Have to prove 2^n >= n^2 for n>=4
I.H.: Works when I plug in four. Show that 2^(n+1) >= (n+1)^2 or show that 2^n * 2 >= (n+1)^2
Since we know 2^n >= n^2 then 2^n * 2 >= 2n^2
So I then go about showing through another induction proof that 2n^2 >= (n+1)^2
...2(n+1)^2 >= (n+2)^2
2n^2 + 4n + 2 >= n^2 + 2n + 4
n^2 + 4n + 2 >= 2n + 4
n^2 + 2n + 2 >= 4
Since we know n must be >= 4 ...16 + 8 + 2 >= 4
So we add to what we know... 2^n >= 2n^2 >= (n+1)^2 therefore 2^n >= (n+1)^2.
Is this valid? Thanks.
This part of your argument makes no sense:
So I then go about showing through another induction proof that 2n^2 >= (n+1)^2
...2(n+1)^2 >= (n+2)^2
2n^2 + 4n + 2 >= n^2 + 2n + 4
n^2 + 4n + 2 >= 2n + 4
n^2 + 2n + 2 >= 4
Since we know n must be >= 4 ...16 + 8 + 2 >= 4
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In prove by induction, you cannot simply plug in numbers.