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So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.

I start of with the "basecase" for n = 3:

RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$

Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$. To prove this inequality, we have to proof that it holds for n = k + 1.

$2^{k} > k^{2} - 2$

We multiply both sides with 2:

$2^{k+1} > 2(k^{2} - 2)$

$2^{k+1} > 2k^{2} - 4$

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

According to our assumption, $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $

Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:

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

Thus:

$2^{k+1} > (k+1)^{2} - 2$

QED.

How does my proof look, does it have any deficits, and how would you prove it?

Thank you!

Tanamas
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    "because $k^2-2\geq 2k+1$" This statement requires proof as well or citation. Overall, it looks mostly fine and what I would expect from a beginning student, but the presentation can be cleaned up and the details you handwaved could be made more rigorous. Recommended reading: How to write a clear induction proof? – JMoravitz Dec 20 '20 at 20:29

3 Answers3

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I think this is a little clearer.

$2^{k+1}\ge2k^2-4$

$=k^2+2k+1-2+k^2-2k-3$

$=(k+1)^2-2+(k-3)(k+1)$

$\ge(k+1)^2-2$ for $k\ge3$

Neil W
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The inductive step uses $2k^2-4\ge(k+1)^2-2$, or equivalently $(k-1)^2\ge4$, which holds for $k\ge3$. With quadratic inequalities, it's always wise to complete the square.

J.G.
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I think it is correct. For me, it is better to write (although simple) why we have $k^2-2\geq 2k-1$.

I write the proof as follows. We need to prove that $2^{k+1} > (k+1)^2-2$.

Indeed, (as you said) from $2^k > k^2-2$, we have $2^{k+1} > 2k^2-4$. We now prove that $2k^2-4\geq(k+1)^2-2$. After expanding and reducing, it leads to prove that $(k+1)(k-3)\geq 0$. This is correct for $k\geq 3$.

I present this way because I do not want to surprise the reader why we need statement $k^2-2\geq 2k-1$. I mean your statement "Because ..." is not natural in a mathematical proof.

TrungDung
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