Thread: 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.

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

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

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.

I'm not entirely sure I understand.

Originally Posted by renolovexoxo

I'm not entirely sure I understand.

Read reply #2 again.
You have at most two more steps to a complete proof.