Prime factorial proof

  • Thread starter kingtaf
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  • #1
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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!.
 

Answers and Replies

  • #2
39
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Consider the prime factors of n! - 1
 
  • #3
CRGreathouse
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Bertrand's postulate, anyone?
 
  • #4
39
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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
 
  • #5
8
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I considered Bertrand's Postulate but as hochs said it got messy.i still cant figure it out
 
  • #6
39
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I considered Bertrand's Postulate but as hochs said it got messy.i still cant figure it out

are there prime factors of n! - 1 that is less than n?
 
  • #7
151
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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.
 
  • #8
8
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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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