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

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- April 19th 2010, 05:31 PMamiv4Mathematical 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 - April 19th 2010, 05:36 PMArchie Meade