Prime Factorial Proof: Existence of a Prime Between n and n!

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    Factorial Prime Proof
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Discussion Overview

The discussion centers around the existence of a prime number p such that n < p < n! for integers n greater than 2. Participants explore various approaches to prove or disprove this statement, including references to Bertrand's postulate and the prime factors of n! - 1.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes the existence of a prime p between n and n! for n > 2 and invites proof or disproof.
  • Another suggests considering the prime factors of n! - 1 as a potential proof method.
  • Several participants reference Bertrand's postulate, with mixed feelings about its complexity in relation to the problem.
  • A participant describes a specific example using 5! to illustrate how the prime factors of n! - 1 relate to primes greater than n.
  • Another participant questions whether there are prime factors of n! - 1 that are less than n.
  • A later reply expresses satisfaction with the understanding gained from the discussion, indicating a resolution for that participant.

Areas of Agreement / Disagreement

There is no clear consensus on the proof or disproof of the statement. Multiple competing views and approaches remain, with some participants finding clarity while others express ongoing confusion.

Contextual Notes

Participants mention the complexity of applying Bertrand's postulate and the potential for simpler proofs involving n! - 1, but do not resolve these complexities or assumptions.

kingtaf
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Prove or disprove: If n is an integer and n > 2, then there exists a prime p such that
n < p < n!.
 
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Consider the prime factors of n! - 1
 
Bertrand's postulate, anyone?
 
CRGreathouse said:
Bertrand's postulate, anyone?

That's way over-kill.

Just consider the prime factors of n! - 1, that's a one-line proof for this problem
 
I considered Bertrand's Postulate but as hochs said it got messy.i still can't figure it out
 
kingtaf said:
I considered Bertrand's Postulate but as hochs said it got messy.i still can't figure it out

are there prime factors of n! - 1 that is less than n?
 
hochs said:
That's way over-kill.

Just consider the prime factors of n! - 1, that's a one-line proof for this problem

Here's one way to think about it, kingtaf...

Take, for example 5! = 1 * 2 * 3 * 4 * 5 = 120

120 (modulo 5) = 0
120 (modulo 4) = 0
120 (modulo 3) = 0
120 (modulo 2) = 0

Then, what does that make 119 with respect to those moduli? -1, right? Meaning that 2, 3 ,4 and 5 cannot divide 119 evenly. But, by the unique factorization theorem we know that 119 has prime divisors, either 119 if 119 is prime (it isn't) or some combination of primes, each of which is greater than n = 5.

In this case, the divisors of 119 are 7 and 17. 7 > 5. 17 > 5.

I'm using 5 here for demonstration purposes, but hopefully it is plain to see that it doesn't matter what n is. The same logic will hold.
 
Thank you everyone I got it.

After Hoch's hint using n! - 1, I figured it out. It's actually pretty simple. I feel really stupid!
 

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