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I've been asked to prove by induction that $n^2\leq 2^n$, and told it is true $ \forall n\in \mathbb{N},n>3$

I think I have found the right way to the proof, but I'm not sure since I get stuck half-way there. What I did was taking a base case of $n=4$ and tested it, and it resulted to be true. Then I assumed it would be true for some number $k$, such that $n=k$ and $k^2\leq 2^k$, and attempted to prove

$(k+1)^2 \leq 2^{k+1}$

And this is how I attempted to prove this. First of all, I started with my assumption.

$=k^2\leq 2^k$

$=2k^2\leq2^{k+1}$

Then I tried to prove that $(k+1)^2 \leq 2k^2$, for this would imply my thesis, i.e. $(k+1)^2\leq2^{k+1}$. So I went forth on my effort:

$(k+1)^2≤2k^2$

$=k^2+2k+1\leq2k^2$

$=2k+1\leq k^2$

(By assumption)

$=2k+1\leq 2^k$

Now that I simplified it, I need to prove this is true; this is, prove that $(k+1)^2 \leq 2k^2$. So I take a base case of $n=4$ and in deed it satisfies the inequality. So I assume it is true for some number $j, k=j$ and try to prove it. Nevertheless I have failed in trying to prove this, I don't really know if my steps so far are right or wrong. Is my reasoning okay? And if its, how can I prove $2k+1\leq 2^k$? Thank you in advance.

Martin Hansen
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lafinur
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  • This question has been asked so many times already. Please make an attempt to search for the problem. https://math.stackexchange.com/questions/319913/proof-that-n2-2n/319938#319938, https://math.stackexchange.com/questions/263825/proof-of-n2-leq-2n?noredirect=1&lq=1, https://math.stackexchange.com/questions/263825/proof-of-n2-leq-2n?noredirect=1&lq=1https://math.stackexchange.com/questions/496892/induction-proof-n2-2n-for-n-4?noredirect=1&lq=1 – JavaMan Mar 26 '19 at 17:30
  • https://math.stackexchange.com/questions/876127/prove-by-induction-that-2n-n2?noredirect=1&lq=1, https://math.stackexchange.com/questions/1489889/proof-by-induction-2n-ge-n2-for-n-ge4?noredirect=1&lq=1, https://math.stackexchange.com/questions/439026/prove-that-nk-2n-for-all-large-enough-n?noredirect=1&lq=1 – JavaMan Mar 26 '19 at 17:30
  • When you realize that you must prove $2k + 1<k^2$ !!!! *DONT* !!! go back to your assumption to prove $2k+1 < 2^k$. For one thing proving $2k + 1< 2^k$ won't prove $k^2 < 2k + 1 < 2^k$ is impossible, but more importantly, you got here via the assumption. Going back will be circular. You have gotten to the "practical world" where proving $2k+1< k^2$ is your direct way of proving your proposition. So prove it directly. Prove $2k + 1 < k^2$. That should be easy. – fleablood Mar 26 '19 at 17:37
  • ... for one thing $k^2 -2k - 1= k^2 - 2k + 1 - 2=(k-1)^2 - 2\ge 3^2 -2 = 7 > 0$ so $k^2 > 2k + 1$. – fleablood Mar 26 '19 at 17:43

1 Answers1

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You need to prove $2k + 1 \le k^2$.

Do it this way: $1 < k$ so $2k + 1 < 2k + k = 3k$. And $3 < k$ so $3k < k^2$.

So to put it together:

Induction step:

If $k^2 < 2^k; k > 3$ then

$(k+1)^2 = k^2 + 2k + 1 < $

$k^2 + 2k + k = k^2 + 3k < $

$k^2 + k*k = 2k^2 < $

$2*2^k = 2^{k+1}$.

.....

.... or simply note...

$2k + 1 < k^2 \iff$

$1 < k^2 - 2k \iff$

$2 < k^2 - 2k + 1 = (k-1)^2$.

And $k-1 \ge 3$ the $(k-1)^2 \ge 9 > 2$.

fleablood
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