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Is there a prime number less than the product of consecutive primes, but greater than the last consecutive prime?

It's been a while since I graduated college, and I haven't used my degree since, but I started thinking about this problem this morning:

Consider a product of consecutive primes

$\Pi_{i=N}^{M}p_i = X$

Is there always a prime between $p_M$ and $X$?

I feel like there should be an elementary proof of this, but I haven't done any real math in almost 6 years now.

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marked as duplicate by Matt Pressland, Cocopuffs, Matthew Conroy, Davide Giraudo, rschwieb Dec 12 '12 at 18:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
This is a more general version of that question. Over there, he has N = 1. –  Ted Dziuba Dec 12 '12 at 17:53
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@TedDziuba This is true, although it follows from Bertrand's postulate in exactly the same way. –  Cocopuffs Dec 12 '12 at 17:54
    
The text of that question differs from the formula (he says product of consecutive primes, but the example assumes the first one is $1$). I would agree that this question is better phrased, maybe the earlier one should be closed as a duplicate? Particularly if the author of the earlier question clarifies that he was allowing any initial prime as in the text. –  Matt Pressland Dec 12 '12 at 17:55

1 Answer 1

up vote 2 down vote accepted

Like Cocopuffs indicated, this is true from Bertrand's postulate.

  • The last prime in the product of a sequence of $M-N+1$ consecutive primes is $p_M$.
  • Bertrand's postulate guarantees a prime between $p_M$ and $2 \cdot p_M$.
  • $2 \cdot p_M < X$ since $2 \cdot p_M$ is clearly less than $\Pi_{i=N}^{M} p_i$ (except in pathological cases, e.g., $M = N$ or $p_N = 2$).
  • Therefore, there is a prime between $p_M$ and $X$.
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