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Unformatted text preview: Week 1 Solutions MATH 233 Section 1.3, Problem# 4 p =A is truthful, p=A is lying q= B is truthful, q=B is lying r=C is truthful, r=C is lying # 4a: A is lying and B or C is truthful = P /\ (q \/ r) #4b : A and B are lying, or A and C are truthful= (p/\q) \/(p/\r) #4c : At least two people are telling the truth = Two or more people are telling the truth = (p/\q) \/(p/\r) \/(q/\r) \/(p/\q/\r) #4d : Exactly two people are telling the truth= [ (p/\q) \/(p/\r) \/(q/\r) ] /\ (p/\q/\r) From [ (p/\q) \/(p/\r) \/(q/\r) ] , you had to exclude the possibility that all three are telling the truth. Section 1.3, Problem# 12a : Either everyone is not hungry at mealtime, or everyone is tired and the snackbar makes a profit. h= everyone is hungry at mealtime t= everyone is tired p= a profit is made by the snackbar h t p  t /\p  h\/( t/\p) T T T _ T _____T T T F _ F _____F T F T _ F _____F T F F _ F _____F F T T _ T _____T F T F _ F _____T F F T _ F _____T F F F _ F _____T Section 1.4, Problem# 8 : # 8a : Every biology major is required to take geometry. Let G(s)= s is required to take geometry (this is the predicate) B = the set of biology majors (this is the domain) quantified predicate: s in B, G(s) 10c : If x is even, then x <=12 This is True , since all the even numbers 8 and 10 are <=12. # 8b : There are computer science majors who do not minor in mathematics. Let M(s)= s minors in mathematics (this is the predicate) C = the set of computer science majors (this is the domain) quantified predicate: s in C, M(s) # 8c : There is no math major who is required to take a business course. Let B(s)= s is required to take a business course (this is the predicate) M = the set of math majors (this is the domain) quantified predicate: s in M, B(s) Alternate answer: s in M, B(s) # 8d : There are puzzles that have no solution. Let S(x)= x has a solution (this is the predicate) P = the set of puzzles (this is the domain) quantified predicate: s in P, S(x) Alternate answer: s in M, B(s) Section 1.5, Problem # 4a : p q p /\q (p /\ q) >q T T _ T _____T T F _ F _____T F T _ F _____T F F _ F _____T Section 1.5, Problem # 10: D={1,3,5,7,8,10,11,12} 10b : If x is odd, then x>7. This is False , since x=1 is odd and 1 is not >7 (Note that x=3,5,7 are counterexamples, too) Section 1.6, Problem #2b: Let = This number is a perfect square. = The equation has a rational solution. llowing form: a converse fallacy . Truth table for (p /\ q)\/ (q >p) p q p /\q  q >p (p /\ q)\/ (q >p) column has all T values....
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 Summer '10
 Li
 Math

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