Math Problem Statement
Assume the conjecture is true for n = n and test the validity for n = n + 1. Write down your conclusion.
Solution
The question you're working on involves proving a matrix relation for Fibonacci numbers using mathematical induction. Let's break down the process.
The conjecture states that:
1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$$ where $$F_n$$ represents the Fibonacci numbers. To prove this by induction, we'll follow these steps: ### Step 1: Base Case $$n = 1$$ For $$n = 1$$, we check whether the conjecture holds true: $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^1 = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$ We need to confirm that this equals: $$\begin{pmatrix} F_2 & F_1 \\ F_1 & F_0 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$$ which is clearly true, as $$F_2 = 1$$, $$F_1 = 1$$, and $$F_0 = 0$$. So, the base case holds. ### Step 2: Inductive Hypothesis Assume that the conjecture holds for some arbitrary $$n = k$$, i.e., $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^k = \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{pmatrix}$$ ### Step 3: Inductive Step We need to prove that the conjecture holds for $$n = k+1$$. This means we must show that: $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{k+1} = \begin{pmatrix} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{pmatrix}$$ Using matrix multiplication: $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{pmatrix}$$ Carry out the multiplication: $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{pmatrix} = \begin{pmatrix} F_{k+1} + F_k & F_k + F_{k-1} \\ F_{k+1} & F_k \end{pmatrix}$$ Using the Fibonacci recurrence relation $$F_{n+1} = F_n + F_{n-1}$$, this simplifies to: $$\begin{pmatrix} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{pmatrix}$$ which is exactly what we needed to show. ### Conclusion: By induction, the conjecture holds for all $$n \geq 1$$. --- Do you want more details or have any questions? Here are 5 related questions: 1. How are Fibonacci numbers defined recursively? 2. Can you apply this matrix form of Fibonacci numbers to compute large Fibonacci numbers more efficiently? 3. What other matrix identities are useful in number theory? 4. How does the Fibonacci sequence relate to the golden ratio? 5. How can this matrix form be used in computer algorithms for fast Fibonacci computation? **Tip:** Matrix exponentiation is a highly efficient method to compute Fibonacci numbers, especially for large indices. It reduces the time complexity to $$O(\log n)$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Matrix Exponentiation
Fibonacci Numbers
Mathematical Induction
Formulas
(1 1; 1 0)^n = (F(n+1) F(n); F(n) F(n-1))
F(n+1) = F(n) + F(n-1)
Theorems
Fibonacci Recurrence Relation
Mathematical Induction
Suitable Grade Level
University level (Advanced Mathematics or Number Theory)