# Math Help - Mathematical Induction Proof

1. ## Mathematical Induction Proof

Prove that for all the natural numbers n that 2^n > n

Base Case is easy
Then the inductive step you have 2^k > k as the inductive hypothesis
show that p[k+1] holds
2^(k+1) > k+1
on the left side 2^(k+1) = 2^k * 2
but idk what else to do

2. Originally Posted by amiv4
Prove that for all the natural numbers n that 2^n > n

Base Case is easy
Then the inductive step you have 2^k > k as the inductive hypothesis
show that p[k+1] holds
2^(k+1) > k+1
on the left side 2^(k+1) = 2^k * 2
but idk what else to do
Hi amiv4,

$2^{k+1}>k+1$ ?

$(2)2^k>k+1$ ?

$2^k+2^k>k+1$ ?

If $2^k>k$ then $2^k+2^k>k+1$

since $2^k>1$