Let $x>0$ and let $n_0$ be the largest integer such that $n_0\leq x$...(Help proving the existence of $n_0$) I'm reading Rudin's book (PMA) and it says Let $x>0$ and let $n_0$ be the largest integer such that $n_0\leq x$ (Note that the existence of $n_0$ depends on the arquimedean property of $\Bbb R$). My question is how can I prove that using the arquimedean property of $\Bbb R$?
 A: Given $x>0$, by the Archimedean property there is a positive integer $n$ with $x < n$. Let $A$ be the set of all positive integers $n$ with $x<n$. Since $A$ is a nonempty subset of the natural numbers, and the natural numbers are well-ordered, $A$ has a smallest element, say $m_0$. Then $n_0 = m_0 - 1$ must satisfy $n_0 \leq x$, and it is also the largest integer with $n_0 \leq x$ because any integer larger than $n_0$ must lie in $A$.
In short, the Archimedean property guarantees that the set $A$ is nonempty.
A: Split two cases. $ x\le 1 $ and $x \gt 1$. First case is obvious. The second one is disposed as follows.  
Note that given $x \in \Bbb R$ by the Archimedian Property there exists $n \in \Bbb N$ such that $ n \gt x $. Then the set $ A =  \{ n \in \Bbb N \ | \ n \gt x\} $ is non-empty. 
Then by the Well-Ordering Principle $A $ has a least element $ n' $. |et $ n_{\circ} = n' - 1 $
Now prove that $n_{\circ}$ is indeed the smallest integer such that $n_{\circ} \le x$. Assume not and reach a contradiction.  
