How to show $\Bbb R$ is Archimedean? Suppose $X$ is a real number such that $X > 0$. We want to show there exists and $n \in \mathbb{N}$ such that $X \geq \frac{1}{n} $.
MY attempt: If $X < \frac{1}{n} \; \; \; \forall n $ then $X \leq 0 $ by passing to the limit. Contradiction. Can someone show me a way to show this from scratch? By actually finding an $n$ that does this job? thanks
 A: The Archimedianity of $\Bbb R$ can be seen as a corollary of the unboundedness of $\Bbb N$ in $\Bbb R$ with the usual order. That is, pick $x\in\Bbb R_{>0}$. Then $x^{-1}>0$ is not an upper bound for $\Bbb N$, so there is $n\in\Bbb N$ such that $x^{-1}<n$. Inverting, gives $x>n^{-1}$.
A: The Archimedian Property of $\mathbb{R}$ can also be described as if $a,b \in \mathbb{R}$, $a,b > 0$, then $\exists n \in \mathbb{N}$ s.t. $na > b$.
Proof:
If $a>b$, let $n=1$. If $a=b$, let $n=2$. If $a<b$, then take the set $X = \{ na | n \in \mathbb{N}\}$. $X$ is nonempty as $a \in X$. Now suppose $X$ has an upper bound, call it $b$. We know from the Least Upper Bound property that any subset of $\mathbb{R}$ that is bounded above has a least upper bound. So let $L = \text{lub}(X)$. Then, since $a>0$, $\exists n_0a \in X$ s.t. $L-a < n_0a$. It follows that $L < (n_0 +1)a$, which is a contradiction.
It is a corollary that for any $\epsilon >0 \in \mathbb{R}$, $\exists n \in \mathbb{N}$ s.t. $\frac{1}{n} < \epsilon$.
A: The archimedan property simply states that there are no infinitesimals or infinite numbers in your system/group/set.
For example, you will always find a rational in between two numbers or a large natural after the maximum of your two numbers.
The set $\mathbb{R}$ is archimedan. ($\mathbb{Q}$ is dense in $\mathbb{R}$).
If the rationals can be embedded into a set such that it is dense within the set, it is archimedan.
