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Unformatted text preview: Math 55: Discrete Mathematics, Fall 2008 Homework 2 Solutions Problems marked * will be corrected fully, out of 10 points. The others will be checked quickly, for 2 points each. 2.3: 16. The following are not the only possible solutions but they are some of the simplest ones. (a) f ( n ) = n + 1. (b) f (0) = 0, f ( n ) = n 1 if n > 0. (c) f (0) = 1, f (1) = 0, f ( n ) = n if n > 1. (d) f (0) = 1, f ( n ) = n if n > 0. 40. (a) f 1 ( S ∪ T ) = { a ∈ A : f ( a ) ∈ S ∪ T } = { a ∈ A : f ( a ) ∈ S or f ( a ) ∈ T } = { a ∈ A : f ( a ) ∈ S } ∪ { a ∈ A : f ( a ) ∈ T } = f 1 ( S ) ∪ f 1 ( T ). (b) Just change ∪ to ∩ and “or” to “and” in the above solution to part (a). A more laborious but also correct way to do this is to verify in each part that the set on the lefthand side is a subset of the one on the right, and conversely. *[5 pts each part] (C) (i) It is true that if f ◦ g is onetoone, then g is onetoone. To prove it, suppose g ( x ) = g ( y ). Then f ◦ g ( x ) = f ◦ g ( y ), which implies x = y since f ◦ g is onetoone. We have shown thatonetoone....
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This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at Berkeley.
 Fall '08
 STRAIN
 Math

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