Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 9: Problem 26
Prove that the statement is true for every positive integer \(n\). \(a+b\) is a factor of \(a^{2 n-1}+b^{2 n-1}\)
Short Answer
Expert verified
The statement is proven by induction.
Step by step solution
01
Understand the Problem
We are asked to show that \(a+b\) is a factor of the expression \(a^{2n-1} + b^{2n-1}\) for every positive integer \(n\). This means we need to express \(a^{2n-1} + b^{2n-1}\) in a form where it is clearly divisible by \(a+b\).
02
Use Induction Basis
Consider the base case: \(n = 1\). In this case, the expression becomes \(a^{2(1)-1} + b^{2(1)-1} = a + b\). We can see directly that \(a+b\) is a factor because it is the same as the expression itself.
03
Assume Inductive Step
Assume that for some positive integer \(k\), \(a+b\) is a factor of \(a^{2k-1} + b^{2k-1}\). This means there exists \(m\) such that \(a^{2k-1} + b^{2k-1} = m(a+b)\).
04
Prove for k+1
We need to prove that \(a+b\) is a factor of \(a^{2(k+1)-1} + b^{2(k+1)-1}\). Simplifying it gives us \(a^{2k+1} + b^{2k+1}\). Using the identity \(a^{2k+1} + b^{2k+1} = (a^{2k-1} + b^{2k-1})(a^2 - ab + b^2)\) and since \(a^{2k-1} + b^{2k-1}\) is divisible by \(a+b\) by our IH (Inductive Hypothesis), it follows that \(a^{2k+1} + b^{2k+1}\) is also divisible by \(a+b\).
05
Verify the Inductive Step
By our identity and the inductive assumption, \((a^{2k-1} + b^{2k-1})(a^2 - ab + b^2)\) ensures the divisibility holds at the next step \(k+1\). Thus, the statement is proven for \(k+1\) based on the assumption it holds for \(k\).
06
Conclusion of Induction
Since both base case and the inductive step are true, by mathematical induction, \(a+b\) is a factor of \(a^{2n-1} + b^{2n-1}\) for all positive integers \(n\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorization
Factorization is a method used in mathematics to express a given quantity as a product of its factors. When dealing with polynomials, factorization helps in simplifying expressions and solving equations.
In context of the problem, we need to show that \( a+b \) is a factor of the expression \( a^{2n-1} + b^{2n-1} \).
This means we break down the expression in such a way that \( a+b \) divides it perfectly without leaving a remainder.
This is similar to factorizing a number by expressing it in terms of its prime components.
In context of the problem, we need to show that \( a+b \) is a factor of the expression \( a^{2n-1} + b^{2n-1} \).
This means we break down the expression in such a way that \( a+b \) divides it perfectly without leaving a remainder.
This is similar to factorizing a number by expressing it in terms of its prime components.
- Recognizing patterns, like the difference of squares or sum of cubes, makes factorization easier.
- These known algebraic identities can be used to break down complex expressions.
Polynomial Division
Polynomial division is a process similar to long division where one polynomial is divided by another, resulting in a quotient and sometimes a remainder.
In this exercise, our goal is to show that \( a+b \) perfectly divides \( a^{2n-1} + b^{2n-1} \).
This is done by expressing the polynomial as a product of \( a+b \) and another polynomial, ensuring there is no remainder.
In this exercise, our goal is to show that \( a+b \) perfectly divides \( a^{2n-1} + b^{2n-1} \).
This is done by expressing the polynomial as a product of \( a+b \) and another polynomial, ensuring there is no remainder.
- The division process involves aligning terms of the same degree and performing operations similar to numeric long division.
- When the remainder is zero, it confirms the division is exact, meaning the divisor is indeed a factor of the dividend polynomial.
Positive Integers
Positive integers are the set of all whole numbers greater than zero, represented as \( \{1, 2, 3, 4, \ldots\} \).
In this problem, the parameter \( n \) is a positive integer, and we are required to prove the given statement for every value of \( n \).
This significance is what makes the method of mathematical induction appropriate for this problem.
In this problem, the parameter \( n \) is a positive integer, and we are required to prove the given statement for every value of \( n \).
This significance is what makes the method of mathematical induction appropriate for this problem.
- Induction relies on establishing a base case, typically the smallest positive integer.
- It then proves that if the proposition holds for an arbitrary positive integer \( k \), it will also hold for \( k+1 \).
Basis and Inductive Step
The basis and inductive step are two critical components of mathematical induction, a technique used to prove statements about positive integers.
The basis step demonstrates that the statement is true for the initial value (usually \( n = 1 \)). For this exercise, the base case with \( n = 1 \) verifies that \( a+b \) is indeed a factor of \( a+b \).
The inductive step involves the assumption that for some integer \( k \), the statement holds true, and then proving it also holds for \( k+1 \).
The basis step demonstrates that the statement is true for the initial value (usually \( n = 1 \)). For this exercise, the base case with \( n = 1 \) verifies that \( a+b \) is indeed a factor of \( a+b \).
The inductive step involves the assumption that for some integer \( k \), the statement holds true, and then proving it also holds for \( k+1 \).
- This is known as the Inductive Hypothesis (IH), where you assume \( a+b \) is a factor for \( a^{2k-1} + b^{2k-1} \).
- By proving that \( a+b \) is also a factor of \( a^{2(k+1)-1} + b^{2(k+1)-1} \), we conclude the induction proof.
