## Math 3000 Sample Exam I

There are seven (7) questions listed below, you may choose to do any five (5) of them. Each
question will count 20 points. If you complete more than 5 questions, you must clearly
indicate which 5 you wish to be graded. There will be no extra credit - don't bother asking
for it. **Good Luck!!.**

**1.** Use a truth table to show that the two statements (P Q) R and (P (Q R)) are
logically equivalent.
*Click here for answer.*

**2.** Use logical symbols to rewrite these statements and determine if the argument is valid (that
is, the premise [first three statements] implies the conclusion [last statement]).
- If turtles can sing, then artichokes can fly.
- If artichokes can fly, then turtles can sing and dogs can't play chess.
- Dogs can play chess if and only if turtles can sing.
- Therefore, turtles can't sing.

*Click here for answer.*

**3.** If P(x) and Q(x) are propositional functions involving x, show that
(x)(P(x)Q(x))((x)P(x) (x)Q(x))

*Click here for answer.*

**4.** Prove by induction that for all natural numbers n, we have:
2 + 2^{2} + 2^{3} + ... + 2^{n} = 2^{n+1} - 2.

*Click here for answer.*

**5.** Determine whether the following statements are true or false. If false, briefly explain why?
*Click here for the answer.*

**6**. Let A and B be subsets of a universal set U. Prove that:
(A - B)(B - A) = (AB) - (AB)

*Click here for answer.*

**7.** Prove that there are an infinite number of primes.
*Click here for answer.*

Answer to question 1:

P | Q | R | PQ | (PQ) R | QR | P(QR) |
---|

T | T | T | T | **T** | T | **T** |

T | T | F | T | **F** | F | **F** |

T | F | T | F | **T** | T | **T** |

T | F | F | F | **T** | T | **T** |

F | T | T | F | **T** | T | **T** |

F | T | F | F | **T** | F | **T** |

F | F | T | F | **T** | T | **T** |

F | F | F | F | **T** | T | **T** |

Since the fifth and seventh columns are identical, the two expressions are logically equivalent.
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Answer to question 2:
Let S = "turtles can sing", A = "artichokes can fly" and D = "dogs can play chess". The given statements can now be written symbolically as:

SA,

A(S~D),

DS. By transitivity, the first two implications give S(S~D). But, since S and D are logically equivalent, this means that S(D~D), which is a contradiction. Since a true statement cannot imply a contradiction, we must have that S is false, i.e., ~S is true - turtles can not sing.
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Answer to question 3:

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Answer to question 4:
Let A be the set of all natural numbers for which the statement is true.

The statement, with n = 1, is 2 = 2^{2} - 2. As the RHS evalutes to 2, the statement is true in this case, and so, 1 is in A. Now, assume that the statement is true for the integer m, and consider n = m + 1. We have:

2^{1} + 2^{2} + 2^{3} + ... + 2^{m} + 2^{m+1} =
2^{m+1} - 2 + 2^{m+1} = 2(2^{m+1}) - 2 = 2^{m+2} -2, showing that
the statement is true for n = m + 1. Thus, A is an inductive set and by PMI the statement is true for all n.

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Answer to question 5:

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Answer to question 6:
x(A - B)(B - A)((xA)(xB))((xB)(xA))

(((xA)(xB))(xB))(((xA)(xB))(xA))

(((xA)(xB))((xB)(xB)))(((xA)(xA))((xB)(xA)))

(((xA)(xB))((xB)(xA)))

(xAB)(xBA)

x(AB) - (BA)

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Answer to question 7:
See returned Special Assignment #1.

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