55PS2 - SOLUTION SET 6DUE 3/12/2008 This solution set is...

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SOLUTION SET 2—DUE 2/6/2008 Please report any errors in this document to Ian Sammis ( [email protected] ). Problem 1 (#2.1.8) . Determine whether these statements are true or false. a) ∅ ∈ {∅} b) ∅ ∈ {∅ , {∅}} c) {∅} ∈ {∅} d) {∅} ∈ {{∅}} e) {∅} ⊂ {∅ , {∅}} f) {{∅}} ⊂ {∅ , {∅}} g) {{∅}} ⊂ {{∅} , {∅}} Solution. a) True—the empty set is actually a member of the set on the right. b) True again—once again, the empty set is actually in the set on the right. c) This one’s false. The empty set is in the set on the right, but the set containing the empty set is not. d) This is true—the set containing the empty set is an element (in fact, the only element) of the set on the right. e) True—since the empty set is in the set on the right, every element of the Frst set is in the containing set. f) True again, for the same reason. g) ±alse. The set on the right is nothing but a deeply stupid way of writing {{∅}} . Thus the two sets are actually equal. In this book (remember— not in all books!) A B is false if A = B . Problem 2 (#2.1.16) . Find two sets A and B such that A B and A B . Solution. ±rom the previous problem, A = {∅} , B = {∅ , {∅}} works. Problem 3 (#2.1.32) . Explain why ( A × B ) × ( C × D ) and A × ( B × C ) × D are not the same. Solution. Let a,b,c,d be members of A,B,C,D , respectively. A typical element of ( A × B ) × ( C × D ) is (( a,b ) , ( c,d ))—an ordered pair of two ordered pairs. A typical element of A × ( B × C ) × D is ( a, ( b,c ) ,d )—an ordered triplet, the center element of which is an ordered pair. Although each contains the same information (one element each from A,B,C, and D ), the information is organized di²erently in each case. Problem 4 (#2.1.34) . Translate each of these quanti±cations into English and determine its truth value. a) x R ( x 3 n = 1) . b) x Z ( x + 1 > x ) . c) x Z ( x 1 Z ) . d) x Z ( x 2 Z ) . Solution. In English: 1
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2 SOLUTION SET 2—DUE 2/6/2008 a) There exists some real number whose cube is not -1. Obviously true (0, for example). b) There exists an integer such that the next integer is Yes–in fact all of them. c) One less than every integer is, in fact, an integer. Again. .. obviously, yes. d) The square of every integer is an integer. Of course it is. Problem 5 (#2.1.36) . Find the truth set of each of these predicates where the domain is the set of integers. a) P ( x ) : “ x 3 1” b) Q ( x ) : “ x 2 = 2” c) R ( x ) : “ x < x 2 Solution. It’s important here to remember that we’re on the integers, not the reals. a) Since every integer except zero has a cube that’s at least one, the truth set is Z \ { 0 } .
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This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.

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55PS2 - SOLUTION SET 6DUE 3/12/2008 This solution set is...

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