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Archimedean Property. For any $x,\varepsilon\in\mathbb{R}$, $\varepsilon>0$, there is an $n\in\mathbb{N}$ such that $n\varepsilon>x$.

Corallary. For any $\varepsilon >0$, there is an $n\in\mathbb{N}$ such that $\dfrac {1} {n} < \varepsilon$.

Proof of Corollary. If we calculate this inequality, we obtain $n\varepsilon >1$. So, by the Archimedean property we have $n\varepsilon >1$. So, we are done.

Can you check my proof of corollary?

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    $\begingroup$ It's not clear, what it is you mean by calculate the inequality. You should also mention explicitly that for $x=1$, and so on... $\endgroup$ – AnotherJohnDoe Oct 8 '17 at 10:16
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    $\begingroup$ @AnotherJohnDoe Okey. Thanks. $\endgroup$ – pozcukushimatostreet Oct 8 '17 at 10:59
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Working backwards is a good technique to figure out how to get your proof to work out, but when presenting your proof, usually you want to present it in the "forward" direction. So something along the lines of,

Let $\varepsilon >0$. Since $\varepsilon > 0$, then by the Archimedian property, there exists an $n \in \mathbb{N}$ such that $n\varepsilon > 1$. From this, one sees that $1/n < \varepsilon$.

Note that you'll need to work through some of the details such as showing that the inequality you get from the Archimedian property is equivalent to the desired inequality. Doing this will depend on your set of axioms and theorems you have at your current disposal.

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  • $\begingroup$ Why do we need to show that $1\gt 0$?? $\endgroup$ – AnotherJohnDoe Oct 8 '17 at 10:14
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    $\begingroup$ My apologies...I'm up a bit later than usual and somehow ended up reading the statement of the Archimedian property as saying "$x, \varepsilon >0$". I've edited my answer. $\endgroup$ – benguin Oct 8 '17 at 10:19

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