1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof: Natural Number

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that for any natural number K, there is an n large enough so that (2^n) > K.


    2. Relevant equations



    3. The attempt at a solution
    K is a natural number -> the smallest possible K would be 0 (lower bound?) and the smallest 2^n is 1 when n = 0, and the upper bound for both sides are infinite. So if I set n=0 and K =0 I get 2^n > K. I'm not sure if this is the right approach.
     
  2. jcsd
  3. Apr 15, 2009 #2
    Take some number K, find the possible n's that satisfy 2^n>K. Ie. solve for n.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook