# Thread: Proof of base 2 between 0 and E

1. ## Proof of base 2 between 0 and E

For E > 0, prove that there exists an N in Natural numbers such that
0 < 1/(2^N) < E:

I'm not really even sure how to start with this one. Proofs always get the better of me.

2. ## Re: Proof of base 2 between 0 and E

Originally Posted by renolovexoxo
For E > 0, prove that there exists an N in Natural numbers such that
0 < 1/(2^N) < E: I'm not really even sure how to start with this one. Proofs always get the better of me.
Where one starts depends on what one has already proved.
Do you know that $\displaystyle \forall c>0, ~\exists n\in\mathbb{N}[c<n]~?$

Do you know that $\displaystyle \forall n\in\mathbb{N}[n<2^n]~?$

Use $\displaystyle \frac{1}{E}=c>0$

3. ## Re: Proof of base 2 between 0 and E

All we've proved relating to this so far is that if x is in R, given any E>0 there exists a rational number r such that the |x-r|<E

4. ## Re: Proof of base 2 between 0 and E

Originally Posted by renolovexoxo
All we've proved relating to this so far is that if x is in R, given any E>0 there exists a rational number r such that the |x-r|<E
That is not going to help here.
But in order to have proven the above, you must have used the first property, Archimedean property
Induction easily proves the second property.
Put the two together.

5. ## Re: Proof of base 2 between 0 and E

I'm not entirely sure I understand.

6. ## Re: Proof of base 2 between 0 and E

Originally Posted by renolovexoxo
I'm not entirely sure I understand.