I looked at a lot of questions and answers about this topic, and none seem to satisfy what I am looking for. Or it just was not obvious to me. Please help me understand.
In class, we partially proved that $\forall x\in R$ there exists a unique $n_0\in Z$ such that $n_0\le x< n_0+1$. We aim to show that $n_0$ is the floor function of $x$. I am having trouble finishing it, and understanding the intuition behind it. Here is how I understand the first part of the proof, and how I try to complete it:
Proof:
First Case: $x>0$
Let $A=\{n\in N;n>x\}\not=\emptyset$ (I assume we know this set exists because of the Archimedean Property which equivalently states that $\forall x>0\in R, \exists n\in N$ such that $n>x$. So the existence of such a "$n$" makes the set not empty)
This set is bounded below by $x$, so by the well-ordering principal of $N$, $min(N)$ exists. Let $m_0=min(A)$. So $m_0\in A$, so $m_0>x$, and $m_0-1\notin A$ as $m_0$ is the minimum of $A$. Since $\forall a\in A \implies a>x$ then $\forall a\notin A \implies a\le x$, so $m_0-1\le x$. Since $m_0-1$ is the biggest integer smaller or equal to $x$, $m_0-1$ is the floor of $x$. Do I understand the first case correctly?
Second Case: $x\le 0$
This is my proof. I assume for the first case we had to juggle around to use the Archimedean Property because it does not directly imply the existence of an $n_0<x$. However for $x\le 0$, we have by the equivalent of the Archimedean Property that $\forall x<0\in R, \exists n\in Z$ such that a $n<x$. So I just take $n$ to be the $max(n\in Z; n\le x)$. Is this correct?
Any help is greatly appreciated!
EDIT: I replaced $N$ with $Z$ in case 2
HINT.-For $x\lt0$ you have $$n_0\le-(x)\lt n_0+1\iff -(n_0+1)\lt x\le -n_0$$