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Algebra question involving prime numbers

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Let the prime numbers, in order of magnitude, be p1, p2 ... Prove that pn+1 ≤ p1p2...pn + 1


    2. Relevant equations



    3. The attempt at a solution
    I have no idea how to start. I think it involves reductio ad absurdum.
     
    Last edited: Apr 18, 2008
  2. jcsd
  3. Apr 18, 2008 #2

    Dick

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    Look up Euclid's proof that there are an infinite number of primes. You can adapt it directly.
     
  4. Apr 18, 2008 #3
    Okay. So we wanna show that there are infinitely many prime numbers. Let's assume that there are only a finity amount of prime numbers which we'll label as p1, p2, ..., pn.

    Now let's consider some number, N, which is obtained by multiplying all prime numbers and adding 1:

    N = p1p2 ... pn + 1.

    N must have a prime factor. This prime factor must be one of p1, p2, ..., pn, because they are all the prime numbers. But this causes a contradiction since N leaves a remainder of 1 when divided by any of the prime numbers.

    Conclusion: there are infinitely many prime numbers.

    ----

    Now let's go to our problem: we want to show that pn+1 ≤ p1p2 ... pn + 1.

    Let's assume it is false, so: pn+1 ≥ p1p2 ... pn + 1.

    Now consider the number N ≥ p1p2 ... pn + 1 where N = pn+1.

    This number must have a prime factor, so we will have to divide by a prime number which is one of p1, p2, ..., pn.

    But we can't do this because this leaves a remainder of 1, which leads to a contradiction.


    Is it something like this? Sorry for the long post but this way I can summarise everything up for myself.
     
  5. Apr 18, 2008 #4

    Dick

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    You're off to a good start. But the end is a little muddled. Call M=(p1p2...pn)+1. M has a prime factor, call it p, that is NOT one of the p1...pn. Good so far. Since p=pj for some j, it must be that j>n. Ok so far? Since you are listing the primes in ascending order it follows that p=pj>=pn+1. Since p is a factor of M, p<=M. So pn+1<=p<=M.
     
  6. Apr 18, 2008 #5
    What is pj, a prime number? And why does it follow that j>n?
     
  7. Apr 18, 2008 #6

    tiny-tim

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    Hi Darkiekurdo! :smile:

    No … N = pn+1 is prime, isn't it (it's the next prime after pn)? :redface:

    You need to consider N - 1. :smile:
     
  8. Apr 18, 2008 #7

    Dick

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    Read what I wrote carefully. pj is a prime number that divides M. It's not one of p1...pn. So j>n.
     
  9. Apr 18, 2008 #8
    Ah I see it now. Thanks to the both of you!
     
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