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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.

  1. (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.

  1. (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.

  1. (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)

T

T

T

F

F

T

T

F

F

F

T

F

T

F

F

T

F

F

F

F

F

T

T

F

F

F

T

F

F

F

F

F

T

F

F

F

F

F

F

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

T

T

F

F

T

T

F

F

F

T

F

T

F

F

T

F

F

F

F

F

T

T

F

F

F

T

F

F

F

F

F

T

F

F

F

F

F

F

F

  1. (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.
  1. (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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