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303a3solns - ASSIGNMENT 3 MATH 303 FALL 2011 If you find...

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ASSIGNMENT 3 MATH 303, FALL 2011 If you find any errors please let me know. Manipulation (M1) (a) 3 = { 0 , 1 , 2 } (b) 3 = {∅ , {∅} , {∅ , {∅}}} . (M2) { 0 , 3 , 4 } + = { 0 , 3 , 4 } ∪ {{ 0 , 3 , 4 }} = { 0 , 3 , 4 , { 0 , 3 , 4 }} (M3) Y X = {{ ( a, c ) , ( b, c ) } , { ( a, c ) , ( b, d ) } , { ( a, c ) , ( b, e ) } , { ( a, d ) , ( b, c ) } , { ( a, d ) , ( b, d ) } , { ( a, d ) , ( b, e ) } , { ( a, e ) , ( b, c ) } , { ( a, e ) , ( b, d ) } , { ( a, e ) , ( b, e ) }} (M4) X Y = {{ ( c, a ) , ( d, a ) , ( e, a ) } , { ( c, a ) , ( d, a ) , ( e, b ) } { ( c, a ) , ( d, b ) , ( e, a ) } , { ( c, a ) , ( d, b ) , ( e, b ) } { ( c, b ) , ( d, a ) , ( e, a ) } , { ( c, b ) , ( d, a ) , ( e, b ) } { ( c, b ) , ( d, b ) , ( e, a ) } , { ( c, b ) , ( d, b ) , ( e, b ) }} (M5) The set { ( a, c ) , (1 , c ) , ( b, 2) , (2 , 3) , ( a, 4) } is not a function because it contains both ( a, c ) and ( a, 4) and thus a maps to both 4 and c , which is not possible for a function. (M6) Define the map f : { a, b, c } → 3 by f ( a ) = 0, f ( b ) = 1, f ( c ) = 2. Then f is onto 3 as 3 = { 0 , 1 , 2 } and f is one-to-one as there are no two distinct elements of { a, b, c } mapping to the same element of 3. Pure Math (P1) (4 points) (a) [ 4 = 0 1 2 3 = ∅ ∪ { 0 } ∪ { 0 , 1 } ∪ { 0 , 1 , 2 } = { 0 , 1 , 2 } = 3 (b) Most of the time the assignment was up this question was posed to that you needed to start your induction at 1 rather than 0. Thus, it is ok if you did that and also ok if you reindexed to start at 0 .
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