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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 $n_{\tiny -}<a$.

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Every number has an additive inverse, so if there is infinitely large positive element, its inverse is an infinitely large negative.

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  • $\begingroup$ But that's not true for $\mathbb{N}$, right? $\endgroup$ – Billy Rubina Apr 15 '13 at 11:00
  • $\begingroup$ No, but the discussion is about $\Bbb R$. $\endgroup$ – Asaf Karagila Apr 15 '13 at 11:01
  • $\begingroup$ The part I'm stuck is when he says that there's a positive integer $n$. $\endgroup$ – Billy Rubina Apr 15 '13 at 11:02
  • $\begingroup$ Apparently I don't understand your question. $\endgroup$ – Asaf Karagila Apr 15 '13 at 11:03
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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Apr 15 '13 at 11:07

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