why $N>\frac{1}{\epsilon}$? i am asking here the most simple and dumbest question ever. once i was reading about convergence yesterday night, i came to the notion which beats me over times: The archimedean axiom
the text was about the convergence of the sequence $\frac{1}{n}$. 
The Sequence $\frac{1}{n}$ is a zero-sequence: Proof: There is $\epsilon>0$. According to AA, there is $N\in\Bbb{N}$ with $N>\frac{1}{\epsilon}$. Then $|\frac{1}{n}-0|=\frac{1}{n}<\epsilon$ for all $n\ge N$
what i dont understand is: why AA and why $N>\frac{1}{\epsilon}$? what does that mean in words? i imagined in a line saying that $\epsilon=2$ then $\frac{1}{\epsilon}$ is the 0.5 part of that line. So the $N$ should be greater than that part of $\epsilon$-neighborhood, why?   
 A: Archimedes Axiom says that you can always find a "big enough" $N$ to beat any real number you are given - the natural numbers are unbounded in the reals.
Basically you imagine you are given an $\epsilon$ to beat - normally $\epsilon$ is thought of as small and positive and you need to get below it.
So if we want $\frac 1 n < \epsilon$ we are dealing with small numbers. To use Archimedes in the form you have it we need to be dealing with large ones, so we convert our target to $n > \frac 1 {\epsilon}$.
$n$ is a natural number, $\frac 1 {\epsilon}$ is a real number, and we have got your problem in a shape where Archimedes can be used.
A: The Archimediaan axiom states that for any real number $x$ there's some $N\in \mathbb{N}$ for which $x<N$.  So let $\epsilon$ be some tiny number, as close to $0$ as you want (without being actually equal to $0$).  Then $1/\epsilon$ is some big number.  By $\sf AA$ there is a $N$ so that $1/\epsilon < N$.  So $1/N < \epsilon$.  Thus there is no smallest number in the sequence $1/n$ for $n\in \mathbb{N}$.
A: There are several equivalent ways to axiomatize the Archimedean property of $\Bbb R$; from what you’ve written it sounds like you’re working with the version that says that for any real number $x$ there is a positive integer $n$ such that $n>x$. Intuitively this axiom just says that there is no real number that is larger than all of the integers: no matter how far out in $\Bbb R$ you go, you’ll still find integers.
In the proof that you’re reading, you have a positive real number $\epsilon$, and you want to show that there is some $N\in\Bbb N$ such that $\frac1n<\epsilon$ for all $n\ge N$. (This is in order to show that $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ is a zero-sequence.) The Archimedean property doesn’t say anything directly about small numbers, but recall that for positive real numbers $a$ and $b$, $a<b$ if and only if $\frac1a>\frac1b$, so we can take reciprocals and deal with large numbers. Since $\epsilon>0$, $\frac1\epsilon>0$ as well, and the Archimedean property ensures that there is a positive integer $N>\frac1\epsilon$. By elementary algebra this implies that $\frac1N<\epsilon$, and a little more algebra shows that $\frac1n\le\frac1N<\epsilon$ for all $n\ge N$. That’s exactly what’s needed to show that $\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$ is a zero-sequence.
A: It slightly depends on your statement of the Archimedean Axiom.  One is

Let $x$ be any real number. Then there exists a natural number $n$
  such that $n$ is greater than $x$.

So if $\epsilon$ is a positive real number then so too is $\frac{1}{\epsilon}$. 
By AA there is a natural number $N$ such that $N$ is greater than $\frac{1}{\epsilon}$. And for any natural number $n$ greater than $N$, $n$ is also greater than $\frac{1}{\epsilon}$ or in symbols $n \gt \frac{1}{\epsilon}$.
For all such $n$ greater than $\frac{1}{\epsilon}$, you have $\frac{1}{n}$ is less than $\epsilon$, and so $\frac{1}{n}$ is closer to zero than $\epsilon$ is, which is what you wanted to prove.     
