Economics 2P30
Foundations of Economic Analysis
Department of Economics
Midterm Examination #1 - Suggested Solutions
Section A: Definitions
∗ ∗ ∗ ∗ ∗ ∗ ∗ Define 4 of the following 5 terms in two sentences or less. ∗ ∗ ∗ ∗ ∗ ∗ ∗
Solution:
Section B: Proofs
∗ ∗ ∗ ∗ ∗ ∗ ∗ Choose 3 of the following 4 questions. ∗ ∗ ∗ ∗ ∗ ∗ ∗ True or false? If true, prove. If false, derive a counterexample.
Solution: False. Let A = {1}, B = {1,2} and C = {1,2,3}. Then we have A ⊂ B and B ⊂ C. However, A ∩ C = {1} and hence B ⊂/ A ∩ C.
Solution: False. For example, x = 3 and y = 5 then x + y = 8 which is even since 8 = 2 ⋅ 4.
Solution: For n = 1 we have 21 = 21+1 −2 and hence, the statement is true for n = 1. Now assume 2 + 22 + ⋯ + 2n = 2n+1 − 2. We need to show that 2 + 22 + ⋯ + 2n + 2n+1 = 2n+2 − 2. Naturally: 2 + 22 + ⋯ + 2n + 2n+1 = 2n+1 − 2 + 2n+1 = 2(2n+1) − 2 = 2n+2 − 2. Therefore, the statement is true.
Solution: True. We need to show that for all sets X, X ⊂ X and ∅ ⊂ X. For the former, for all x, x ∈ X ⇒ x ∈ X is a tautology. Therefore, ∀X, X ∈ P(X). For the latter, x ∈ ∅ is always false. Hence x ∈ ∅ ⇒ x ∈ X is always a true statement. Therefore, ∅ ⊂ X.
Section C: Analytical
∗ ∗ ∗ ∗ ∗ ∗ ∗ Choose 2 of the following 3 questions. ∗ ∗ ∗ ∗ ∗ ∗ ∗
Solution:
(a) The truth tables are:
P R Q ∼ [(P ∧ R) ⇒ (∼ Q ∨ R)] (Q∧ ∼ R) ∧ (P ∧ R)
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P Q R ∼ [(P ∧ R) ⇒ (∼ Q ∨ R)] [(P∧ ∼ P) ∧ R] ∧ Q
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Solution:
Solution: An infinite number of such examples exist. For example, let U = {1,2,3,4},
X = {1}, Y = {3} and Z = {1,2,3}. It follows that X ∩ Y = ∅ since there are no elements that are common to both sets ((a) is satisfied). Since X,Y,Z ⊂ U, it naturally follows that (X ∪ Y ) ∪ Z ⊂ U ((b) is satisfied). X ∩ Z = {1} ≠ ∅ ((c) is satisfied). Y ∩ Z = {3} ≠ ∅ ((d) is satisfied). Lastly, since X ∩ Y = ∅, X ∪ Y c immediately follows. One can verify since Y c = {1,2,4,5,6} and since X = {1} it is obvious that X ⊂ Y c ((e) is satisfied).
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