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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
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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. ) Therefore, we have a₁ = ... = an, and an = an+1, SO A₁ = ··· = An = an+1.
Transcribed Image Text: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. ) Therefore, we have a₁ = ... = an, and an = an+1, SO A₁ = ··· = An = an+1.
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