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Chapter 5: Problem 20

Use mathematical induction to prove the inequalities in Exercises \(18-30 .\) Prove that \(3^{n}

Short Answer

Expert verified
By induction, \(3^n < n!\) is true for all \(n > 6\).

Step by step solution

01

- Base Case

Verify the inequality for the smallest integer greater than 6, which is 7. Calculate both sides of the inequality when n = 7:We need to prove that \(3^7 < 7!\).Calculate \(3^7\): \(3^7 = 2187\)Calculate \(7!\): \(7! = 5040\)Since \(2187 < 5040\), the inequality holds for the base case.
02

- Inductive Hypothesis

Assume the inequality is true for some integer \(k > 6\). Thus, assume \(3^k < k!\) is true.
03

- Inductive Step

Prove the inequality is true for \(k + 1\). Consider the integer \(k + 1\):Starting from the inductive hypothesis \(3^k < k!\), we need to show that \(3^{k+1} < (k+1)!\)Rewrite \(3^{k+1}\) using the properties of exponents:\(3^{k+1} = 3 \times 3^k\)Now we have:\(3 \times 3^k < 3 \times k!\) Since \(k > 6\), then \(3 < k + 1\) (for all integers \(k > 6\)). Therefore, we also have:\(3 \times k! < (k+1) \times k!\)Combining these, we get \(3^{k+1} < (k+1)!\)This completes the inductive step.
04

- Conclusion

By mathematical induction, the inequality \(3^n < n!\) is true for all integers \(n > 6\).

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Key Concepts

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

Base Case
When using mathematical induction, we start with the base case. This step involves verifying the given inequality for the smallest integer. In this exercise, the smallest integer greater than 6 is 7.
We check whether the inequality \(3^n < n!\) holds true when \(n = 7\).
First, calculate \(3^7\):
\[ 3^7 = 2187 \]
Next, calculate \(7!\) (factorial of 7):
\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]
Since \(2187 < 5040\), the base case holds. The inequality is true for \(n = 7\). This confirmation serves as the foundation for proceeding with the induction process.
Inductive Hypothesis
After the base case, the next step in mathematical induction is establishing the inductive hypothesis. Here, we assume the inequality holds for some integer \(k\) where \(k > 6\).
So, we assume:
\[ 3^k < k! \]
This assumption is crucial because it lets us use this supposed truth to prove it for the next integer \((k+1)\). Keep in mind that we are only assuming this for the purpose of deriving what happens at \(k+1\).
Inductive Step
In the inductive step, we need to prove that the inequality holds for \(k + 1\) given that it holds for \(k\). Essentially, if \(3^k < k!\) is true, we must show that \(3^{k+1} < (k+1)!\).
Start from the inductive hypothesis:
\[ 3^k < k! \]
Multiply both sides by 3 to express \(3^{k+1}\):
\[ 3 \times 3^k < 3 \times k! \]
We know that for all integers \(k > 6\), \((k + 1) > 3\). Therefore, we also have:
\[ 3 \times k! < (k + 1) \times k! = (k + 1)! \]
Combining these inequalities, we get:
\[ 3^{k+1} < (k+1)! \]
This completes the inductive step, thus proving that if the inequality holds for \(k\), it holds for \(k+1\) as well.
Factorial Inequality
The factorial inequality we are dealing with here is \(3^n < n!\) for integers \(n > 6\). Factorial functions grow much faster than exponential functions.
Let's see some key properties:
  • Factorial of a number \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\).
  • Exponential functions like \(3^n\) grow at a fixed rate, while factorial functions grow at an accelerating rate.
  • For numbers greater than 6, the factorial values rapidly outpace the corresponding exponential values. This is why \(3^n < n!\) holds true for integers greater than 6.

Understanding this growth pattern helps explain why our proved inequality stands strong and continues to hold as \(n\) increases.

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Most popular questions from this chapter

Use strong induction to show that every positive integer can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers \(2^{0}=1,2^{1}=2,2^{2}=4\) and so on. [Hint: For the inductive step, separately con- sider the case where \(k+1\) is even and where it is odd. When it is even, note that \((k+1) / 2\) is an integer. \(]\)

A partition of a positive integer \(n\) is a way to write \(n\) as a sum of positive integers where the order of terms in the sum does not matter. For instance, \(7=3+2+1+1\) is a partition of \(7 .\) Let \(P_{m}\) equal the number of different partitions of \(m,\) and let \(P_{m, n}\) be the number of different ways to express \(m\) as the sum of positive integers not exceeding \(n\). a) Show that \(P_{m, m}=P_{m}\) . b) Show that the following recursive definition for \(P_{m, n}\) is correct: $$P_{m, n}=\left\\{\begin{array}{ll}{1} & {\text { if } m=1} \\ {1} & {\text { if } n=1} \\ {P_{m, m}} & {\text { if } m< n} \\ {1+P_{m, m-1}} & {\text { if } m=n>1} \\ {P_{m, n-1}+P_{m-n, n}} & {\text { if } m>n>1}\end{array}\right.$$ c) Find the number of partitions of 5 and of 6 using this recursive definition.

Exercises 22 and 23 present examples that show inductive loading can be used to prove results in computational geometry. Let \(P(n)\) be the statement that when non intersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two vertices of the polygon are not endpoints of any of these diagonals. a) Show that when we attempt to prove \(P(n)\) for all integers \(n\) with \(n \geq 3\) using strong induction, the in- ductive step does not go through. b) Show that we can prove that \(P(n)\) is true for all inte- gers \(n\) with \(n \geq 3\) by proving by strong induction the stronger assertion \(Q(n),\) for \(n \geq 4,\) where \(Q(n)\) states that whenever nonintersecting diagonals are drawn inside a convex polygon with \(n\) sides, at least two nonadjacent vertices are not endpoints of any of these diagonals.

Let \(S\) be the subset of the set of ordered pairs of integers defined recursively by Basis step: \((0,0) \in S\) . Recursive step: If \((a, b) \in S,\) then \((a, b+1) \in S\) \((a+1, b+1) \in S,\) and \((a+2, b+1) \in S\) a) List the elements of \(S\) produced by the first four ap- plications of the recursive definition. b) Use strong induction on the number of applications of the recursive step of the definition to show that \(a \leq 2 b\) whenever \((a, b) \in S .\) c) Use structural induction to show that \(a \leq 2 b\) whenever \((a, b) \in S .\)

Use a loop invariant to prove that the following program segment for computing the \(n\) th power, where \(n\) is a positive integer, of a real number \(x\) is correct. $$ \begin{array}{c}{\text { power } :=1} \\ {i :=1} \\ {\text { while } i \leq n} \\ {\text { power } :=\text { power } * x} \\ {i :=i+1}\end{array} $$
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