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Unformatted text preview: 34 Chapter 2 Basic Structures: Sets, Functions, Sequences, and Sums CHAPTER 2
Basic Structures: Sets, Functions, Sequences, and Sums SECTION 2.1 Sets 2. 10. 12. 14. There are of course an inﬁnite number of correct answers.
a) {371  n:0,1,2,3,4} or {3: I :t' is a multiple of BAD g a: S 12}.
b) {3: i —3 S a: S 3 }, where we are assuming that the domain (universe of discourse) is the set of integers. c) {3:  3; is a letter of the word monopoly other than 1' or y . Each of the sets is a subset of itself. Aside from that, the only relations are B Q A, C Q A, and C g D. a) Since the set contains only integers and {2} is a set, not an integer, is not an element.
b) Since the set contains only integers and {2} is a set, not an integer, {2} is not an element.
0) The set has two elements. One of them is patently {2}.
d) The set has two elements. One of them is patently e) The set has two elements. One of them is patently {2}. f) The set has only one element, {{2}}; since this is not the same as {2} (the former is a set containing a
set, whereas the latter is a set containing a number), {2} is not an element of a) true b) true 0) false—sec part (a) d) true I e) trueithe one element in the set. on the left is an element of the set on the right, and the sets are not equal
1') true—similar to part (e) g) false—the two sets are equal The numbers 1, 3, 5, 7, and 9 form a subset of the set of all ten positive integers under discussion, as shown here. “7e allow B and C to overlap, because we are told nothing about their relationship. The set A must be a
subset of each of them, and that forces it to be positioned as shown. “7e cannot actually show the properness
of the subset relationships in the diagram, because we don’t know where the elements in B and C that are Section 2.}. 16. 18. 20. 22. 24. 26. 28. 30. 32. Sets 35
not in A are located—there might be only one (which is in both B and C), or they might be located in
portions of B and/or C outside the other. Thus the answer is as shown, but with the added condition that
there must be at least. one element of B not in A and one element of C not in A. G) Since the empty set is a subset of every set, we just need to take a set B that contains (3! as an element. Thus
we can let A : Q7 and B = {(2)} as the simplest example. The cardinality of a set is the number of elements it has. a) The empty set has no elements, so its cardinality is 0. b) This set has one element (the empty set), so its cardinality is '1.
c) This set has two elements, so its cardinality is 2. d) This set has three elements, so its cardinality is 3. The union of all the sets in the power set of a set X must be exactly X. In other words, we can recover X
from its power set, uniquely. Therefore the answer is yes. a) The power set. of every set includes at least the empty set, so the power set cannot be empty. Thus (3 is
not the power set of any set. b) This is the power set of c) This set has three elements. Since 3 is not a power of ‘2, this set cannot be the power set of any set. d) This is the power set of {(1,1)}.
By deﬁnition it is the set of all ordered pairs (c,p) such that e is a course and p is a. professor. "We can conclude that A : Q) 01' B : Q). To prove this, suppose that neither A nor B were empty. Then
there would be elements a E A and b E B. This would give at last one element, namely (tub), in A X B, so
A x B would not be the empty set. This contradiction shows that either A or B (or both, it goes without saying) is empty. In each case the answer is a set of 3—tuples. C ) {
d) {inn"6.413).(way).(mm).(acuity).(their).(y,:v.:y).(y.y.$)a{yew} Suppose A 7E B and neither A nor B is empty. "We must prove that A x B gé B x A. Since A 7% B, either
we can ﬁnd an element .1' that is in A but not B, or vice versa. The two cases are similar, so without loss of
generality, let us assume that :1: is in A but not B. Also, since B is not empty, there is some element y E B.
Then (:r,y) is in A x B by deﬁnition, but it is not in B X A since .1: ¢ B. Therefore A X B 7% B x A. The only difference between (A x B) X (C x D) and A x (B X C) x D is parentheses, so for all practical purposes
one can think of them as essentially the same thing. By Deﬁnition 9, the elements of (A X B) X (C x D)
consist of ordered pairs y), where a: E A X B and y E C X D, so the typical element of (A X B) X (C x D) 36 34. 36. 38. Chapter 2 Basic Structures: Sets, ﬂinctions, Sequences, and Sums
looks like ((a, b), (c, By Deﬁnition 10, the elements of A x (B x C) x D consist of 3—tuples ((1,3),d), where a E A, d E D, and .1; E B x C, so the typical element of A x (B x C) x D looks like (a,(b,c),d). The structures ((e,b), (c,c£)_) and (n, (b, c),d) are different, even if they convey exactly the same information (the ﬁrst is a pair, and the second is a 3—tuple). To be more precise, there is a natural onetoone correspondence between (A x B) X (C x D) and A X (B x C) x D given by ((a,b), (c, (1)) H (n, (b,c),d). 3) There is a real number whose cube is —1. This is true, since a: = —1 is a solution. b) There is an integer such that the number obtained by adding 1 to it is greater than the integer. This is
truePin fact, every integer satisﬁes this statement. c) For every integer, the number obtained by subtracting 1 is again an integer. This is true. (1) The square of every integer is an integer. This is true. In each case we want the set of all values of m in the domain (the set of integers) that satisfy the given equation
or inequality. 3) It is exactly the positive integers that satisfy this inequality. Therefore the truth set is E Z l x3 Z 1} 2
{51:6 le2 1} : {1,2,3,...}. b) The square roots of 2 are not integers, so the truth set is the empty set, E). (3) Negative integers certainly satisfy this inequality, as do all positive integers greater than 1. However, 0 7i 02
and lyﬁ 12. Thus the truth set is {cc 6 Z  .1; < $2} = E Z  30% 0A3: 545 1} z {...,—3,72,—1,2,3,...}. a) If S E S, then by the deﬁning condition for S we conclude that .S' 9! S, a contradiction.
b) If S {515 S, then by the deﬁning condition for S we conclude that it is not the case that S Q! 8 (otherwise 3 would be an element of 5'), again a contradiction. ...
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This note was uploaded on 02/18/2012 for the course MATH 640:244 taught by Professor Ming during the Spring '09 term at Rutgers.
 Spring '09
 Ming

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