Beginning Activity Beginning Activity 1: Exploring a Proposition about Factorials
Definition.
If \(n\) is a natural number, we define \(\boldsymbol{n}\) factorial, denoted by \(n!\) , to be the product of the first \(n\) natural numbers. In addition, we define \(0!\) to be equal to 1.
Using this definition, we see that
In general, we write \(n! = 1 \cdot 2 \cdot 3 \cdots \left( {n - 1} \right) \cdot n\) or \(n! = n \cdot \left( {n - 1} \right) \cdots 2 \cdot 1\text{.}\) Notice that for any natural number \(n\text{,}\) \(n! = n \cdot (n-1)!\text{.}\)
1.
Compute the values of \(2^n\) and \(n!\) for each natural number \(n\) with \(1 \leq n \leq 7\text{.}\)
Now let \(P(n)\) be the open sentence, “\(n! > 2^n\text{.}\)”
2.
Which of the statements \(P(1)\) through \(P(7)\) are true?
3.
Based on the evidence so far, does the following proposition appear to be true or false? For each natural number \(n\) with \(n \geq 4\text{,}\) \(n! > 2^n\text{.}\)
Let \(k\) be a natural number with \(k \geq 4\text{.}\) Suppose that we want to prove that if \(P(k)\) is true, then \(P(k+1)\) is true. (This could be the inductive step in an induction proof.) To do this, we would be assuming that \(k! > 2^k\) and would need to prove that \((k+1)! > 2^{k+1}\text{.}\) Notice that if we multiply both sides of the inequality \(k! > 2^k\) by \((k + 1)\text{,}\) we obtain
4.
In the inequality in (27), explain why \((k + 1) \cdot k! = (k + 1)!\text{.}\)
5.
Now look at the right side of the inequality in (27) Since we are assuming that \(k \geq 4\text{,}\) we can conclude that \((k+1) > 2\text{.}\) Use this to help explain why \((k + 1)2^k > 2^{k+1}\text{.}\)
6.
Now use the inequality in (1) and the work in Exercise 4 and Exercise 5 to explain why \((k+1)! > 2^{k+1}\text{.}\)