My Math Forum Mathematical Induction: Prove that n^2 < 2^n then n > 4

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 October 6th, 2017, 08:52 AM #1 Member   Joined: Sep 2014 From: Sweden Posts: 75 Thanks: 0 Mathematical Induction: Prove that n^2 < 2^n then n > 4 I am struggling with mathematical induction and have watched a lot of videos on youtube about it but still doesn't really get it. The first step is clear but the second isn't. According to a PDF-file from my highschool I'm supposed to follow these steps: $\displaystyle \displaystyle P(1) \ \ \ \ (\ P(5) \ in \ this \ case, \ i \ guess\ )\\\\ \displaystyle \forall n \in \mathbb{N} : P(n) \Rightarrow P(n+1)\\\\ \displaystyle \forall n \in \mathbb{N} : P(n)\\\\$ The assignment I have is: Show that $\displaystyle \displaystyle n^2 < 2^n$ then $\displaystyle n > 4$ (1) Assume that P(5) is true and prove that it is true $\displaystyle \displaystyle Assume \ that \ P(5) \ is \ true\\\\ \displaystyle P(5) = 5^2 < 2^5 = 25 < 32 \ true\\\\ \displaystyle P(n) \Rightarrow P(n+1)\\\\ \displaystyle P(n+1)=(n+1)^2<2^{n+1}\\\\ \displaystyle =(n+1)^2<2^n \cdot 2^1\\\\$ And from here I don't get any further . . . Last edited by DecoratorFawn82; October 6th, 2017 at 09:03 AM.
 October 6th, 2017, 09:31 AM #2 Global Moderator   Joined: May 2007 Posts: 6,436 Thanks: 562 $\displaystyle (n+1)^2=n^2+2n+1,\ n^2<2^n,\ 2n+1  October 6th, 2017, 09:47 AM #3 Member Joined: Sep 2014 From: Sweden Posts: 75 Thanks: 0 How can you know that$\displaystyle \displaystyle 2n+1 < n^2$? I get that$\displaystyle (n+1)^2 = n^2+2n+1$But the following lines below can't be equal? So where does the second line come from?$\displaystyle (n+1)^2<2^n \cdot 2^1 =\displaystyle 2n+1
 October 6th, 2017, 11:56 AM #4 Global Moderator   Joined: Dec 2006 Posts: 18,568 Thanks: 1483 Let $P(n)$ be the proposition that $n^2 < 2^n$ for $n > 4$, where $n \in \mathbb{N}$. As $5^2 < 2^5, \,P(5)$ is true. If $k > 4$ and $P(k)$ is true, $k^2 < 2^k$. Also, $2k + 1 < 2k + 3 + (k - 3)(k + 1) = k^2$, so $2k + 1 < 2^k$. Hence $(k + 1)^2 = k^2 + 2k + 1 < 2^k + 2^k = 2^{k+1}$, i.e. $P(k + 1)$ is true. By induction, $P(n)$ is always true. The proposition given in the title is slightly different, and untrue for n = 1. Thanks from greg1313 and DecoratorFawn82
 October 7th, 2017, 04:26 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,944 Thanks: 797 You have "The assignment I have is: Show that $\displaystyle n^2< 2^n$ then n> 4" I presume you mean "Show that $\displaystyle n^2< 2^n$ when n> 4".

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