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I'm trying to prove that $n^2 < 2^n$. I see this proof question from the example of 2.5.3 here.

I understand that when you prove an inductive case where $P(k+1)$ is true, you have first to assume the base case where $k^2 < 2^k$ is true.

Based on the base case assumption, you have to consider the case where $(k+1)^2 < 2^{k+1}$. In the linked article, it says, "To prove such an inequality, start with the left-hand side and work towards the right-hand side:"

And I can't understand the calculation flow. I understand $(k+1)^2$ to the second power is expanded, and you get $k^2 + 2k + 1$. And the article says by the inductive hypothesis, you get "${} < 2^k + 2k + 1$, which confuses me: I can't understand what is happening here. I can't understand why the $=$ sign changes into $<$. And the following flow makes me confused further: ${} < 2^k + 2^k$, which is derived from "since $2k + 1 < 2^k$ for $k \geq 5$," to which I wonder where $2k + 1$ came from. And the calculation flow in the article reaches "${} = 2^{k+1}$".

I am sure that my description here is unclear so I will attach the part of the article below. The explanation of the article says $(k+1)^2 < 2{k+1}$, which I can't follow. Could you please explain why and how you can prove that $P(k+1)$ is true?

Image of page from textbook.

Sammy Black
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Drftsgfhtt ur gfhh
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    You do know that $(k+1)^2 = k^2 + 2k + 1$, don't you? Becasue that is an equality you use an equal sign. Induction hypothesis is $k^2 < 2^k$. SO that means $k^2 + \text{Barbar the elephant} < 2^k + \text{Barbar the elephant}$. If instead of $\text{Barbar the elephant}$ we used $2k + 1$, we'd get $k^2 + 2k+1 < 2^k + 2k+1$. You aske where the $2k+1$ came from. It came from $(k+1)^2 = k^2 + \color{red}{2k+1}$. So we have $(k+1)^2 = (k+1)(k+1)=(k+1)k + (k+1)1 = k^2 +k + k + 1 = k^2 + 2k+1=k^2 +\color{red}{2k+1}< 2^k +\color{red}{2k+1}$ – fleablood Mar 13 '24 at 03:16
  • You mean $n^2<2^n?$ Hard to read your question – Thomas Andrews Mar 13 '24 at 03:18
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    I guess the second part is how do we know that $\color{red}{2k + 1} < \color{blue}{2^k}$. Well, $k \ge 5$ so ... actually, the link you post to just assumes that is obvious. – fleablood Mar 13 '24 at 03:21
  • Thank you for fixing the difficult-to-read parts of my post.
    Yes, I know that (k+1)² = k ² + 2k + 1.     – Drftsgfhtt ur gfhh Mar 13 '24 at 04:11
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    So we are assuming $\color{red}{k^2}<\color{red}{2^k}$ and we need to prove $(\color{red}k\color{green}{+1})^2<2^{\color{red}k\color{green}{+1}}$. And $(\color{red}k\color{green}{+1)}^2=\color{red}{k^2}+\color{green}{2k+1}<\color{red}{2^k}+\color{green}{2k+1}$. Now $\color{blue}2\color{orange}k+\color{purple}1<\color{blue}2\color{orange}k+\color{purple}k=\color{green}3\color{orange}k<\color{green}k\color{orange}k=\color{red}{k^2}$ and we know $\color{red}{k^2}<\color{red}{2^k}$. so $\color{green}{2k+1<2^k}$ and $\color{red}{k^2}+\color{green}{2k+1}<\color{red}{2^k}+\color{green}{2^k}=2^{k+1}$ – fleablood Mar 13 '24 at 04:28
  • I still find it difficult to understand how you can prove that p(k+1) is true. From (k+1)², you get k ² + 2k + 1. To me, suddenly, “< 2 ᴷ + 2k + 1” appears. Could you explain why you can show the inequality sign < and how to get “2 ᴷ + 2k + 1”? – Drftsgfhtt ur gfhh Mar 13 '24 at 04:32
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    You do understand that the induction hypothesis is nothing more nor less than that we are assuming we know that $k^2 < 2^k$, right. SO we KNOW that $k^2 < 2^k$. That means $k^2 + \text{anything we want} < 2^k + \text{anything we want}$, right? SO $k^2 + (2k + 1)< 2^k + (2k+1)$. – fleablood Mar 13 '24 at 04:34
  • I understand the meaning of the induction hypothesis at last. Thank you for the clarification. So, if the base case is true, then you can extend it further to developed cases: you can add the same value to both the left and right sides of the equation. – Drftsgfhtt ur gfhh Mar 13 '24 at 04:41
  • I now find it difficult to understand where “2 ᴷ + 2 ᴷ” comes from. fleablood explained the calculation flow, but I still don’t understand. Could you explain it? – Drftsgfhtt ur gfhh Mar 13 '24 at 04:48
  • Well, you know $2^{k+1} =2\times 2^k= 2^k + 2^k$. – fleablood Mar 13 '24 at 05:25
  • ... and you should know that $k^2 > 2k+1$ – fleablood Mar 13 '24 at 05:31
  • I understand how you can get 2 ᴷ + 2 ᴷ: because it is about an exponent of base 2, the + 1 of the exponent part k means the initial 2 ᴷ × 2. And the 2 is 2 ᴷ, so you get 2 ᴷ + 2 ᴷ. Thank you for explaining it clearly. – Drftsgfhtt ur gfhh Mar 13 '24 at 05:42
  • Thank you for showing me the relevant question in the past, peterwhy. I need more time to understand this. – Drftsgfhtt ur gfhh Mar 13 '24 at 05:56

1 Answers1

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The claim they are using is $2k+1 < 2^k$ for $k\geq 5$. And they assume it is clear. You can prove it with induction too. The induction step in that goes like (induction hypothesis: $2k+1<2^k$)

$$\begin{align} 2(k+1)+1 &= 2k+1 + 2 \\ &< 2^k + 2 \\ & \leq 2^k + 2^k \\ &= 2^{k+1} \end{align}$$

Here in the last inequality we used $2 \leq 2^k$, which is clear when $k$ is positive.

ploosu2
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  • Thank you for explaining, ploosu2. I feel like I understand, but I am not sure. The inductive case is set based on an arbitrary assumption: in this case, 2k + 1 < 2 ᴷ. But other than this, all goes over my head. I will think about it. – Drftsgfhtt ur gfhh Mar 13 '24 at 09:00