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

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?

• 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... – AnotherJohnDoe Oct 8 '17 at 10:16
• @AnotherJohnDoe Okey. Thanks. – pozcukushimatostreet Oct 8 '17 at 10:59

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 $1/n < \varepsilon$.
• Why do we need to show that $1\gt 0$?? – AnotherJohnDoe Oct 8 '17 at 10:14
• 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. – benguin Oct 8 '17 at 10:19