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Chapter 5: Problem 36

Use mathematical induction in Exercises \(31-37\) to prove divisibility facts. Prove that 21 divides \(4^{n+1}+5^{2 n-1}\) whenever \(n\) is a positive integer.

Short Answer

Expert verified
21 divides \[ 4^{n+1} + 5^{2n-1} \] for all positive integers n.

Step by step solution

01

Base Case

Check if the statement is true for the smallest positive integer, which is typically 1. When =1, evaluate the expression: \[ 4^{1+1} + 5^{2 \times 1 - 1} \] This simplifies to \[ 4^2 + 5^1 = 16 + 5 = 21 \] Since 21 is divisible by 21, the base case holds.
02

Inductive Hypothesis

Assume that the statement is true for some positive integer k. This means that \[ 21 \,|\, (4^{k+1} + 5^{2k-1}) \] That is, \[ 4^{k+1} + 5^{2k-1} = 21m \] for some integer m.
03

Inductive Step

Prove that the statement is true for k + 1. We need to show that \[ 21 \,|\, (4^{k+2} + 5^{2(k+1)-1}) \] Rewrite the expression: \[ 4^{k+2} + 5^{2k+1} \]
04

Use the Inductive Hypothesis

Express \[ 4^{k+2} \] as \[ 4 \cdot 4^{k+1} \] and \[ 5^{2k+1} \] as \[ 5^2 \cdot 5^{2k-1}. \] Thus, \[ 4^{k+2} + 5^{2k+1} = 4 \cdot 4^{k+1} + 25 \cdot 5^{2k-1} \] Rewrite 4 as \[ 21 + 1 \] and 25 as \[ 21 + 4 \]: \[ (21 + 1) \cdot 4^{k+1} + (21 + 4) \cdot 5^{2k-1} \]
05

Expand and Simplify

Expand the expression: \[ 21 \cdot 4^{k+1} + 4^{k+1} + 21 \cdot 5^{2k-1} + 4 \cdot 5^{2k-1} \] Combine the terms with 21: \[ 21 \cdot 4^{k+1} + 21 \cdot 5^{2k-1} + 4^{k+1} + 4 \cdot 5^{2k-1} \] Factor out 21: \[ 21 (4^{k+1} + 5^{2k-1}) + 4^{k+1} + 4 \cdot 5^{2k-1} \] Since the inductive hypothesis tells us that \[ 4^{k+1} + 5^{2k-1} = 21m, \] we get: \[ 21 (21m) + 4^{k+1} + 4 \cdot 5^{2k-1} = 21 (m + 1) + 4(4^{k+1} + 5^{2k-1}) \] Clearly, this is divisible by 21.
06

Conclusion

Since both the base case and the inductive step have been proved, by mathematical induction, \[ 21 \,|\, (4^{n+1} + 5^{2n-1}) \] for all positive integers n.

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Key Concepts

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

Divisibility
Divisibility refers to the ability of one integer to be evenly divided by another without leaving a remainder. In this exercise, we are proving that 21 divides the expression \(4^{n+1} + 5^{2 n-1}\) for any positive integer \(n\). This means when the expression is divided by 21, the remainder is zero. Divisibility is a key concept in algebra and number theory because it helps simplify problems and proofs, especially when using techniques like mathematical induction.
Base Case
The base case is the starting point for mathematical induction. It involves checking whether the statement you aim to prove is true for the initial value, usually \(n = 1\). In this exercise, we verify that 21 divides \(4^{1+1} + 5^{2 \times 1 - 1} = 21\). Since 21 is clearly divisible by 21, the base case holds true, providing a solid foundation for the induction process.
Inductive Step
The inductive step involves proving that if the statement is true for some integer \(k\), it is also true for \(k+1\). This typically requires assuming the statement for \(k\) (this assumption is known as the inductive hypothesis) and then showing that it must hold for \(k+1\). In this problem, assuming \(21 | (4^{k+1} + 5^{2k-1})\), we aim to prove \(21 | (4^{k+2} + 5^{2(k+1)-1})\). We use algebra to express and transform the expressions into a form that will demonstrate divisibility.
Inductive Hypothesis
The inductive hypothesis is the assumption that the statement holds for some positive integer \(k\). In this exercise, we assume that \(21 | (4^{k+1} + 5^{2k-1}) \), or in other words, \(4^{k+1} + 5^{2k-1} = 21m \) for some integer \(m\). This hypothesis is used in the inductive step to prove that the statement holds for \(k+1\) by manipulating the expression \(21 | (4^{k+2} + 5^{2(k+1)-1})\) and showing it satisfies the same divisibility. Without the hypothesis, the inductive step cannot be performed effectively.

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Most popular questions from this chapter

Let \(P(n)\) be the statement that a postage of \(n\) cents can be formed using just 4 -cent stamps and 7 -cent stamps. The parts of this exercise outline a strong induction proof that \(P(n)\) is true for all integers \(n \geq 18\) . a) Show that the statements \(P(18), P(19), P(20),\) and \(P(21)\) are true, completing the basis step of a proof by strong induction that \(P(n)\) is true for all integers \(n \geq 18\) . b) What is the inductive hypothesis of a proof by strong induction that \(P(n)\) is true for all integers \(n \geq 18 ?\) c) What do you need to prove in the inductive step of a proof that \(P(n)\) is true for all integers \(n \geq 18 ?\) d) Complete the inductive step for \(k \geq 21\) . e) Explain why these steps show that \(P(n)\) is true for all integers \(n \geq 18\) .

Use structural induction to prove that \(\left(w_{1} w_{2}\right)^{R}=w_{2}^{R} w_{1}^{R}\)

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 \((a+2, b+3) \in S\) and \((a+3, b+2) \in S\) a) List the elements of \(S\) produced by the first five appli- cations of the recursive definition. b) Use strong induction on the number of applications of the recursive step of the definition to show that \(5 | a+b\) when \((a, b) \in S .\) c) Use structural induction to show that \(5 | a+b\) when \((a, b) \in S .\)

Verify that the program segment $$ \begin{array}{c}{x :=2} \\ {z :=x+y} \\ {\text { if } y>0 \text { then }} \\\ {z :=z+1} \\ {\text { else }} \\ {z :=0}\end{array} $$ is correct with respect to the initial assertion \(y=3\) and the final assertion \(z=6\)

Suppose that both the program assertion \(p\\{S\\} q_{0}\) and the conditional statement \(q_{0} \rightarrow q_{1}\) are true. Show that \(p\\{S\\} q_{1}\) also must be true.
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