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Unformatted text preview: Modern Analysis 1 Homework 02 1. Let F be a complete ordered field and Q its prime subfield. Fix a > 1 in F and for x X define A ( x ) = { a q : Q 3 q < x } . (i) Show that A ( x ) is nonempty and bounded above, whence it is possible to define a x = sup A ( x ) . (ii) Show that if also y F then a x + y = a x a y . Solution For convenience, we shall simply write X = A ( x ) and Y = A ( y ). (i) The fact that F is nonempty and bounded above only needs F to be Archimedean. Indeed, we may choose a positive integer m such that m > x (in which case m Q and m < x so a m X ) and a positive integer n such that x < n (in which case if Q 3 q < x then certainly q < n so that a q < a n and a n is an upper bound for X ). As F is complete, sup X exists and we may call it a x ; we address below the fact that if x happens to lie in Q then this supremum coincides with a x as defined algebraically....
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This note was uploaded on 11/09/2011 for the course MAA 5228 taught by Professor Robinson during the Fall '09 term at University of Florida.
 Fall '09
 Robinson

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