Prove that if $ n > 2 $ then between $n$ and $n!$ is at least one prime.
Ok I can see that it's obviously true, but what to use to prove it?
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2how can it be obviously true if you do not know how to prove it?! – Mariano Suárez-Álvarez Feb 14 '15 at 16:45
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You could also use this and note that $2n \leq n!$ for $n > 2$. – Mike Pierce Feb 14 '15 at 16:46
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1In fact, there is always a prime between $n$ and $2n$, $\forall$ $n \ge 1$. See: Bertrand's Postulate. – Feb 14 '15 at 16:47
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1@mapierce271 there is a difference between $n!+1$ and $n!$ Most of the answers use that $+1$, I think (the others rely on Bertrand's postulate, which is sort of absurd as this is a basic exercise...) – Mariano Suárez-Álvarez Feb 14 '15 at 16:47
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@AhmedHussein, but the proof of that is actually considerably harder —indeed, it is a famous theorem! – Mariano Suárez-Álvarez Feb 14 '15 at 16:48
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@AhmedHussein yes I've seen this, but this proof's too hard for me. – Bonifacy Pankracy Feb 14 '15 at 16:50
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Alright. I'll prove this one. – Feb 14 '15 at 17:00
2 Answers
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Let $n \in \Bbb N$, such that $n \ge 2$.
We have that any prime $p < n$ is an element of $\{1,2,...,n-1\}$, hence $p|n!$ $(*)$.
Now, we have that $n!$ and $n! -1$ are coprime. If there is no prime divisor $p$ of $n! - 1$, such that $p > n$, then the prime divisors of $n! - 1$ are $<n$, but this means that $n!$ and $n! - 1$ share a common divisor by $(*)$, which is impossible. Hence $n! - 1$ must have a prime divisor $>n$, say $q$. It's clear that $q \le n! - 1 < n!$, but $q > n$. Therefore:
There is a prime number $q$, such that $n < q < n!$.
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Let $p$ be any prime divisor of $n!-1$. Then $p\leq n!-1<n!$. We easily see that $p\nmid n!$, so $p$ doesn't divide any number up to $n$, so we don't have $p\leq n$, so $p>n$.
Thus $n<p<n!$.
Wojowu
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