nphw2 - natively, for the sake of variety, one could apply...

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Numbers and Polynomials Homework 2 1. Show that if a, b R satisfy a < b then a < a + b 2 < b . 2. Show that if a, b R satisfy 0 < a < b then a < 2 1 a + 1 b < b . Solutions : 1. Note first that a + a < a + b by OA . Here, a + a = 1 a +1 a = (1+1) a = 2 a by MID and D so that 2 a < a + b . Now 2 > 0 by Sigmon 1.15(a)(f) so that 1 / 2 > 0 by Sigmon 1.15(h) and therefore (1 / 2)2 a < (1 / 2)( a + b ) by OM . It follows that a < a + b 2 (why?). Similarly, a + b 2 < b . 2. One could, of course, essentially repeat the argument just given. Alter-
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Unformatted text preview: natively, for the sake of variety, one could apply the conclusion. Thus, from < a < b it follows (by Sigmon 1.15(j)) that 1 b < 1 a whence 1 b < 1 b + 1 a 2 < 1 a and therefore a < 2 1 a + 1 b < b ( Sigmon 1.15(j) again). 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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