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## Homework Statement

Show that for any natural number K, there is an n large enough so that (2^n) > K.

## Homework Equations

## 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.