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 âˆ¶ x ∈ X ∨ x ∈ Y }.
- A statement which is either true or false.
- The set of all subsets of X.
- X ∩ Y = {x âˆ¶ 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 + â‹¯ + 2n = 2n+1 −
Solution: For n _{= }1 we have 2^{1 }_{= }2^{1}^{+1 }_{−}2 and hence, the statement is true for n _{= }1. Now assume 2 + 22 + â‹¯ + 2n = 2n+1 − 2. We need to show that 2 + 22 + â‹¯ + 2n + 2n+1 = 2n+2 − 2. Naturally: 2 + 22 + â‹¯ + 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)
T
F
T
F
T
F
T
F
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T |
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T |
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F |
- 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
T
F
T
F
T
F
T
F
T |
T |
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T |
T |
F |
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T |
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F |
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F |
- (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 X^{c }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 }_{= }X^{c }_{∪ }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).
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