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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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,Quote:
Originally Posted by amiv4
$\displaystyle 2^{k+1}>k+1$ ?
$\displaystyle (2)2^k>k+1$ ?
$\displaystyle 2^k+2^k>k+1$ ?
If $\displaystyle 2^k>k$ then $\displaystyle 2^k+2^k>k+1$
since $\displaystyle 2^k>1$
