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I attempted to solve the proof but looking at another proof but I got stuck. Here's what I have so far.

Let $n = 1$, $10 > 1$. Assume valid for $k$ greater than or equal to $1$. Then $n = k + 1$.

$10^k > k^2$ .... Case(1) = $100 > 10$

And then what is after? Am I even doing the proof properly? I feel that I did a step incorrectly which led me to a dead-end.

EDIT: There was a hint in the back of my book, "$10n^2 = n^2 + 2n^2 + 7n^2$, now prove that $2n^2 \geq 2n$ and $7n^2 \geq 1$.

Alan Muniz
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Mark Park
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1 Answers1

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You have the right general concept, but you should be much clearer about your strategy. You want to show that if the statement holds for some $k \geq 1$, then it also holds for $k+1$. So assume that $10^k > k^2$ for some $k \geq 1$. Then, $10^{k+1} = 10^k \cdot 10 > 10k^2$. All you have left to do is check that $10k^2 > (k+1)^2$ for $k \geq 1$, and I'll leave this part to you.

JavaMan
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  • I see now! This was my first attempt at proof by induction and my question goes to show my lack of intuition. Thank you for the help JavaMan. – Mark Park Mar 09 '19 at 22:59