Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 10 down vote accepted

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 $x>n^{-1}$.

share|cite|improve this answer
why cant we have $\geq $ ? – ILoveMath Oct 27 '13 at 23:24
If $x^{-1}\leqslant n$ then $x^{-1}<n+1$. Thus, there is no harm in putting $<$ directly. – Pedro Tamaroff Oct 27 '13 at 23:26
gotta wait $8$ minutes to accept the answer . sorry – ILoveMath Oct 27 '13 at 23:27
"Unboundedness of $\mathbb N$" is one way to say it, but really you're talking about a property of both $\mathbb N$ and $\mathbb R$. There are some other ordered sets containing $\mathbb N$ in which $\mathbb N$ ** is ** bounded. – Robert Israel Oct 27 '13 at 23:28
thanks for your time che and for your answer – ILoveMath Oct 27 '13 at 23:29

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. $na > b$.


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. $\frac{1}{n} < \epsilon$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.