Prove that, for every natural number n greater than 2, there is a prime number between n and n!.
Proof: For every natural number $n > 2$, there is a prime number between $n$ and $n!$.
Step 1: Understand the problem.
We are asked to show that for any integer $n > 2$, there exists at least one prime $p$ such that:
$$ n < p < n! $$
Step 2: Use a classical approach based on divisor arguments.
Consider the number:
$$ N = n! - 1 $$
Let's analyze the prime factors of $N$.
Step 3: Key observation.
- For any integer $k$ with $2 \leq k \leq n$, $k$ divides $n!$ (by the definition of factorial).
- But $k$ does not divide $n! - 1$ because:
$$ n! \equiv 0 \pmod{k} \implies n! - 1 \equiv -1 \not\equiv 0 \pmod{k} $$
This means none of the integers $2, 3, \dots, n$ divides $N = n! - 1$.
Step 4: Consequence.
- Any prime divisor of $N$ must be greater than $n$.
Step 5: Conclusion.
- $N = n! - 1$ is either prime itself (in which case it's a prime greater than $n$) or composite.
- If it's composite, it must have a prime factor greater than $n$, since none of the primes $\leq n$ divide it.
Thus, there exists a prime $p > n$ such that $p$ divides $n! - 1$.
Since $n! - 1 < n!$, this prime $p$ satisfies:
$$ n < p < n! $$
✅ Therefore, for every integer $n > 2$, there is a prime number between $n$ and $n!$.
This proof uses an idea similar to Euclid's proof of the infinitude of primes but applied to the interval $(n, n!)$.
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