write the latex of this solution of the math questions [LATEXquestion: submit a printout of your answer, typeset in LTEX]Let A = {" |n € N, m e N, m > n}.m(a) Prove that inf A = 0.(b) Prove that sup A = 1.Solution: 0 is a lower bound of A since > 0. We need to show that there is nolarger lower bound. So for anye>0 there exist n, m eN with m >n and 2 with m > el. Such an mexists because of the Archimedean property of R. Since in this case 0is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now sincen 0 thenumber 1 - e is not an upper bound. We do this by finding n, m eN with n < mand >1-e. One chooses (after trying a bit) an m eN with m > e and m > 1.Then choose n = m - 1. In this case one obtains1>1-e,m-1which is what we had to show. We have used here that < e and therefore ->

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[I\TEXquestion: submit a printout of your answer, typeset in INTEX] Let A = {|ne N, m e N, m > n}. (a) Prove that inf A = 0. (b) Prove that sup A = 1. Solution: 0 is a lower bound of A since > 0. We need to show that there is no larger lower bound. So for any e> 0 there exist n, m eN with m >n and 2 with m >e. Such an m exists becatuse of the Archimedean property of R. Since in this case 0 is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now since n0 the number 1-e is not an upper bound. We do this by finding n, m e N with n < m and >1-e. One chooses (after trying a bit) an m e N with m>e and m > 1. Then choose n= m - 1. In this case one obtains m-1 1 >1-e, which is what we had to show. We have used here that

[I\TEXquestion: submit a printout of your answer, typeset in INTEX] Let A = {|ne N, m e N, m > n}. (a) Prove that inf A = 0. (b) Prove that sup A = 1. Solution: 0 is a lower bound of A since > 0. We need to show that there is no larger lower bound. So for any e> 0 there exist n, m eN with m >n and 2 with m >e. Such an m exists becatuse of the Archimedean property of R. Since in this case 0 is not a lower bound. Hence, inf A = 0, as 0 is the greatest lower bound. Now since n0 the number 1-e is not an upper bound. We do this by finding n, m e N with n < m and >1-e. One chooses (after trying a bit) an m e N with m>e and m > 1. Then choose n= m - 1. In this case one obtains m-1 1 >1-e, which is what we had to show. We have used here that

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter4: Eigenvalues And Eigenvectors

Section4.6: Applications And The Perron-frobenius Theorem

Problem 17EQ

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