Essentials of Discrete Mathematics
3rd Edition
ISBN: 9781284056242
Author: David J. Hunter
Publisher: Jones & Bartlett Learning
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Question
Chapter 2.5, Problem 4E
To determine
To find:
A pair of incomparable element for poset
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If the set A has 3 elements and the set B = {3, 4, 5}, then find the number ofelements in (A×B).
Let A, B and C be sets. Show that
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A= {2, 3, 5, 8}
B = {1, 3, 5, 7, 9}
C = {3, 5}
D is a set containing 2 elements and D ⊂ A' and D ∩ B = Ø.
(n) List the elements of D.
Chapter 2 Solutions
Essentials of Discrete Mathematics
Ch. 2.1 - Prob. 1ECh. 2.1 - Prob. 2ECh. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10E
Ch. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.2 - Prob. 1ECh. 2.2 - Prob. 2ECh. 2.2 - Prob. 3ECh. 2.2 - Prob. 4ECh. 2.2 - Prob. 5ECh. 2.2 - Prob. 6ECh. 2.2 - Prob. 7ECh. 2.2 - Prob. 8ECh. 2.2 - Prob. 9ECh. 2.2 - Prob. 10ECh. 2.2 - Prob. 11ECh. 2.2 - Prob. 12ECh. 2.2 - Prob. 13ECh. 2.2 - Prob. 14ECh. 2.2 - Prob. 15ECh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - Prob. 18ECh. 2.2 - Prob. 19ECh. 2.2 - Prob. 20ECh. 2.2 - Prob. 21ECh. 2.2 - Prob. 22ECh. 2.2 - Prob. 23ECh. 2.2 - Prob. 24ECh. 2.2 - Prob. 25ECh. 2.2 - Prob. 26ECh. 2.2 - Prob. 27ECh. 2.2 - Prob. 28ECh. 2.2 - Prob. 29ECh. 2.2 - Prob. 30ECh. 2.2 - Prob. 31ECh. 2.2 - Prob. 32ECh. 2.3 - Prob. 1ECh. 2.3 - Prob. 2ECh. 2.3 - Prob. 3ECh. 2.3 - Prob. 4ECh. 2.3 - Prob. 5ECh. 2.3 - Prob. 6ECh. 2.3 - Prob. 7ECh. 2.3 - Prob. 8ECh. 2.3 - Prob. 9ECh. 2.3 - Prob. 10ECh. 2.3 - Prob. 11ECh. 2.3 - Prob. 12ECh. 2.3 - Prob. 13ECh. 2.3 - Prob. 14ECh. 2.3 - Prob. 15ECh. 2.3 - Prob. 16ECh. 2.3 - Prob. 17ECh. 2.3 - Prob. 18ECh. 2.3 - Prob. 19ECh. 2.3 - Prob. 20ECh. 2.3 - Prob. 21ECh. 2.3 - Prob. 22ECh. 2.3 - Prob. 23ECh. 2.3 - Prob. 24ECh. 2.3 - Prob. 25ECh. 2.3 - Prob. 26ECh. 2.3 - Prob. 27ECh. 2.3 - Prob. 28ECh. 2.3 - Prob. 29ECh. 2.3 - Prob. 30ECh. 2.3 - Prob. 31ECh. 2.3 - Prob. 32ECh. 2.3 - Prob. 33ECh. 2.3 - Prob. 34ECh. 2.3 - Prob. 35ECh. 2.3 - Prob. 36ECh. 2.4 - Prob. 1ECh. 2.4 - Prob. 2ECh. 2.4 - Prob. 3ECh. 2.4 - Prob. 4ECh. 2.4 - Prob. 5ECh. 2.4 - Prob. 6ECh. 2.4 - Prob. 7ECh. 2.4 - Prob. 8ECh. 2.4 - Prob. 9ECh. 2.4 - Prob. 10ECh. 2.4 - Prob. 11ECh. 2.4 - Prob. 12ECh. 2.4 - Prob. 13ECh. 2.4 - Prob. 14ECh. 2.4 - Prob. 15ECh. 2.4 - Prob. 16ECh. 2.4 - Prob. 17ECh. 2.4 - Prob. 18ECh. 2.4 - Prob. 19ECh. 2.4 - Prob. 20ECh. 2.4 - Prob. 21ECh. 2.4 - Prob. 22ECh. 2.4 - Prob. 23ECh. 2.4 - Prob. 24ECh. 2.4 - Prob. 25ECh. 2.4 - Prob. 26ECh. 2.4 - Prob. 27ECh. 2.4 - Prob. 28ECh. 2.4 - Prob. 29ECh. 2.4 - Prob. 30ECh. 2.4 - Prob. 31ECh. 2.4 - Prob. 32ECh. 2.4 - Prob. 33ECh. 2.4 - Prob. 34ECh. 2.4 - Prob. 35ECh. 2.4 - Prob. 36ECh. 2.4 - Prob. 37ECh. 2.4 - Prob. 38ECh. 2.4 - Prob. 39ECh. 2.4 - Prob. 40ECh. 2.5 - Prob. 1ECh. 2.5 - Prob. 2ECh. 2.5 - Prob. 3ECh. 2.5 - Prob. 4ECh. 2.5 - Prob. 5ECh. 2.5 - Prob. 6ECh. 2.5 - Prob. 7ECh. 2.5 - Prob. 8ECh. 2.5 - Prob. 9ECh. 2.5 - Prob. 10ECh. 2.5 - Prob. 