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. ∗ ∗ ∗ ∗ ∗ ∗ ∗

1. (3%) Tautology
2. (3%) Union of X and Y
3. (3%) Proposition
4. (3%) Power set of X
5. (3%) Intersection of X and Y

Solution:

1. A propositional form that is always true.
2. X ∪ Y = {x ∶ x ∈ X ∨ x ∈ Y }.
3. A statement which is either true or false.
4. The set of all subsets of X.
5. 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.

1. (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.

2. (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.

3. (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.

4. (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. ∗ ∗ ∗ ∗ ∗ ∗ ∗

1. (20%) Suppose P, Q and R are atomic propositions.
   • Derive the truth tables for the following two propositional forms.

1. ∼ [(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)

• 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

2. (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.

3. (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).