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Magazine 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 \( n = 2 \).
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 \((k+1)! < (k+1)^{k+1}\).
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:
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:
- \((k+1) \times k^k < (k+1) \times (k+1)^k \)
- \((k+1) \times (k+1)^k = (k+1)^{k+1} \)
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.
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.
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.
