Question

Let P(n) be the statement that n! < nn where n is an integer greater than 1.

Answer 1

Answer: See the step by step explanation.

Step-by-step explanation:

a) First, Let P(n) be the statement that n! < n^n

where n ≥ 2 is an integer (This is because we want the statement of P(2).

In this case the statement would be (n = 2): P(2) = 2! < 2^2

b) Now to prove this, let's complet the basis step:

We know that 2! = 2 * 1 = 2

and 2^2 = 2 * 2 = 4

Therefore: 2 < 4

c)  For this part, we'll say that the inductive hypothesis would be assuming that k! < k^k for some k ≥ 1

d) In this part, the only thing we need to know or prove is to show that P(k+1) is also true, given the inductive hypothesis in part c.

e) To prove that P(k+1) is true, let's solve the inductive hypothesis of k! < k^k:

(k + 1)! = (k + 1)k!  

(k + 1)k!  < (k + 1)^k  < (k + 1)(k + 1)^k

Since k < k+1 we have:

= (k + 1)^k+1

f) Finally, as the base and inductive steps are completed, the inequality is true for any integer for any n ≥ 1. If we had shown P(4)

as our basis step, then the inequality would only be proven for n ≥ 4.