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So I have already read this page with the solution:

For all $n>2$ there exists a prime number between $n$ and $ n!$

Now I was able to reason that $p < n!$ Because I was given the hint that $(n! - 1)$ has a prime divisor. But I am still a little hazy on the reasoning provided why $n < p$.

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D.C. the III
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2 Answers2

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A (prime) divisor of $n!-1$ can't be a divisor of $n!$ (if it were, it would divide $1$). Furthermore $n!$ is divisible by all of $1, 2, \ldots n$.

mrf
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  • Ok i think i see it. Are we kinda appealing to the same idea used in the "no largest prime proof", where since $(n! - 1)$ has a prime divisor it cannot be any of the $n! $ terms because that woukd produce a remainder so then it means our prime divisor has to be larger than n? – D.C. the III Aug 27 '15 at 20:50
  • @dc3rd Yes, that's very similar. – mrf Aug 27 '15 at 20:56
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From Bertrand's postulate , there exist always a prime number between $n$ and $2n$ but for $n>1$, $n!>2n$

Domenico Vuono
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