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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?

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

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    $\begingroup$ Or don't use logarithm and prove $2^n>n$. $\endgroup$ – Jonas Meyer Sep 25 '13 at 4:20
  • $\begingroup$ Kind of think you had it right with "Property of natural numbers" proofwiki.org/wiki/Archimedean_Principle $\endgroup$ – user66360 Sep 25 '13 at 4:24
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Twink Sep 25 '13 at 4:26
  • $\begingroup$ I think both names are accepted. $\endgroup$ – Twink Sep 25 '13 at 4:28
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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 $\forall x \in \mathbb{R}, \, \lceil x \rceil \ge x$.

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