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

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    Factorial Prime Proof
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SUMMARY

The discussion centers on proving the existence of a prime number \( p \) such that \( n < p < n! \) for any integer \( n > 2 \). Participants highlight the utility of examining the prime factors of \( n! - 1 \) as a straightforward proof method. The example of \( 5! = 120 \) illustrates that \( 119 \) has prime factors \( 7 \) and \( 17 \), both greater than \( 5 \). This reasoning confirms that the logic applies universally for any integer \( n \).

PREREQUISITES
  • Understanding of factorial notation and properties, specifically \( n! \)
  • Familiarity with prime numbers and their properties
  • Knowledge of modular arithmetic
  • Concept of unique factorization theorem
NEXT STEPS
  • Study Bertrand's Postulate and its implications in number theory
  • Explore modular arithmetic applications in prime factorization
  • Learn about the unique factorization theorem in detail
  • Investigate other proofs of the existence of primes between \( n \) and \( n! \)
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Mathematicians, number theorists, and students interested in prime number theory and factorial properties.

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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