nphw6 - + . As (rearranging) z- < , there exists a A...

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Numbers and Polynomials Homework 6 Let the nonempty subsets A and B of R be bounded above. Let S = { a + b : a A, b B } , P = { a b : a A, b B } . (i) Verify that S has a least upper bound. (ii) Compare lub S to lub A + lub B . (iii) What about P ? Solution Write α = lub A and β = lub B for convenience. (i) S is nonempty, of course. If a A and b B then a + b 6 α + β so S is bounded above, too. LUB applies. (ii) We have just shown that α + β is an upper bound for S ; we claim that it is least (so the two numbers to be compared are equal). For this, let z < α
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Unformatted text preview: + . As (rearranging) z- < , there exists a A such that z- < a (recall that = lub A ). As (rearranging) z-a < , there exists b B such that z-a < b (recall that = lub B ). The resulting inequality z < a + b shows that z is not an upper bound of S . (iii) Though nonempty, P need not be bounded above: indeed, A and B may contain arbitrarily large negative numbers. 1...
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This note was uploaded on 11/12/2011 for the course MAS 3300 taught by Professor Staff during the Fall '08 term at University of Florida.

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