Unformatted text preview: p k u k p ≤ k a k p = max {k a k p , k b k p } . (ii) For all rationals a , b , prove δ p ( a , b ) ≥ 0 and δ p ( a , b ) = 0 if and only if a = b . Solution. δ p ( a , b ) ≥ 0 because k c k p ≥ 0 for all c . If a = b , then δ p ( a , b ) = k a − b k p = k k p = 0; conversely, if δ p ( a , b ) = 0, then a − b = 0 because 0 is the only element c with k c k p = 0. (iii) For all rationals a , b , prove that δ p ( a , b ) = δ p ( b , a ) . Solution. δ p ( a , b ) = δ p ( b , a ) because k − c k p = k − 1 k p k c k p = k c k p ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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