Problem 5. (ii) please! G show that sup a 0, then a < b.(ii) Use (i) to show that if |a – b| < e for all e > 0, then a = b.3. Let ACR. Define –A = {-a : a € A}. Suppose that A is non-empty and bounded below. Show thatinf(A) = –sup(-A)4. Let A= {-" :n€ N}. Prove that sup(A) = 1, inf(A) = }.5. (i) Let A, BCR be sets which are bounded above, such that AC B. Show that sup(A) < sup(B).(ii) Let A, B CR such that sup(A) < sup(B). Show that there exists b E B that is an upper bound of A.Show that this result does not hold if we instead assume that sup(A) < sup(B).6. For A, B C R, defineA +B = {a+b : a E A,b € B}A ·B = {a ·b: a E A,b e B}(i) Determine {3, 1,0} + {2,0, 2, 1} and {3, 1,0} · {2,0, 2, 1}(ii) Assume that sup(A) and sup(B) exist. Prove that sup(A + B) = sup(A) + sup(B).(iii) Give an example of sets A, B where sup(A · B) + sup(A) · sup(B)Warm-up Problems, Not for credit:1. Let F be any field. Prove that both the additive and multiplicative identities in F are unique.2. Given an ordered field F, we saw that it has a setpositive elements P satisfying certain two conditions.러면 여기에 입력하십시오.2 24°C 8A D G qxLG1 镜hao

Answered: Image /qna-images/answer/bd3982e8-3e71-4edf-8ad5-961928641441.jpg

Prove that sup(A) =1, inf(A) = . 5. (i) Let A, BCR be sets which are bounded above, such that AC B. Show that sup(A) < sup(B). (ii) Let A, B CR such that sup(A) < sup(B). Show that there exists b e B that is an upper bound of A. Show that this result does not hold if we instead assume that sup(A) < sup(B). 6. For A. B CR define

Prove that sup(A) =1, inf(A) = . 5. (i) Let A, BCR be sets which are bounded above, such that AC B. Show that sup(A) < sup(B). (ii) Let A, B CR such that sup(A) < sup(B). Show that there exists b e B that is an upper bound of A. Show that this result does not hold if we instead assume that sup(A) < sup(B). 6. For A. B CR define

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter1: Fundamentals

Section1.7: Relations

Problem 23E

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