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

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

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?

• As the Archimedean property gives you "... then there exists a natural number $n$ ..." you cannot put $n=2^m$. – Hagen von Eitzen May 16 at 14:41
• @Hagen I also doubted that. So, how should I proceed then? – user587389 May 16 at 14:46

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

• Ok Sir, I got it. Thank you. – user587389 May 16 at 14:53

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 $$m$$.

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} .

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