Is n! Always Greater Than n^2? An Induction Proof

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Homework Help Overview

The discussion revolves around proving the inequality n! > n^2 using mathematical induction, specifically for n ≥ 4. Participants are examining the validity of the statement and the steps involved in the proof process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of proving the base case for n = 4 and the inductive step for n + 1. There is confusion regarding the correct inequality to prove, with some participants mistakenly referencing n! > 2^n instead of n! > n^2. Questions arise about the transitions made in the proof steps and the validity of assumptions.

Discussion Status

The conversation is ongoing, with some participants providing insights into the induction process while others express confusion about specific steps. A participant claims to have resolved their issue, indicating some progress in understanding.

Contextual Notes

There is a noted discrepancy in the initial problem statement, as one participant points out that n! > n^2 is not true for all positive integers, prompting a re-evaluation of the problem's parameters.

ristin
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I have to do an induction proof that n!>n^2
 
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You have a very fundamental problem here: n!> n^2 is NOT true for all positive integers. Please recheck the statement of the problem.
 
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yes it is true when n>3, how do you prove it when n>3
 
Okay, that's a different problem. To use induction to prove that P(n) is true for all [itex]n\ge a[/itex] you must prove two things:
1) That P(a) is true.
2) That if P(k) is true, for some integer [itex]k\ge a[/itex] then P(k+1) is true.

Here, P(n) is "[itex]n!> n^2[/itex]" for [itex]n\ge 4[/itex].
(1) So P(a)= P(4) is "[itex]4!> 4^2[/itex]". 4!= 4(3)(2)(1)= 24 while [itex]4^2= 16[/itex]. Yes, 24> 16.

(2) Now assume that [itex]k!> 2^k[/itex] for some k> 3. Then
[tex](k+1)!= (k+1)(k!)> (k+1)2^k[/tex]
Now we need to show that [itex](k+1)(k^2)> (k+1)^2[/itex].

Since k+1> 0 we really only need to show that [itex]k^2> k+1[/itex]. Let's do a separate induction to prove that [itex]n^2> n+ 1[/itex] for n> 1. If n= 2, [itex]2^2= 4[/itex] while [itex]2+ 1= 3[/itex]. Yes, [itex]4> 3[/itex].

Now assume that [itex]k^2> k+1[/itex] for some k. Then
[tex](k+1)^2= k^2+ 2k+ 1> (k+1)+ 2k+ 1= 3k+ 2[/tex]
[tex]= (k+1+ 1)+ 2k> (k+1+ 1)[/tex].
That is, (k+1)^2> (k+1)+ 1 so the statement "[itex]n^2> n+1[/itex]" is true for all n larger than 1.

Now that we know [itex](k+1)k^2> (k+1)^2[/itex], it follows that [itex](k+1)!> (k+1)k^2> (k+1)^2[/itex] and we are done.
 
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yes i do know how to do them, this one is just stumping me, to how to prove it, i know my basis step is when n=4 and then we assume P(n) is true and prove that P(n+1) is also true. I multiplied both sides of my inequality by (n+1) to get n!(n+1)>n^2(n+1) which is equal to (n+1)!>n^2(n+1) then i got stuck
 
I need to prove n!>n^2 not 2^n i have done this proof before
 
do you see what i mean? that what you solved was n!>2^n where I need to solve n!>n^2?
 
on (2) when you went from (k+1)!= (k+1)(k!)> (k+1)2^k to (k+1)(k^2)> (k+1)^2 why did you get rid of the k!
 
how in (2) when you went from (k+1)!=(k+1)(k!)>(k+1)k^2 to (k+1)(k+2)>(k+1)^2
 
  • #10
Thank you for all of your help! I figured it out :)
 

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