Archimedes property of real numbers Suppose $x \in \mathbb{R}$ such that $0 < x < 1 $ and $x \notin \{ \frac{1}{n} : n \in \mathbb{N} \} $. We want to show that there exists $n \geq 0$ such that 
$$ \frac{1}{n+1} < x < \frac{1}{n} $$
MY attempt:
Given real number $x,y$, then there exists $n$ such that $nx >y $. This is the archimidean property. To obtain our result, we can put $y = z $, $x = 1$and $n = \frac{1}{N}$, then $\frac{1}{N} > z $. Similarly, we can put $n = N +1 $, $x = 1$ and $y = 1$ and $x = z$, then $(1+N)z > 1 \implies z > \frac{1}{N+1} $. Therefore, we have obtained $N$ such that 
$$ \frac{1}{N} > z > \frac{1}{N+1} $$.
this is what we wanted.
is this correct?
 A: Like mathematics2x2life suggests the standard Archimedean property is that given any $x \in \Bbb R, \; \exists \ n_x \in \Bbb N$ such that $x \lt n_x$.
Through this we can find a natural number $n$ such that $\frac 1 n \lt x$. Therefore the set $A = \{n \in \Bbb N \ | \ \frac 1 n \lt x \}$ is non-empty. By the Well-Ordering principle this set has a least element say $m$.
Suppose  $x \ge \frac 1 {m-1} $. Then $x \gt \frac 1 {m-1} $ due to the stipulation that $x$ is not of the form $\frac 1 n $ for $n \in \Bbb N$ contradicting the fact that $m$ is the least element in $A$. 
$\implies x \lt \frac 1 {m-1}$. 
Here we have assumed that $m$ is not $1$. But $m =1 \implies x \gt 1$ leading to a contradiction..
A: Your proof does not work. For one, it is not true that given $x,y \in \mathbb{R}$, there is an $n$ such that $xn>y$. For example, take $x=-2$ and $y=3$. The rest of the proof is not clear. I am unsure if you are saying $z=y$ or $z=1$. But in either case you did not prove it for any $0<z<1$.
Finally, the Archimedean Property is that given any $x \in \mathbb{R}$, there is an $n\in \mathbb{N}$ such that $x \leq n$. What you are referring to is true only for positive reals and is typically stated as a corollary to the Archimedean Property. In any case, I shall give a hint how to prove your statement. 
HINT. Let $x \in \mathbb{R}$ such that $1<x$ and $x$ is not an integer. Show that there is an $n \in \mathbb{N}$ such that $n<x<n+1$ (use the Archimedean Property). Moreover, this integer $n$ is unique (not that uniqueness is needed). Then take the reciprocal in your inequality to obtain the result.
