# Does this definition of archimedean property guarantee that a set has no infinitely large or infinitely small elements?

I've just read the archimedean property:

1. Archimedean Property: If $a\in \mathbb{R}$, then there's a positive integer $n$ such that:

$$n>a$$

Remark: The archimedean property is sometimes expressed in the following equivalent way: for any positive real number $a$, there is a positive integer $n$ such that $\frac{1}{n}<a$.

And at first contact, I wasn't understanding the purpose of the concept, and then I searched wikipedia and found:

Roughly speaking, it is the property of having no infinitely large or infinitely small elements.

I guess that the first given property guarantees only that there are infinite positive real numbers, I'm thinking like: for all $a\in \mathbb{R}$, there is always a positive $n$ that is bigger than $a$ and independently of what real number is chosen, there will be always a bigger positive natural number but the property does not enforce that given $a\in \mathbb{R}$, there will always be a negative integer $n_{\tiny -}$ such that $n_{\tiny -}<a$.

I guess that it would be much better stated to say:

If $a\in \mathbb{R}$, then there's a positive integer $n_{\tiny +}$ and a negative integer $n_{\tiny -}$ such that $n_{\tiny +}>a$ and $n_{\tiny -}<a$.

• But that's not true for $\mathbb{N}$, right? – Billy Rubina Apr 15 '13 at 11:00
• No, but the discussion is about $\Bbb R$. – Asaf Karagila Apr 15 '13 at 11:01
• The part I'm stuck is when he says that there's a positive integer $n$. – Billy Rubina Apr 15 '13 at 11:02
• Gustavo, the definition is given for positive elements only. But if there were $a\in\Bbb R$ such that for all $k\in\Bbb Z$, $a<k$ then $-a$ would be a positive real number with the property that $k<-a$ for any positive integer $k$. Therefore it is enough to require the property for positive numbers. – Asaf Karagila Apr 15 '13 at 11:07