Proving an Inequality for n≥4: 2n<n! and 2n≤2((n-1)!)

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Homework Help Overview

The discussion revolves around proving inequalities involving factorials and powers of two for natural numbers n greater than or equal to 4. Specifically, the inequalities 2n < n! and 2n ≤ 2((n-1)!) are being examined, with an emphasis on using mathematical induction as a potential approach.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers using mathematical induction to prove the first inequality, questioning whether this proof can be extended to the second inequality. Some participants express uncertainty about the requirements of the exercise and whether the second inequality can be derived directly from the first.

Discussion Status

Participants are actively engaging with the problem, sharing thoughts on the validity of the inductive proof and its implications. There is a mix of interpretations regarding the exercise's expectations, and some guidance has been offered regarding the structure of the proof.

Contextual Notes

There is mention of the professor's usual allowance for assuming the truth of statements following an inductive proof, which raises questions about the specific requirements for this exercise. Additionally, the discussion includes a focus on the clarity of reasoning and the need for careful consideration of the inequalities involved.

silvermane
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Homework Statement


Prove that if n is a natural number and is greater/equal to 4, then 2n<n!,
and show that 2n is less/equal to 2((n-1)!) follows.

The Attempt at a Solution


I'm thinking I just need to use induction on n for the first part, where I get the inequality
(n+1)! = n!*(n+1) > 2n(n+1) > 2n*(2) =2n+1.

After this conclusion, can I just say that 2n is less/equal to 2((n-1!) follows from my proof of induction or is there something else that I need to do?

Thanks in advance for your help! No answers please :)
 
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Yes, you can say that this follows from your induction proof.

But I think that the point of your exercise is to show that 2^n\leq 2(n-1)!, while using only that 2^n&lt;n!. So while saying that this follows from your induction is correct, I don't think that your exercise wants you to do that :smile: Just my interpretation, I could be wrong tho...
 
micromass said:
Yes, you can say that this follows from your induction proof.

But I think that the point of your exercise is to show that 2^n\leq 2(n-1)!, while using only that 2^n&lt;n!. So while saying that this follows from your induction is correct, I don't think that your exercise wants you to do that :smile: Just my interpretation, I could be wrong tho...

Yes, this isn't what I was too sure of, though the professor usually let's us just assume it true after our initial induction proof. If I need to, how should I start? :)
(it's late and I can't even think clearly.)
Thanks so much for your help and advice! :blushing:
 
Try this:

2^n=2.2^{n-1}\leq 2(n-1)!
 
I think I figured it out! :blushing:

From my inductive proof, I showed that 2n is less than n! or values of n greater/equal to 4.
So by my inductive proof, I can say that 2n-1 is less/equal to (n-1)! for values of n greater/equal to 5. (which is bold because I want to make sure that statement is correct)

It follows that if I multiply both sides by 2, we have that
2n is less/equal to 2(n-1)! for values of n greater/equal to 5.

That was so cool!
 

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