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Discrete Mathematics Math 6A

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Discrete Mathematics Math 6A
Homework 1

1.1-1 Proposition: A statement that is either true (T) or false (F).
a) Boston is the capital of Massachusetts. (True proposition) b) Mianmi is the capital of Florida (False proposition Tallahassee is the capital) c) 2+3=5 (True) d) 5+7=10 (False) e) x+2 = 11 (not a proposition) f) Answer this question (not a proposition; since it does not assert anything) g) x+y = y+x for every pair of real number x and y (True; commutative law holds for real numbers; x and y are not variables in this sentence)

1.1-8 Express as an English sentence.
Let p, q, and r be the propositions p: You have the flu. q: You miss the final examination. r: You pass the course pq: If you have the flue, then you miss the final exam. ~q > r: You do not miss the finam exam iff you pass the course. q ~r: If you mss the final exam, then you do not pass the course p  q  r: You have the flue, or you miss the final exam, or you pass the course. (p ~r)  (q  ~r): It is either case that If you have the flue, then you do not pass the course, or the case that if you miss the final exam, then you do not pass the course(, or both) (p  q)  (~q  r): Either you have flue and miss the final exam, or you do not miss the final exam and do pass the course

1.1-9 Let p and q be the propositions
p: You drive over 65 miles per hour. q: You get a speeding ticket. Write these propositions using p and q and logical connectives. You do not drive over 65 miles per hour: ~p You drive over 65 miles per hour, but you do not get a speeding ticket: p~q You will get a speeding ticket if you drive over 65 miles per hours: pq If you do not drive over 65 miles per hour, then you will not get a speeding ticket: ~p ~q Driving over 65 miles per hour is sufficient for getting a speeding ticket: pq You get a speeding ticket, but you do not drive over 65 miles per hours: q  ~p Whenever you get a speeding ticket, you are driving over 65 miles per hour: qp

1.1-12 Biconditional are true/false? P  Q: T,F,F,T
2+2=4 iff 1+1=2: T  T, which is true 1+1=2 iff 2+3=4: T  F, which is false It is winter iff it it not spring, summer, or fall.: In the winter, T  T, and in the other seasons it is F  F, both of which are true 1+1=3 iff pigs can fly: F  F, which is true 0>1 iff 2>1: F  T, which is false

1.1-31 Construct a truth table for (p  q)  (r  s)

1.2-4 Use the truth tables to verify the associative law.
(pq)r  p(qr) b) (pq) rp(qr) p q r pq (pq)r qr p(qr) T F p q r pq (pq)r qr p(qr) T F

Tautology: Proposition that is always true. for example: P OR (NOT P).
Contradiction: Proposition that is always false. for example: P AND (NOT P). Determine whether (~q  (pq))  ~p is a tautology (~q  (pq))  ~p = ~{~q  (pq)}  ~p = {q  ~(pq)  ~p} = {q  ~(~p  q)}  ~p = {q  (p  ~q)}  ~p = {(q  p)  (q  ~q)}  ~p = {(q  p)  T}  ~p = (q  p)  ~p = q  (p  ~p) = q  T = T

1.2-14 Show that p  q and (p q)  (~p  ~q) are equivalent.
= (pq)(qp) = (~p  q)  (~q  p) = {(~p  q)  ~q }  {(~p  q)  p} = (~q  q)  (~q  ~p)  (p  ~p)  (p  q) = F  (~q  ~p)  F  (p  q) = (~q  ~p)  (p  q) Show that (pr)  (qr) and (p  q)  r are logically equivalent. (pr)  (qr) = (~p  ~q)  r) = ~(p  q)  r = (p  q)  r

let N(x) be the statement "x has visited North Dakota," where the universe of discourse(domain) consistes of the students in your school. Express each of these quantifications in English. xN(x): Some student in the school has visited North Dakota (=There exists a student in the school who has visited North Dakota) b) xN(x): Every student in the school has visited North Dakota (= All students in the school have visited North Dakota) c) ~xN(x): No student in the school has visited North Dakota (equiv with (f), ~(a)) (= There does not exit a student in the school who has visited North Dakota) d) x~N(x): Some student in the school has not visited North Dakota (equiv with (e),~(b)) (= There exists a student in the school who has not visited north Dakota) e) ~xN(x): It is not true that every student in the school has vistied North DAkota (= not all students i the school have visited North Dakota) f) x~N(x): All students in the school have not visited North Dakota (=Every student in this school has failed to visited North Dakota, = no student has visited North Dakota)

1.3-9 Let P(x) be the statement "x can speak Russian" and let Q(x) be the statement "x knows the computer language C++". Express each of these sentences in terms of P(x), Q(x), quantifiers, and logical connectives. the universe of discourse of quantifiers consists of all students at your school. There is a student at your school who can speak Russian and who knows C++: x(P(x)  Q(x)) b) There is a student at your school who can speak Russian but who doesn't know C++: x(P(x)  ~Q(x)) c) Every student at your school either can speak Russian or knows C++: x(P(x)  Q(x)) d) No student at your school can speak Russian or knows C++: ~x(P(x)  Q(x)) = x~(P(x)  Q(x)) = x(~P(x)  ~Q(x))

1. 3-12 Let Q(x) be the statement "x+1 > 2x
Let Q(x) be the statement "x+1 > 2x." If the universe of discourse consists of all integers, what are these truth values? a) Q(0): True b) Q(-1): True c) Q(1): False d) xQ(x): True e) xQ(x): False f) x~Q(x): True g) x~Q(x): False

Suppose that the universe of discourse of the propositional function P(x) consists of the integers 1,2,3,4, and 5. Express these statements without using quantifiers, instead using only negations, disjunctions, and conjuctions. a) xP(x): P(1)  P(2)  P(3)  P(4)  P(5) b) xP(x): P(1)  P(2)  P(3)  P(4)  P(5) c) ~xP(x): ~(P(1)  P(2)  P(3)  P(4)  P(5)) d) ~xP(x): ~(P(1)  P(2)  P(3)  P(4)  P(5)) e) x(x ≠ 3)  P(x))  x~P(x): ((1 ≠ 3)  P(1))  (2 ≠ 3)  P(2))  (3 ≠ 3)  P(3))  (4 ≠ 3)  P(4)) (5 ≠ 3)  P(5))  (~P(1)  ~P(2)  ~P(3)  ~P(4)  ~P(5)) = (P(1)  P(2)  P(4)  P(5))  (~P(1)  ~P(2)  ~P(3)  ~P(4)  ~P(5))

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