AnalysisHW2 125A

# AnalysisHW2 125A - MAT 125a, HW2 Solutions 7 (a) (max { a,b...

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Unformatted text preview: MAT 125a, HW2 Solutions 7 (a) (max { a,b } = a + b + | a- b | 2 ): There are two cases: a b and a &lt; b . If a b then we have a + b + | a- b | 2 = a + b + a- b 2 = 2 a 2 = a = max { a,b } . If a &lt; b then we have a + b + | a- b | 2 = a + b- ( a- b ) 2 = 2 b 2 = b = max { a,b } . In either case our equality holds. (b) (min { a,b } =- max {- a,- b } = a + b-| a- b | 2 ): First, by part (a) we have- max {- a,- b } =-- a- b + | - a + b | 2 = a + b- | (- 1)( a- b ) | 2 = a + b- | a- b | 2 And so the right hand equality holds. To show that min { a,b } =- max {- a,- b } we may assume, without loss of generality, that a b . Since a b we have min { a,b } = a . Furthermore a b implies that- a - b hence max {- a,- b } =- a . So we have min { a,b } = a =- max {- a,- b } . 10 (a) The l.u.b. of { 1 n : n N } is 1 since 1 n 1 for all n N and 1 is in the set. The g.l.b. of { 1 n : n N } is 0. To see that the g.l.b. is 0 note that 1 /n &gt; 0 for all n N and by LUB2 no number greater then 0 will be a lower bound for this set.LUB2 no number greater then 0 will be a lower bound for this set....
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## AnalysisHW2 125A - MAT 125a, HW2 Solutions 7 (a) (max { a,b...

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