Inequality Mathematical Induction

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Homework Help Overview

The discussion revolves around proving the inequality n! > 2^n for integer values of n, specifically for n ≥ 4, using mathematical induction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish the base case for n=4 and assumes the statement holds for n=k. They express uncertainty about the next steps in proving the case for k+1. Some participants question the reasoning behind multiplying by 2 and suggest focusing on the factorial expression (k+1)! instead. Others clarify that proving (k+1)! > 2^{k+1} is the goal, and they explore the implications of rewriting the inequality.

Discussion Status

Participants are actively engaging with the problem, offering hints and clarifications regarding the approach to take. There is a recognition of the need to prove (k+1)! > 2^{k+1}, and some guidance has been provided on how to manipulate the factorial expression. Multiple interpretations of the steps are being explored without a clear consensus on the next best move.

Contextual Notes

The discussion is constrained by the requirement to prove the inequality specifically for n ≥ 4, and participants are navigating the implications of this constraint in their reasoning.

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Homework Statement


Prove the Inequality for the indicated integer values of n.

[tex]n!>2^n,n\geq4[/tex]

Homework Equations



The Attempt at a Solution



For n=4 the formula is true because

[tex]4!>2^4[/tex]

Assume the n=k

[tex]k!>2^k[/tex]

Now I need to prove the equation for k+1

I can multiply both sides by 2 and have

[tex]2(k!)=2(k!)>(2)2^k=2^{k+1}[/tex]

Is this what you would do next? I'm not quite sure what to do past this point.

[tex]2(k!)>2^{k+1}>k+1[/tex]
 
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Why would you multiply by 2?
It's not 2k! you want to make a statement about, but (k + 1)! = (k + 1) k!
 
CompuChip said:
Why would you multiply by 2?
It's not 2k! you want to make a statement about, but (k + 1)! = (k + 1) k!

I don't know. When dealing with regular equations you would just set k=(k+1) but that's not the case here I think.

Wouldn't you be trying to prove that [tex](k+1)!>2^{k+1}[/tex]? Why would I be making a statement about '(k + 1)! = (k + 1) k!'? Where is the 2 raised to the power of (k+1) and why is there 2 factorials when there isn't in the original problem?
 
Sorry, you are approaching the problem from a different angle than I expected.

Once you have 2 (k!) > 2k + 1, you only need to prove that (k + 1)! > 2 k! to be done with it.

The hint I gave, that (k + 1)! is equal to (k + 1) times k!, is still useful.
 
CompuChip said:
Sorry, you are approaching the problem from a different angle than I expected.

Once you have 2 (k!) > 2k + 1, you only need to prove that (k + 1)! > 2 k! to be done with it.

The hint I gave, that (k + 1)! is equal to (k + 1) times k!, is still useful.

Ahh, so you could rewrite

(k + 1)! > 2 k!

as

k!(k+1) > 2 k!

Here you can see the when k > 1, k!(k+1) > 2 k! and since the original formula is only supposed to work for numbers greater than or equal to 4 its fine.

Thanks.
 
We are required to show that [tex]n!>2^{n}[/tex], for the k+1 term, multiply by k+1 to obtain:
[tex] (k+1)k!=(k+1)!>2^{k}(k+1)>2^{k+1}[/tex]
Since k+1>2 for k>2 and we are done
 

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