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C__work_280sol1

# C__work_280sol1 - Solutions to Homework 1 Note If you dont...

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Solutions to Homework 1 February 10, 2007 Note: If you don’t understand how a question was graded then (1) ﬁrst read the solutions, (2) check with the grader who graded it, and then (and only then) see Prof. Halpern. There is no statute of limitations on complaints about grading. General comment: You should try to make their proofs clear and suc- cint, and make sure that each step follows logically. Remember, the graders can’t read your minds! You have to lay out your reasoning clearly for them. Also, writing neatly and laying things out neatly helps. Remember, they’re all grading about 85 solutions! 0.1, 11 No justiﬁcation was required for this problem because of its simplic- ity. However, points were deducted for an incorrect or unclear justiﬁcation (sometimes less is more!). For those who are interested, here is a formal proof: A B = A : If x ( A B ), then x A and x B (thus x A ). If x A , then x B since A B . So x A and x B or x ( A B ). Therefore A B = A if A B . A B = B : If x ( A B ), then x A or x B . If x B , then we are done. If x A , then again, we are done. Since A is a subset of B , x A x B . Conversely, if x B , then from the deﬁnition of union, x ( A B ). Therefore A B = B if A B . 1

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Note: in general, to prove set equality, show that for every x A , x B , and vice versa. 0.1, 14 A B S . 0.1, 27 (b) { ( A,B ) } | A B } (this is the superset relation) (c) { ( x,y ) | x y } Comments: Note that the reverse of { ( x,y ) | x y } is not { ( y,x ) | y x } . They are just the same sets. Also, many students seem to have problems with specifying relations formally. I tried to be lenient with this, but some mistakes were penalized. For example, some students failed to write a relation as a SET of ORDERED PAIRS. For example, in 27(c) students often wrote: { x,y | x y } or x y or just . I considered the last of these minimally acceptable, but the ﬁrst two were not. Section 0.1, 33 (a) Injective; not surjective since 1 is not an image of anything. (b) Injective; not surjective since 0 is not an image of anything. (c) Bijective.
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C__work_280sol1 - Solutions to Homework 1 Note If you dont...

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