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Unformatted text preview: Section 6.1 71 Chapter 6: Set Theory The first section of this chapter introduces additional terminology for sets and the concept of an element argument to prove that one set is a subset of another. The aim of this section is to provide a experience with a variety of types of sets and a basis for deriving the set properties discussed in the remainder of the chapter. The second and third sections show how to prove and disprove various proposed set properties of union, intersection, set difference and (general) complement using element arguments, algebraic arguments, and counterexamples. Section 5.4 introduces the concept of Boolean algebra, which generalizes both the algebra of sets with the operations of union and intersection and the properties of a set of statements with the operations of or and and . The section goes on to discuss Russell’s paradox and shows that reasoning similar to Russell’s can be used to prove an important property of computer algorithms. Section 6.1 3. c . Yes. Every element in T is in S because every integer that is divisible by 6 is also divisible by 3. To see why this is so, suppose n is any integer that is divisible by 6. Then n = 6 m for some integer m . Since 6 m = 3(2 m ) and since 2 m is an integer (being a product of integers), it follows that n = 3 · (some integer), and, hence, that n is divisible by 3. 6. a . A * B because 2 ∈ A (because 2 = 5 · 0 + 2) but 2 / ∈ B (because if 2 = 10 b- 3 for some integer b , then 10 b = 5 , so b = 1 / 2 , which is not an integer). b. B ⊆ A Proof : Suppose y is a particular but arbitrarily chosen element of B . [We must show that y is in A . By definition of A, this means that we must show that y = 5 · (some integer) + 2 .] By definition of B , y = 10 b- 3 for some integer b . [Scratch work: Is there an integer, say a, such that y = 5 a +2? If so, then 5 a +2 = 10 b- 3 , which implies that 5 a = 10 b- 5 , or, equivalently, that a = 2 b- 1 . So give this value to a and see if it works.] Let a = 2 b- 1. Then a is an integer and 5 a + 2 = 5(2 b- 1) + 2 = 10 b- 5 + 2 = 10 b- 3 = y . Thus y is in A [as was to be shown] . c . B = C Proof : Part 1, Proof That B ⊆ C : Suppose y is a particular but arbitrarily chosen element of B . [We must show that y is in C . By definition of C, this means that we must show that y = 10 · (some integer) + 7 .] By definition of B , y = 10 b- 3 for some integer b . [Scratch work: Is there an integer, say c, such that y = 10 c + 7? If so, then 10 c + 7 = 10 b- 3 , which implies that 10 c = 10 b- 10 , or, equivalently, that c = b- 1 . So give this value to c and see if it works.] Let c = b- 1. Then c is an integer and 10 c + 7 = 10( b- 1) + 7 = 10 b- 10 + 7 = 10 b- 3 = y ....
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