# Show that for every real number x, there exists a natural number n such that $x < 2^n$

Let x be a positive real number. I want to prove that $\forall$ x, $\exists$ n $\in$ N such that x < $2^n$ .

To me it seems that as x increases, I can just pick larger and larger values for n to satisfy this property. Since n goes to infinity, I should be able to do this process forever. Any idea how to prove this?

HINT: Use logarithm and the Archimedean Property of the real numbers.

• Or don't use logarithm and prove $2^n>n$. – Jonas Meyer Sep 25 '13 at 4:20
• Kind of think you had it right with "Property of natural numbers" proofwiki.org/wiki/Archimedean_Principle – user66360 Sep 25 '13 at 4:24
• en.wikipedia.org/wiki/… – Twink Sep 25 '13 at 4:26
• I think both names are accepted. – Twink Sep 25 '13 at 4:28

Hint:

First, restrict $x$ to be a natural number, and prove $\forall x\, \exists n \in \mathbb{N},\, x < 2^n$ by doing "assume that there is a natural number $x$ which does not satisfy above proposition..."

Then, use $\forall x \in \mathbb{R}, \, \lceil x \rceil \ge x$.