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

R and (P

R)) are logically equivalent.

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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.

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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))

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4. Prove by induction that for all natural numbers n, we have:

2 + 2² + 2³ + ... + 2ⁿ = 2ⁿ⁺¹ - 2.

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5. Determine whether the following statements are true or false. If false, briefly explain why?

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6. Let A and B be subsets of a universal set U. Prove that:

(A - B)(B - A) = (AB) - (AB)

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7. Prove that there are an infinite number of primes.

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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:

A,
A

~D),
D

S. By transitivity, the first two implications give S

~D). But, since S and D are logically equivalent, this means that S

~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 = 2² - 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:
2¹ + 2² + 2³ + ... + 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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