This document provides a proof by contradiction for the statement "for all positive real numbers x, 2^x > x+1". It assumes 2^x < x+1 is true, and shows for x = 2 this leads to a contradiction (4 < 3), disproving the assumption. Therefore, the original statement, that 2^x > x+1 for all positive real x, must be true.

1. Provide either a proof or a counterexample for the following statements: For all positive real
numbers x, 2^x>x+1. (We have not covered proofs by induction, so is it possible to prove by
contradiction or contraposition?)
Solution
2^x > x+1 Let 2^x < x+1 For x = 2, 2^x = 2^2 = 4 x+1 = 2+1 = 3 2^x < x+1 => 4 <
3 ...Which is contradicting our assumption. So 2^x > x+1......hence proved.