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Let $\epsilon\gt0$. Determine how large $n\in\mathbb{N}$ must be to ensure that the given inequality is satisfied, and use the Archimedean Property to establish that such n exist.

a. $\frac 1n\lt\epsilon$

b. $\frac{1}{n^2}\lt\epsilon$

c. $\frac{1}{\sqrt{n}}\lt\epsilon$

I'm not sure where to begin solving this problem. Help would be greatly appreciated.

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  • $\begingroup$ "the given inequality"? $\endgroup$ – TMM Apr 28 '17 at 16:13
  • $\begingroup$ Can you write down what does it mean by "$\Bbb{R}$ has the Archimedean property?" $\endgroup$ – Li Chun Min Apr 28 '17 at 16:37
  • $\begingroup$ You know that given any real number $x$ there is always $n\in\Bbb N$ such that $n>x$, right? $\endgroup$ – Juniven Apr 28 '17 at 16:53
  • $\begingroup$ You cannot answer that simple question? $\endgroup$ – Juniven Apr 28 '17 at 16:56
  • $\begingroup$ I figured my answer was too simple. If $\epsilon$ and M are both positive, then there exists n $\in$ $\Bbb{N}$ such that n$\epsilon$ $\gt$ M. If M=1, then n $\gt$ $\frac{1}{\epsilon}$. Because $\epsilon$ is too small, it can be approximated to be 0. Therefore n can be infinitely large as long as $\epsilon$ $\gt$ 0. But is the answer really that simple? $\endgroup$ – ErinA Apr 28 '17 at 17:15
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Turn them upside-down. If $b>0$ then $(0<a<b\iff 1/a>1/b).$

So $1/n<\epsilon \iff n>1/\epsilon.$

And $1/n^2<\epsilon \iff n^2>1/\epsilon>0 \iff n>1/\sqrt {\epsilon}.$

And $1/\sqrt n\;<\epsilon \iff \sqrt n\;>1/\epsilon>0 \iff n>1/\epsilon^2.$

If $x$ and $y$ are positive there exists $n\in \mathbb N$ with $nx>y.$ With $x=1,$ we see that any positive real $y$ is less than some $n\in \mathbb N.$ So in each of your 3 Q's, the existence of $n$ is assured.

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