discrete math worksheet, Assignments of Discrete Mathematics

discrete math worksheet Worksheet for Sets (Discrete Math) Answers A = {a, b, c, d, e, f} B = {a, c, f}, and C = {b, c, d, g}. Find each of the following: 1. Let (a) A∪B={a,b,c,d,e,f} (b) A∩B = {a, c, f} (c) A∪C={a,b,c,d,e,f,g} (d) A∩C = {b, c, d} (e) A−B = {b, d, e} (f) B−A=∅ (g) B∪C={a,b,c,d,f,g} (h) B∩C = {c} the universal set be the set R of all real numbers and let A = {x ∈ R | -1 < x < 3}, B = 2. Let {x ∈ R | 0 ≤ x ≤ 5}, and C = {x ∈ R | 1≤ x < 7}. Find each of the following: (a) A∪B = { x ∈ R | -1 < x ≤ 5} (b) A∩C = { x ∈ R | 1 ≤ x < 3} (c) C−A={x∈R|3≤x <7} (d) Ac = { x ∈ R | x ≤ -1 or x ≥ 3} (e) Bc = { x ∈ R | x < 0 or x > 5} (f) Ac ∪Bc = { x ∈ R | x < 0 or x ≥ 3} (g) (A∪B)c = { x ∈ R | x ≤ -1 or x > 5} 3. True or False? (a) Z+ ⊆ Q True (b) Z− ∪ Z+ = Z False since zero is an integers but is not an element of (Z− ∪ Z+) (c) Q ∩ R = Q True (d) ∅∈{∅} True (e) 0∈∅ False since an empty set, ∅, means that there is no element in it. 4. LetsetA={x,y,z}.

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2019/2020

Uploaded on 04/29/2023

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Worksheet for Sets (Discrete Math) Answers
1. Let A = {a, b, c, d, e, f} B = {a, c, f}, and C = {b, c, d, g}. Find each of the following:
(a) AB = {a, b, c, d, e, f}
(b) AB = {a, c, f}
(c) AC = {a, b, c, d, e, f, g}
(d) AC = {b, c, d}
(e) A−B = {b, d, e}
(f) B−A =
(g) BC = {a, b, c, d, f, g }
(h) BC = {c}
2. Let the universal set be the set R of all real numbers and let A = {x R | -1 < x < 3}, B =
{x R | 0 x 5}, and C = {x R | 1 x < 7}. Find each of the following:
(a) AB = { x R | -1 < x 5}
(b) AC = { x R | 1 x < 3}
(c) C−A = { x R | 3 x < 7}
(d) Ac = { x R | x -1 or x 3}
(e) Bc = { x R | x < 0 or x > 5}
(f) Ac Bc = { x R | x < 0 or x 3}
(g) (AB)c = { x R | x -1 or x > 5}
3. True or False?
(a) Z+ Q True
(b) Z Z+ = Z False since zero is an integers but is not an element of (Z Z+)
(c) Q R = Q True
(d) {} True
(e) 0 False since an empty set, , means that there is no element in it.
4. Let set A = {x, y, z}.
(a) List all the elements in the power set of (A).
(A) = {, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}}
(b) How many elements in the power set of (A). 8
(c) Let a set X has n elements. How many proper subsets the set X has? 2n-1
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Worksheet for Sets (Discrete Math) Answers

  1. Let A = {a, b, c, d, e, f} B = {a, c, f}, and C = {b, c, d, g}. Find each of the following: (a) A∪B = {a, b, c, d, e, f} (b) A∩B = {a, c, f} (c) A∪C = {a, b, c, d, e, f, g} (d) A∩C = {b, c, d} (e) A−B = {b, d, e} (f) B−A = ∅ (g) B∪C = {a, b, c, d, f, g } (h) B∩C = {c}
  2. Let the universal set be the set R of all real numbers and let A = {x ∈ R | - 1 < x < 3 }, B = {x ∈ R | 0 ≤ x ≤ 5 }, and C = {x ∈ R | 1 ≤ x < 7 }. Find each of the following: (a) A∪B = { x ∈ R | - 1 < x ≤ 5} (b) A∩C = { x ∈ R | 1 ≤ x < 3 } (c) C−A = { x ∈ R | 3 ≤ x < 7 } (d) Ac^ = { x ∈ R | x ≤ - 1 or x ≥ 3 } (e) Bc^ = { x ∈ R | x < 0 or x > 5 } (f) Ac^ ∪Bc^ = { x ∈ R | x < 0 or x ≥ 3 } (g) (A∪B)c^ = { x ∈ R | x ≤ - 1 or x > 5 }
  3. True or False? (a) Z+^ ⊆ Q True (b) Z−^ ∪ Z+^ = Z False since zero is an integers but is not an element of (Z−^ ∪ Z+) (c) Q ∩ R = Q True (d) ∅∈{∅} True (e) 0 ∈∅ False since an empty set, ∅, means that there is no element in it.
  4. Let set A = {x, y, z}. (a) List all the elements in the power set of (A). (A) = {∅, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}} (b) How many elements in the power set of (A). 8 (c) Let a set X has n elements. How many proper subsets the set X has? 2n-
  1. Prove that for all sets A, B, and C, if A ⊆ B and B ⊆ Cc, then A ∩ C = ∅. Proof: Assume that A ∩ C ≠ ∅, which means that there is an element x in A ∩ C. So, x ∈ A and x ∈ C by definition of intersection. Since A ⊆ B (Given) and x ∈ A, then x ∈ B by definition of subset. Since B ⊆ Cc^ (Given) and x ∈ B, then x ∈ Cc^ by definition of subset. So, x ∉ C by definition of complement. However, there is a contradiction where x ∉ C and x ∈ C. Therefore, the assumption is false. Thus, A ∩ C = ∅. Q. E. D.
  2. Prove by using element proof: for all sets A and B, if A ⊆ B, then Bc^ ⊆ Ac. Proof: Let x ∈ Bc. Next show that x ∈ Ac. Since x ∈ Bc, x ∉ B by definition of complement. Since A ⊆ B (Given), if y ∈ A, then y ∈ B by definition of subset. If y ∉ B, then y ∉ A (A conditional statement is logically equivalent to its contrapositive) Since x ∉ B, x ∉ A. So, x ∈ Ac^ by definition of complement. Therefore, Bc^ ⊆ Ac^ by definition of subset. Q. E. D.
  3. Prove using the set identities: (A−B)∪(B−A)∪(A∩B)=A∪B Proof: (A−B)∪(B−A)∪(A∩B) Given = (A ∩ Bc) ∪ (B∩ Ac) ∪ (A∩B) Set Difference Law = [(A ∩ Bc) ∪ (A∩B)] ∪ (B∩ Ac) Commutative and Associative Laws = [A ∩ (Bc^ ∪ B)] ∪ (B∩ Ac) Distributive Laws = A ∩ U ∪ (B∩ Ac) Complement Laws = A ∪ (B∩ Ac) Identity Laws = (A∪B) ∩ (A∪Ac) Distributive Laws = (A∪B) ∩ U Complement Laws = A∪B Identity Laws Q. E. D.