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I have attached the solution as well. The part I don’t understand in the solution is how did we get n>1-E/2E from the original equation. Please explain that part and thank you

I have attached the solution as well. The part I don’t understand in the solution is how did we get n>1-E/2E from the original equation. Please explain that part and thank you

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I have attached the solution as well. The part I don’t understand in the solution is how did we get n>1-E/2E from the original equation. Please explain that part and thank you
Sets: solved exercises Exercise 1. Tell whether the following sets are bounded from below and/or from above, specifying infimum, supremum and minimum and maximum, if they exist. A={2+1: neN}. a) A =
Transcribed Image Text:Sets: solved exercises Exercise 1. Tell whether the following sets are bounded from below and/or from above, specifying infimum, supremum and minimum and maximum, if they exist. A={2+1: neN}. a) A =
SOLUTIONS Exercise 1. la) For all n E N we have 2n + 1 ≥ 1 and 0< For n= 1 2n + 1 Therefore A is bounded from above and from below. 0, we have 1 2n + 1 We prove now that inf A = 0. Clearly 0 is a lower bound of A, and 0 A. We must prove that 0 is the greatest of the lower bounds of A, that is, for all a > 0 there exists n E N such that = 1. Therefore max A = sup A = 1. 1 2n + 1 <1. property¹ there exists n E N such that n > doesn't exist. 1b) If n E Z with n ≥ 0, then 2n + 1 ≥ 1 and 0 < 1-8 The solutions of the equation are the numbers n E R such that n > 2€ Therefore 0 = inf A. Since 0 A, min A By Archimedean 1- E 2€ <0+ ε = E, 1
Transcribed Image Text:SOLUTIONS Exercise 1. la) For all n E N we have 2n + 1 ≥ 1 and 0< For n= 1 2n + 1 Therefore A is bounded from above and from below. 0, we have 1 2n + 1 We prove now that inf A = 0. Clearly 0 is a lower bound of A, and 0 A. We must prove that 0 is the greatest of the lower bounds of A, that is, for all a > 0 there exists n E N such that = 1. Therefore max A = sup A = 1. 1 2n + 1 <1. property¹ there exists n E N such that n > doesn't exist. 1b) If n E Z with n ≥ 0, then 2n + 1 ≥ 1 and 0 < 1-8 The solutions of the equation are the numbers n E R such that n > 2€ Therefore 0 = inf A. Since 0 A, min A By Archimedean 1- E 2€ <0+ ε = E, 1
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