Consider the following statement. For every integer n, if  n > 2  then there is a prime number p such that  n < p < n!. The statement is true, but the following proposed proof is incorrect. Proposed proof: Let n be any integer that is greater than 2. By the infinitude of the primes theorem (Theorem 4.8.4), there is a prime number p such that  n < p. Since  n > 2, n! − 1  is an integer that is greater than 1. So by the divisibility by a prime theorem (Theorem 4.4.4), there is a prime number p such that  p | (n! − 1). Thus,  p ≤ n! − 1,  and so  p < n!. Therefore, by steps 2 and 5, there is a prime number p such that  n < p < n!. Identify the error(s) in the proposed proof. (Select all that apply.) p | (n! − 1) does not imply that p is less than or equal to n! − 1.The statement can only be proved by contradiction.The divisibility by a prime theorem is incorrectly applied in step 4.The p in step 2 may not be the same as the p in step 5.The proof does not work if n is prime.The proof does not work if n! − 1 is prime.The infinitude of the primes theorem is incorrectly applied in step 2.

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