Economics 2P30 Examination
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. ∗ ∗ ∗ ∗ ∗ ∗ ∗
- (3%) Tautology
- (3%) Union of X and Y
- (3%) Proposition
- (3%) Power set of X
- (3%) Intersection of X and Y
Solution:
- A propositional form that is always true.
- X ∪ Y = { x ∶ x ∈ X ∨ x ∈ Y } .
- A statement which is either true or false.
- The set of all subsets of X.
- X ∩ Y = { x ∶ x ∈ X ∧ x ∈ Y } .
Section B: Proofs
∗ ∗ ∗ ∗ ∗ ∗ ∗ Choose 3 of the following 4 questions. ∗ ∗ ∗ ∗ ∗ ∗ ∗ True or false? If true, prove. If false, derive a counterexample.
- (16%) If A ⊂ B and B ⊂ C then B ⊂ A ∩ C.
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.
- (16%) If x and y are both odd, then x + y is odd.
Solution: False. For example, x = 3 and y = 5 then x + y = 8 which is even since 8 = 2 ⋅ 4.
- (16%) ∀n ∈ N, 2 + 22 + 23 + ⋯ + 2n = 2n+1 −
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.
- (16%) For every set X, X ∈ P(X) and ∅ ∈ P(X).
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. ∗ ∗ ∗ ∗ ∗ ∗ ∗
- (20%) Suppose P, Q and R are atomic propositions.
- Derive the truth tables for the following two propositional forms.
- ∼ [(P ∧ R) ⇒ (∼ Q ∨ R)] ii. (Q∧ ∼ R) ∧ (P ∧ R)
- Are the two propositional forms equivalent? Why or why not?
- Find another propositional form which is equivalent to (i) above.
Solution:
(a) The truth tables are:
P R Q ∼ [(P ∧ R) ⇒ (∼ Q ∨ R)] (Q∧ ∼ R) ∧ (P ∧ R)
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- Yes they are equivalent since their truth tables are identical.
- For example, the proposition [(P∧ ∼ P) ∧ R] ∧ Q is equivalent since:
P Q R ∼ [(P ∧ R) ⇒ (∼ Q ∨ R)] [(P∧ ∼ P) ∧ R] ∧ Q
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- (20%) Let the Universe be U = { 1,2,3,4,5,6} and let X = { 2,4,5,6} and Y = { 1,2,3,4} .
- Is X ⊆ Y ? Why or why not?
- Find X ∩ Y .
- Find Xc and Y c.
- Find (X ∩ Y )c.
- Is there a relationship between (c) and (d)? Explain in detail.
Solution:
- No since 5 ∈ X but 5 ∉ Y .
- X ∩ Y = { 2,4} .
- Xc = { 1,3} and Y c = { 5,6} .
- (X ∩ Y )c = { 1,3,4,5} .
- By De Morgan’s Law, (X ∩ Y )c = Xc ∪ Y c.
- (20%) Let X, Y , and Z be sets. Assume that all three sets are nonempty. Find a single example for sets X, Y , and Z so that all of the following properties are true. Be clear and make sure you specify the Universe, denoted U. Clearly demonstrate that your example is true.
- X ∩ Y = ∅
- ((X ∪ Y ) ∪ Z) ⊂ U (c) X ∩ Z ≠ ∅
- Y ∩ Z ≠ ∅
- X ⊂ Y c
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).
