a1-22-08

a1-22-08 - Annotated answers to the January 22 assignment...

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Annotated answers to the January 22 assignment 1. (Page 57) (a) f 0 ; 3 ; 6 ; 9 g 3 j 0 because 0 = 3 & 0 . (b) f 2 g de±nition is on page 3. You can ±nd that out by looking up ²prime³in the index. (c) f 2 ; ± 2 g . Note that x is allowed to be negative because x 2 Z . (d) fg , because there are no integers whose square is 5 . You could also write ; , but they speci±ed that you should list the elements between curly braces. (e) f;g . This is not the empty set, but a one-element set (whose element happens to be the empty set). The given set, 2 ; , is the set of all subsets of the empty set. There is exactly one subset of the empty set: the empty set. (f) 100 ; ± 50 ; ± 20 ; ± 10 ; 10 ; 20 ; 50 ; 100 g . (g) f; ; f 1 g ; f 2 g ; f 3 g ; f 4 g ; f 5 gg is the set of subsets of f 1 ; 2 ; 3 ; 4 ; 5 g of cardinality at most 1 . The ±rst one has cardinality 0 , the other four have cardinality 1 . 2. (Page 57)
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This note was uploaded on 07/13/2011 for the course MAD 2104 taught by Professor Staff during the Spring '08 term at FAU.

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a1-22-08 - Annotated answers to the January 22 assignment...

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