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Chapter 5: Problem 18

Use mathematical induction to prove the inequalities in Exercises \(18-30 .\) Let \(P(n)\) be the statement that \(n !

Short Answer

Expert verified
Using induction, we proved \(n! < n^n\) for all integers \(n > 1\).

Step by step solution

01

State the Base Case

Identify the base case for the inductive proof. Let the base case be for the smallest integer greater than 1, which is when n = 2. Therefore, the statement is given by \(P(2): 2! < 2^2\).
02

Prove the Base Case

Calculate both sides of \(P(2)\): \(2! = 2\) and \(2^2 = 4\). Clearly, \(2 < 4\). Hence, \(P(2)\) is true.
03

State the Inductive Hypothesis

Assume that \(P(k)\) is true for some arbitrary integer \(k\) greater than 1. That is, assume \(k! < k^k\).
04

Determine What to Prove in the Inductive Step

To complete the inductive step, we need to show that \(P(k+1)\) is true, which means showing \((k+1)! < (k+1)^{k+1}\).
05

Use the Inductive Hypothesis

Start from \((k+1)!\) and use the fact that \(k! < k^k\). This leads to \((k+1)! = (k+1)k! < (k+1)k^k\).
06

Simplify the Inductive Step

Since \(k+1 > 1\), we know that \(k^k < (k+1)^k\). Hence, \((k+1)k^k < (k+1)(k+1)^k = (k+1)^{k+1}\). Therefore, \((k+1)! < (k+1)^{k+1}\).
07

Conclusion

By mathematical induction, because the base case is true and the inductive step holds, \(n! < n^n\) is true for all integers \(n > 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case
In mathematical induction, the base case is the starting point for your proof. Here, you need to show that the statement is true for the initial value, usually the smallest integer for which the statement is claimed. For our exercise, we are proving the inequality \( n! < n^n \) for all integers greater than 1. Thus, the smallest integer is 2.
The base case becomes \( P(2): 2! < 2^2 \).
We compute: \(2! = 2 \) and \(2^2 = 4 \). Clearly, \(2 < 4 \), so the base case is true. This step confirms that our statement holds for \( n = 2 \).
Inductive Hypothesis
The inductive hypothesis is a crucial part of your proof. It involves assuming that the statement you're proving holds for some arbitrary integer \( k \). This assumption serves as a stepping stone to prove that the statement is also true for the next integer, \( k+1 \).
For our exercise, we assume that \(P(k)\) is true, which translates to \(k! < k^k\). This assumption lets us move forward to prove that \((k+1)! < (k+1)^{k+1}\).
Inductive Step
The inductive step involves using the inductive hypothesis to show that the statement is true for the next integer. This means proving that if \(P(k)\) is true, then \(P(k+1)\) must also be true.
Starting with the inductive hypothesis, \(k! < k^k\), we aim to show \((k+1)! < (k+1)^{k+1}\).
We know \((k+1)! = (k+1) \times k! \). Using our hypothesis, we get \((k+1)! < (k+1) \times k^k \). Since \(k^k < (k+1)^k \) due to \( k+1 > k \), it follows that:
  • \((k+1) \times k^k < (k+1) \times (k+1)^k \)
  • \((k+1) \times (k+1)^k = (k+1)^{k+1} \)
Hence, \((k+1)! < (k+1)^{k+1}\) , completing the inductive step.
Factorials
A factorial, denoted as \(n! \), represents the product of all positive integers up to \( n \). For example, \(4! = 4 \times 3 \times 2 \times 1 = 24 \). Factorials grow very quickly as \( n \) increases.
Factorials are fundamental in combinatorics, algebra, and calculus, frequently appearing in problems involving permutations, series, and integrals. Familiarity with factorial properties is crucial in many areas of mathematics and practical problem-solving.
Inequality Proof
Proof by induction often involves inequalities. In our exercise, we're proving \( n! < n^n \) for integers greater than 1. Inequality proofs involve showing that one quantity is consistently larger or smaller than another.
In the inductive step of our proof, we use inequalities to transition from \( k \) to \( k+1 \). By breaking down expressions and using properties of numbers, we can show one side of the inequality is less than the other, thus proving our statement.
Inequality proofs require attention to detail and a solid grasp of mathematical properties and logical reasoning.

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Most popular questions from this chapter

Show that if \(A_{1}, A_{2}, \ldots, A_{n}\) are sets where \(n \geq 2,\) and for all pairs of integers \(i\) and \(j\) with \(1 \leq i

Suppose that \(P(n)\) is a propositional function. Determine for which nonnegative integers \(n\) the statement \(P(n)\) must be true if a) \(P(0)\) is true; for all nonnegative integers \(n,\) if \(P(n)\) is true, then \(P(n+2)\) is true. b) \(P(0)\) is true; for all nonnegative integers \(n,\) if \(P(n)\) is true, then \(P(n+3)\) is true. c) \(P(0)\) and \(P(1)\) are true; for all nonnegative integers \(n,\) if \(P(n)\) and \(P(n+1)\) are true, then \(P(n+2)\) is true. d) \(P(0)\) is true; for all nonnegative integers \(n,\) if \(P(n)\) is true, then \(P(n+2)\) and \(P(n+3)\) are true.

When does a string belong to the set \(A\) of bit strings defined recursively by $$ \begin{array}{l}{\lambda \in A} \\ {0 x 1 \in A \text { if } x \in A}\end{array} $$ where \(\lambda\) is the empty string?

Suppose there are \(n\) people in a group, each aware of a scandal no one else in the group knows about. These people communicate by telephone; when two people in the group talk, they share information about all scandals each knows about. For example, on the first call, two people share information, so by the end of the call, each of these people knows about two scandals. The gossip problem asks for \(G(n),\) the minimum number of telephone calls that are needed for all \(n\) people to learn about all the scandals. Exercises \(69-71\) deal with the gossip problem. Use mathematical induction to prove that \(G(n) \leq 2 n-4\) for \(n \geq 4 .\) [Hint: In the inductive step, have a new person call a particular person at the start and at the end. \(]\)

Exercises \(49-51\) present incorrect proofs using mathematical induction. You will need to identify an error in reasoning in each exercise. What is wrong with this "proof"? "Theorem" For every positive integer \(n,\) if \(x\) and \(y\) are positive integers with max \((x, y)=n\) , then \(x=y\) . Basis Step: Suppose that \(n=1 .\) If \(\max (x, y)=1\) and \(x\) and \(y\) are positive integers, we have \(x=1\) and \(y=1\) . Inductive Step: Let \(k\) be a positive integer. Assume that whenever max \((x, y)=k\) and \(x\) and \(y\) are positive integers, then \(x=y .\) Now let max \((x, y)=k+1,\) where \(x\) and \(y\) are positive integers. Then \(\max (x-1, y-1)=k,\) so by the inductive hypothesis, \(x-1=y-1 .\) It follows that \(x=y\) completing the inductive step.
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