1<x\leq q$ follows immediately from this definition. – Peter Franek Dec 21 '14 at 18:06
  • Yes, I suppose with this I can skip the uniqueness part. Thanks! – Stefan Perko Dec 21 '14 at 18:40
  • I apologise for bringing up an old post, but there is a tiny issue with your proof (first paragraph), and that has to do with the case $x=0$. If $x=0$, $q=0$, but $q-1$ isn't a nonnegative integer, so you can't conclude that $q-1 \in A$ in your contradiction argument. You need to argue separately for $x=0$. Just a slight nitpick, but this error is in the spirit of "all horses are black" proofs due to lack of care with base cases in inductive/minimality proofs, so one might better be careful... – Maxis Jaisi Aug 16 '17 at 10:18

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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 $\lceil x \rceil = -\lfloor -x \rfloor.$

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David K
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