Can the ceiling function be used to prove the Archimedean property? Recall the following definition of the Archimedean property:

For each $x \in \Bbb{R}$, there exists some $n \in \Bbb{N}$ such that $n>x$.

My textbook proves this by invoking the completeness axiom. My first instinct however was to use the ceiling function:

Proof: Choose any $x \in \Bbb{R}$. Then consider $n=1+\lceil x \rceil$. By definition, $n>x$, as desired.

This seems like cheating though. Does the ceiling function implicitly depend on the completeness axiom? Is my proof circular?
 A: The ceiling function depends on the Archimedean property.
$$\lceil x \rceil = \min \lbrace n \in \mathbb{Z} \colon n \geqslant x \rbrace$$
If the Archimedean property didn't hold, you'd take the minimum of an empty set for some $x$.
Hence your proof is circular.
A: The ceiling function does not depend on completeness: its restriction to $\Bbb Q$ is well-defined, and the rationals are not complete. It does not, strictly speaking, depend on the Archimedean property either: it is perfectly possible to define $\lceil x\rceil$ to be $\min\{n\in\Bbb Z:x\le n\}$ when that minimum exists, and then to ask what the domain of this ceiling function is. It’s that domain that depends on the Archimedean property: the Archimedean property is equivalent to the statement that the domain of the ceiling function is $\Bbb R$.
Since your argument depends on the fact that the domain of the ceiling function is all of $\Bbb R$, and that fact is equivalent to the fact that $\Bbb R$ has the Archimedean property, your argument is indeed circular.
