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!.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
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