Why is Completeness needed to demonstrate the Archimedean principle? Why is Completeness needed to demonstrate the Archimedean principle?
Could someone criticize the following proof.
Thanks in advance.

Proof
Considering any $x \in \mathbb{R},$ $\lfloor{x}\rfloor \in \mathbb{N},$ and $$0 \leq x - \lfloor{x}\rfloor < 1$$
and 
$$x < \lfloor{x}\rfloor + 1.$$
Given that $\lfloor{x}\rfloor \in \mathbb{N},$ $\lfloor{x}\rfloor + 1 \in \mathbb{N}.$ Hence, $\exists n \in \mathbb{N}$ such that $x < n.$ 
 A: As for why completeness is needed? Here are two ordered fields that aren't complete, one with the Archimedean property and one without:
$\mathbb{Q}$ is an ordered field with the property.
Construct an order on $\mathbb{R}(x)$ (the field of rational functions over $\mathbb{R}$) as follows: a fraction $\frac{P(x)}{Q(x)}$ with $Q$ monic is positive when the leading coefficient of $P$ is positive. In terms of function values, a rational function $f$ is positive when there is some $M$ such that $f(x)>0$ for all $x>M$.
With this order, $\mathbb{R}(x)$ is an ordered field that doesn't have the Archimedean property. $x$ is an element that's greater than every integer, and $\frac1x$ is an element that's smaller than the reciprocal of every positive integer.
If we can construct examples without that completeness axiom in which the property doesn't hold, that means we need that axiom to prove the property.
A: The reals inherit the Archimedean property from the rational numbers.
Fix real $\varepsilon>0$. Then there exists a rational $\frac{p}{q}$ as close as we wish to $\varepsilon$. For example, we could choose one so that the following holds (Here we use that $\mathbb{R}$ is the completion of $\mathbb{Q}$).
$$0<\frac{p}{q}\le\varepsilon$$
Then define $N:=q$ and observe that by multiplying it on all sides we obtain
$$0<p\le \varepsilon N$$
Which is to say, for each real $\varepsilon>0$, there exists a natural number $N$ such that $1<\varepsilon N$. This is precisely the Archimedean property.
