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 exercisesExercise 1. Tell whether the following sets are bounded from below and/or from above, specifyinginfimum, supremum and minimum and maximum, if they exist.A={2+1: neN}.a) A = SOLUTIONSExercise 1.la) For all n E N we have 2n + 1 ≥ 1 and0<For n=12n + 1Therefore A is bounded from above and from below.0, we have12n + 1We prove now that inf A = 0. Clearly 0 is a lower bound of A, and 0 A. We must provethat 0 is the greatest of the lower bounds of A, that is, for all a > 0 there exists n E N suchthat= 1. Therefore max A = sup A = 1.12n + 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 and0 <1-8The solutions of the equation are the numbers n E R such that n >2€Therefore 0 = inf A. Since 0 A, min ABy Archimedean1- E2€<0+ ε = E,1

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Answered: I have attached the solution as well.…

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

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 =

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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