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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 < 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 < 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 > 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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 Fall '10
 Mulase

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