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Introductory Analysis

by: Mrs. Lenore King

Introductory Analysis MTH 4101

Mrs. Lenore King
Florida Tech
GPA 3.91


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This 2 page Class Notes was uploaded by Mrs. Lenore King on Monday October 12, 2015. The Class Notes belongs to MTH 4101 at Florida Institute of Technology taught by Staff in Fall. Since its upload, it has received 40 views. For similar materials see /class/221672/mth-4101-florida-institute-of-technology in Mathematics (M) at Florida Institute of Technology.

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Date Created: 10/12/15
MTH 41015101 Introduction to Analysis Fall 2008 Practice problems ll 9102008 2 9 4 TL 71 2 a 221 k3 b c 2 1 lt n lt n for all n 34 d Bernoulli7s Inequality If x gt 71 then 1 x 2 1 71 for all n E IN Solution For n 1 We have equality and so the assertion is valid Suppose the inequality is valid for k E IN Thus we have 1 zk 21 kx and 1 z gt 0 Now7 consider If x E 071 is a xed real nurnber7 then 0 lt x lt 1 for all n E IN 1 90 1 1 zk1 z Ok l11lkhxk 1lknm l l Thus7 the inequality holds for k 1 Hence7 it holds for all n E IN Let S be a bounded set in IR and let So be a nonernpty subset of SS Show that inf S S inf So 3 supSo S sup S Solution Since S is a bounded set and So is a nonernpty subset of S7 infS7 infSo7 supSo and supS exist From the definition7 we have7 inf S S s for all s E S Thus inf S is a lower bound for So Since inf So is the greatest lower bound of So7 we have inf S S inf So Let so 6 S We then have7 inf So 3 so S supSo So7 inf So 3 sup So Lastly7 if sup S lt sup So7 there exists an element so 6 So such that sup S lt so7 which is a contradiction Hence7 sup So sup S Suppose that A and B are nonernpty subsets of IR satisfying the property a S b for all a E A and all b E B Prove that supA S inf B Solution Let bo E B We have a S bo for all a E A Thus7 bo is an upper bound of A leading to sup A S bo Since7 bo is any arbitrary element of B7 we have sup A S b for all b E B Hence7 sup A is a lower bound of B leading to sup A S inf B Use Archirnedean property to prove the following a If t gt 0 show that there exists nt E N such that 0 lt nit lt t Solution For a given t gt 0 we have by Archimedian principle a 71 E N such that tn gt 1 gt 0 Cf b If y gt 0 show that there exists my E IN such that my 71 g y lt my Solution Let E m E Wig lt By Archirnedian property E is nonempty choose x 1 in the property discussed in the lecture and bounded below by y Hence in rnurn of E exists Let my inf Ey Clearly my 6 Ey why Further my 7 1 does not belong to E and hence we have my 71 S y lt my If x and y are real numbers such that z lt y prove that there exists an irrational number 2 such that z lt z lt y Solution We know that between any two real numbers there exists a rational number So there exists a 2 E Q such that x lt 2 lt Dividing throughout by xE we get the desired result


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