Prove that $0< \frac{1}{2^{m}} <y$

Question

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.

Answer 1

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, $$ y>\frac1{2^m}>0.$$

Answer 2

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 $m$.

Answer 3

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.