Consider the following statement. For every integer n, if n > 2 then there is a prime number p such that n 2, n! - 1 is an integer that is greater than 1. 4. So by the divisibility by a prime theorem (Theorem 4.4.4), there is a prime number p such that p|(n! – 1). . 5. Thus, p < n! - 1, and so p < n!. 6. Therefore, by steps 2 and 5, there is a prime number p such that n < p < n!.
Consider the following statement. For every integer n, if n > 2 then there is a prime number p such that n 2, n! - 1 is an integer that is greater than 1. 4. So by the divisibility by a prime theorem (Theorem 4.4.4), there is a prime number p such that p|(n! – 1). . 5. Thus, p < n! - 1, and so p < n!. 6. Therefore, by steps 2 and 5, there is a prime number p such that n < p < n!.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text: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:
1. Let n be any integer that is greater than 2.
2. By the infinitude of the primes theorem (Theorem 4.8.4),
there is a prime number p such that n < p.
3. Since n > 2, n! - 1 is an integer that is greater than 1.
4. So by the divisibility by a prime theorem
(Theorem 4.4.4), there is a prime number p such that
p|(n! - 1)..
5. Thus, p < n! - 1, and so p < n!.
6. 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.)
O The divisibility by a prime theorem is incorrectly applied in step 4.
O The p in step 2 may not be the same as the p in step 5.
O The statement can only be proved by contradiction.
O The proof does not work if n is prime.
O The infinitude of the primes theorem is incorrectly applied in step 2.
O p|(n! – 1) does not imply that p is less than or equal to n! – 1.
O The proof does not work if n! - 1 is prime.
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