Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
The following proposition is clearly wrong. Provide a counterexample, and explain what is wrong with the proof. Proposition. For all n ≥ 1, for all a₁,..., an € R, we have a₁ = A2 = ... = an. Proof. We will prove by induction on n. Base Case: For n = 1, let a₁ € R. Then a₁ = a₁. 1 Inductive Step: Assume that for all a₁,... an ER, we have a₁ = = an. We will prove that for all a₁,..., an+1 € R, we have an+1. a2 = a₁ Let a₁,..., an+1 be in R. Observe that • Since a₁,... an are n real numbers, by inductive hypothesis, we know that a₁. = An. • Since a2,...an+1 are n real numbers, by inductive hypothesis, we know that a2 = ··· = An = An+1. (Note that the inductive hypothesis is a for-all statement, so we can apply it to any n real numbers.

The following proposition is clearly wrong. Provide a counterexample, and explain what is wrong with the proof. Proposition. For all n ≥ 1, for all a₁,..., an € R, we have a₁ = A2 = ... = an. Proof. We will prove by induction on n. Base Case: For n = 1, let a₁ € R. Then a₁ = a₁. 1 Inductive Step: Assume that for all a₁,... an ER, we have a₁ = = an. We will prove that for all a₁,..., an+1 € R, we have an+1. a2 = a₁ Let a₁,..., an+1 be in R. Observe that • Since a₁,... an are n real numbers, by inductive hypothesis, we know that a₁. = An. • Since a2,...an+1 are n real numbers, by inductive hypothesis, we know that a2 = ··· = An = An+1. (Note that the inductive hypothesis is a for-all statement, so we can apply it to any n real numbers.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question
The following proposition is clearly wrong. Provide a counterexample, and explain what is wrong with the proof. **Proposition.** For all \( n \geq 1 \), for all \( a_1, \ldots, a_n \in \mathbb{R} \), we have \( a_1 = a_2 = \cdots = a_n \). **Proof.** We will prove by induction on \( n \). **Base Case:** For \( n = 1 \), let \( a_1 \in \mathbb{R} \). Then \( a_1 = a_1 \). --- **Inductive Step:** Assume that for all \( a_1, \ldots, a_n \in \mathbb{R} \), we have \( a_1 = a_2 = \cdots = a_n \). We will prove that for all \( a_1, \ldots, a_{n+1} \in \mathbb{R} \), we have \( a_1 = \cdots = a_{n+1} \). Let \( a_1, \ldots, a_{n+1} \) be in \( \mathbb{R} \). Observe that: - Since \( a_1, \ldots, a_n \) are \( n \) real numbers, by inductive hypothesis, we know that \( a_1 = \cdots = a_n \). - Since \( a_2, \ldots, a_{n+1} \) are \( n \) real numbers, by inductive hypothesis, we know that \( a_2 = \cdots = a_n = a_{n+1} \). (Note that the inductive hypothesis is a for-all statement, so we can apply it to any \( n \) real numbers.) Therefore, we have \( a_1 = \cdots = a_n \), and \( a_n = a_{n+1} \), so \( a_1 = \cdots = a_n = a_{n+1} \). --- This proof is incorrect. Can you find the error and provide a counterexample?
Transcribed Image Text:The following proposition is clearly wrong. Provide a counterexample, and explain what is wrong with the proof. **Proposition.** For all \( n \geq 1 \), for all \( a_1, \ldots, a_n \in \mathbb{R} \), we have \( a_1 = a_2 = \cdots = a_n \). **Proof.** We will prove by induction on \( n \). **Base Case:** For \( n = 1 \), let \( a_1 \in \mathbb{R} \). Then \( a_1 = a_1 \). --- **Inductive Step:** Assume that for all \( a_1, \ldots, a_n \in \mathbb{R} \), we have \( a_1 = a_2 = \cdots = a_n \). We will prove that for all \( a_1, \ldots, a_{n+1} \in \mathbb{R} \), we have \( a_1 = \cdots = a_{n+1} \). Let \( a_1, \ldots, a_{n+1} \) be in \( \mathbb{R} \). Observe that: - Since \( a_1, \ldots, a_n \) are \( n \) real numbers, by inductive hypothesis, we know that \( a_1 = \cdots = a_n \). - Since \( a_2, \ldots, a_{n+1} \) are \( n \) real numbers, by inductive hypothesis, we know that \( a_2 = \cdots = a_n = a_{n+1} \). (Note that the inductive hypothesis is a for-all statement, so we can apply it to any \( n \) real numbers.) Therefore, we have \( a_1 = \cdots = a_n \), and \( a_n = a_{n+1} \), so \( a_1 = \cdots = a_n = a_{n+1} \). --- This proof is incorrect. Can you find the error and provide a counterexample?
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS