induction inequality

**bmp05** · February 3rd 2010, 04:09 AM

$ (\forall n)(n \geq n_0) \wedge n^2 \leq 2^{n - k}, k \in \mathbb{N} $

Identify the smallest $n_0$ in relation to $k$ so that the proposition can be proved using mathematical induction (over n).

Fiddling around with the numbers $n_0 \geq 4$ is a least condition for the inequality, but how can I find out the relationship between k and n for all n?

Just as a beginning:
hypothesis:
$ n^2 \leq 2^{n - k} $
re-write the hypothesis:
$ n \leq 2^{\frac{1}{2}(n - k)} $
for $(n - 1)$ :
$ (n - 1)^2 \leq 2^{n - 1 - k} $
$ n^2 - 2n + 1 \leq 2^{n - 1 - k} $

but from this we see that:
$ n^2 < n^2 - 2n + 1 \wedge 2^{n - k - 1} < 2^{n - k} $

Is it possible to conclude that:
$ 2^{n - k} - 2^{n - k - 1} \leq 2n - 1 $
$ = 2^{n - k - 1} \leq 2n -1 $
which is just nonsense. I think? Can someone please help?!

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