Question

Calculus:
If x > 0, show that there exists a n in the naturals such that 1/(2^n) < n.

My work will be included in the following comment.

Answer 1

Here is my attempt at a proof:
Proof by contradiction:
Let \[x \in \mathbb{R} \mbox{ s.t } x > 0\]. Assume there is not a \[n \in \mathbb{N} \mathbb{ s.t. } \frac{1}{2^{n}} < x\].
That is assume \[\forall n \in n, \frac{1}{2^{n}} \geq x \mbox{ for } x > 0\]
Pick any \[x \in \mathbb{R} > 0\]
Then, \[\frac{1}{2^{n}} \geq x \] for some n. This is a contradiction.
\[\therefore \exists n \in \mathbb{N} \mbox{ s.t. } \frac{1}{2^{n}} < x \]
\[\square\]

Is this looking accurate?

Answer 2

Do we need to Prove this: 1/2^n <x for x>0 and n>0

Answer 3

use whatever that axiom is called, the one that says there is no largest integer

Answer 4

for all \(x\) there is an \(n\) such chat \(2^n>\frac{1}{x}\)