11ECh. 2.5 - Prob. 12ECh. 2.5 - Prob. 13ECh. 2.5 - Prob. 14ECh. 2.5 - Prob. 15ECh. 2.5 - Prob. 16ECh. 2.5 - Prob. 17ECh. 2.5 - Prob. 18ECh. 2.5 - Prob. 19ECh. 2.5 - Prob. 20ECh. 2.5 - Prob. 21ECh. 2.5 - Prob. 22ECh. 2.5 - Prob. 23ECh. 2.5 - Prob. 24ECh. 2.5 - Prob. 25ECh. 2.5 - Prob. 26ECh. 2.5 - Prob. 27ECh. 2.5 - Prob. 28ECh. 2.5 - Prob. 29ECh. 2.5 - Prob. 30ECh. 2.5 - Prob. 31ECh. 2.6 - Prob. 1ECh. 2.6 - Prob. 2ECh. 2.6 - Prob. 3ECh. 2.6 - Prob. 4ECh. 2.6 - Prob. 5ECh. 2.6 - Prob. 6ECh. 2.6 - Prob. 7ECh. 2.6 - Prob. 8ECh. 2.6 - Prob. 9ECh. 2.6 - Prob. 10ECh. 2.6 - Prob. 11ECh. 2.6 - Prob. 12ECh. 2.6 - Prob. 13ECh. 2.6 - Prob. 14ECh. 2.6 - Prob. 15ECh. 2.6 - Prob. 16ECh. 2.6 - Prob. 17ECh. 2.6 - Prob. 18ECh. 2.6 - Prob. 19ECh. 2.6 - Prob. 20ECh. 2.6 - Prob. 21ECh. 2.6 - Prob. 22ECh. 2.6 - Prob. 23ECh. 2.6 - Prob. 24ECh. 2.6 - Prob. 25ECh. 2.6 - Prob. 26ECh. 2.6 - Prob. 27ECh. 2.6 - Prob. 28ECh. 2.6 - Prob. 29ECh. 2.6 - Prob. 30E
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- Must two different points be collinear? Must three or more points be collinear? Can three or more points be collinear?arrow_forward11. (See Exercise 10.) According to Definition 5.29, is defined in by if and only if . Show that if and only if . 10. An ordered field is an ordered integral domain that is also a field. In the quotient field of an ordered integral domain define by . Prove that is a set of positive elements for and hence, that is an ordered field. Definition 5.29 Greater than Let be an ordered integral domain with as the set of positive elements. The relation greater than, denoted by is defined on elements and of by if and only if . The symbol is read “greater than.” Similarly, is read “less than.” We define if and only if. As direct consequences of the definition, we have if and only if and if and only if . The three properties of in definition 5.28 translate at once into the following properties of in . If and then . If and then . For each one and only one of the following statements is true: . The other basic properties of are stated in the next theorem. We prove the first two and leave the proofs of the others as exercises.arrow_forwardFor the sets given in Exercise 9, is there a distributive relationship for intersection with respect to union? That is, does ABC=ABAC? A=1,2,3,4, B=2,4,6,8, and C=1,3,5,7,9.arrow_forward
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- Let U = {0,1,2,3,4,5,6,7,8,9,10}; A = {0,1,2,3,5,8}; B = down the elements in the following sets. (a) AUB (b) BUC (c) BC (d) (A' n C) UB (e) n((AUC)n B) {0,2,4,6}; C = {1,3,5,7). Writearrow_forward(2c) Let A and B be sets. Prove that A U B = An B if and only if A = B(arrow_forwardThe difference of two sets, A – B, is defined as the set containing all elements of A except those in B. That is, A - B = AN B'. Find A - B for the pair of sets if U = {1, 2, 3, 4, 5, 6, 7, 8, 9}. (Enter your answer in roster notation. Enter EMPTY or Ø for the empty set.) A = {3, 5, 9} and B = {2, 4, 6, 7, 8, 9} A - B =arrow_forward
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