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If $y$ be a positive real number, show that there exists a natural number $m$ such that $0< \frac{1}{2^{m}} <y$

I think I have to use Archimedean property to prove it. The Archimedean property is, if $x$ is a real number and $y$ is a positive real number then there exists a natural number $n$ such that $ny > x$. So, shall I just put $x=1$ and $n=2^m$? Or is there any other method to prove the above statement?

Please anyone help me solve this problem. Thanks in advance.

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    $\begingroup$ As the Archimedean property gives you "... then there exists a natural number $n$ ..." you cannot put $n=2^m$. $\endgroup$ – Hagen von Eitzen May 16 at 14:41
  • $\begingroup$ @Hagen I also doubted that. So, how should I proceed then? $\endgroup$ – user587389 May 16 at 14:46
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From the Archidean propertry for $x=1$ and $y$ as given, there exists $n\in \Bbb N$ such that $ny>1$. You may already know that $2^n>n$ for all $n\in\Bbb N$. Hence by letting $m=n$, we obtain $2^m-n>0$ and after multiplication with the positive $y$, $2^my-ny>0$, or after rearranging, $2^my>ny>1$. As $2^m>0$, we also have $\frac1{2^m}>0$ and after multiplication with this, $$ y>\frac1{2^m}>0.$$

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  • $\begingroup$ Ok Sir, I got it. Thank you. $\endgroup$ – user587389 May 16 at 14:53
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You can apply $\log$ in order to get into a situation where you can use Archimedes in a clean fashion: $$ 2^m>\frac 1y\implies m\log 2>\log(1/y) $$ and then apply the principle to find $m$.

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Option:

$m \in \mathbb{N}.$

$2^m =(1+1)^m \ge 1+m;$

$\dfrac{1}{2^m} \le \dfrac{1}{1+m};$

$y>0$, real.

Archimedean principle:

There is a $m > 1/y$ , $m$ positive Integer.

Then $m+1 >m> 1/y$, and $y>\dfrac{1}{m+1}$

$0<\dfrac {1}{2^m} \le \dfrac{1}{m+1} <y$.

Note:In your proof setting $n=2^m$ needs a bit of clarification.

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