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The problem follows

"Using the Completeness Axiom for R, prove the Archimedian property of the real numbers : for any x in R, there is an integer n>0 such that n>x "

I tried to prove it in a reductio ad absurdumd. But I can't....

(I tried in this way) : for a fixed x in R, assume n<=x for all interger n. since left side diverges to infinity so, x cannot be fixed. therefore the assumption is not the case, there exists n s.t n>x.

I think my proof has some problems.. and I don't know how I can prove it using the Completeness Axiom.

Thanks you!

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3 Answers

I'm not sure what form of the completeness axiom you're used to, but I guess we could use here the following: if a subset of the real numbers has an upper [lower] bound then it has a least [greatest] upper [lower] bound (supremum) (infimum).

Thus suppose the claim is false, so

$$\exists\,x\in\Bbb R\;\;s.t.\;\;\forall\,n\in\Bbb N\;,\;n<x$$

By the above, let

$$x_0:=\sup\Bbb N\;\implies \forall\,\epsilon>0\;\exists\,n_\epsilon\in\Bbb N\,\,\;s.t.\,\,\;x_0-\epsilon<n_\epsilon\le x_0$$

This means $\,x_0\in\Bbb N\,$ (why?) , but then the successor of $\;x_0\;$ is not longer bounded by $\,x_0\;$ (I'm assuming here Peano's Axioms for the naturals)

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How do you know the left side diverges to infinity?

(one method is to assume it doesn't, then apply completeness to show that assumption is absurd!)

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Without loss of generality let $x > 0$. For the sake of contradiction, let $n\leqslant x\forall n\in\mathbb N$. Let$S=\{{n\over x}: n\in\mathbb N\}$. Since S is bounded above, by completeness of R, it has a least upper bound, say$\alpha$. Then, $\alpha -(1/x) < \alpha$, so $\exists n_0\in\mathbb N$ such that $\alpha -(1/x) \leqslant (n_0/x)\Rightarrow \alpha\leqslant(n_0+1)/x$, a contradiction. Hence we must have an $n\in\mathbb N$ such that $n>x$.

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