MATH3030 Homework
Hong Yue,
Georgia College, Department of Mathematics,
Milledgeville, Georgia, 31061, USA, [email protected]
September 10, 2013
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Week 1 Homework
P12-13:
Exercise 1.2.1. Which of the following expressions are statements?
(3) Is it going to snow tomorrow?
(4) The U.S. has 49 states.
(5) I like to eat fruit, and you often think about traveling to Spain.
(7) Call me on Thursday if you are home.
Solution: (4) and (5) are statements. (3) and (7) are not.
Exercise 1.2.2. Which of the following expressions are statements?
(2) If x ≥ 2 then x3 ≥ 1.
(4) x + y = z.
(6) a2 + b2 = c2 .
Solution: (2) is a statement. (4) and (6) are not.
Exercise 1.2.3. Let P =I like fruit, let Q = I do not like cereal and R =
I know how to cook an omelette. Translate the following statements into
words.
(1) P ∧ Q: I like fruit and do not like cereal.
(2) Q ∨ R: I do not like cereal or I know how to cook an omelette.
(3) ¬R: I don’t know how to cook an omelette.
(4) ¬(P ∨ Q): It is not the case that I like fruit or I do not like cereal.
Or: I don’t like fruit and I do like cereal.
(5) ¬P ∨ ¬Q: I don’t like fruit or I do like cereal.
(6) ¬P ∨ Q: I don’t like fruit or I don’t like cereal.
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(7) (R ∧ P ) ∨ Q: I know how to cook an omelette and I like fruit, or I
do not like cereal.
(8) R ∧ (P ∨ Q): I know how to cook an omelette and, I like fruit or I
don’t like cereal.
Exercise 1.2.5. Let X = Fred has red hair, let Y = Fred has a big nose and
R = Fred likes to eat figs. Translate the following statements into symbols.
(1) Fred does not like to eat figs: ¬R
(3) Fred has red hair or he likes to eat figs: X ∨ R
(5) Fred likes to eat figs and he has red hair, or he has a big nose:
(R ∧ X) ∨ Y
(7) It is not the case that Fred has a big nose, or he has red hair: ¬Y ∨X
Exercise 1.2.15. Let P be a statement, let T be a tautology and let C be a
contradic- tion.
(1) Show that P ∨ T is a tautology:
P
T
F
1
T
T
T
2
P ∨T
T
T
3
We see in column 3 that the statement P ∨ T is always true, regardless
of whether P is true or false. Therefore, P ∨ T is a tautology.
(2) Show that P ∧ C is a contradiction:
P
T
F
1
C
F
F
2
P ∧C
F
F
3
We see in column 3 that the statement P ∧ C is always false, regardless
of whether P is true or false. Therefore, P ∧ C is a contradiction.
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P22.
Exercise 1.3.1. Let P, Q, R and S be statements. Show that the following
are true. (1) ¬(P → Q) ⇒ P.
Proof : We use the truth table to show that ¬(P → Q) → P is a tautology:
P
T
T
F
F
1
Q
T
F
T
F
2
P →Q
T
F
T
T
3
¬(P → Q)
F
T
F
F
4
→
T
T
T
T
5
P
T
T
F
F
6
We see in Column 5 that the statement ¬(P → Q) → P is always true,
and hence it is a tautology. Therefore, ¬(P → Q) ⇒ P .
Exercise 1.3.3. (1) Let P be a statement, let T be a tautology and let C be
a contradiction. Show that (1) P ∧ T ⇔ P ; (2) P ∨ C ⇔ P .
P
T
F
1
T
T
T
2
P ∧T
T
F
3
Columns 1 and 3 are identical, which means that P ∧ T ⇔ P .
P
T
F
1
C
F
F
2
P ∨C
T
F
3
Columns 1 and 3 are identical, which means that P ∨ C ⇔ P .
Exercise 1.3.4. For each pair of statements, determine whether or not the
first implies the second.
(1) If you will kiss me I will dance a jig, and I will dance a jig; and you
will kiss me.
Let P =you will kiss me, Q =I will dance a jig, Translate the implication
from the first statement to the second into symbols, we need to determine
whether or not (P → Q) ∧ Q ⇒ P , or, whether or not (P → Q) ∧ Q → P is
a tautology.
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P
T
T
F
F
1
Q
T
F
T
F
2
P →Q
T
F
T
T
3
(P → Q) ∧ Q
T
F
T
F
4
→
T
T
F
T
5
P
T
T
F
F
6
We see in Column 5 that the statement (P → Q) ∧ Q → P is NOT
always true, hence it is NOT a tautology. Therefore, the first statement
doesn’t imply the second.
Actually, we only need one example to show the implication doesn’t hold:
P
F
1
Q
T
2
P →Q
T
3
(P → Q) ∧ Q
T
4
→
F
5
P
F
6
(3) If cars pollute then we are in trouble, and cars pollute; and we are
in trouble.
Let P =cars pollute, Q =we are in trouble, Translate the implication
from the first statement to the second into symbols, we need to determine
whether or not (P → Q) ∧ P ⇒ Q. It is true by rule 1, Modus Ponens, in
Fact 1.3.1.
(5) Vermeer was a musician or a painter, and he was not a musician; and
Vermeer was a painter.
The answer for this one is also yes by rule 7, Modus Tollendo Ponens.
in Fact 1.3.1.
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Exercise 1.3.5. For each pair of statements, determine whether or not the
two state- ments are equivalent.
(2) This shirt has stripes, and it has short sleeves or a band collar; and
this shirt has stripes and it has short sleeves, or it has a band collar.
Solution: Let P =This shirt has stripes, Q =it has short sleeves, and R =it
has a band collar. Translate the equivalence between the two statements into
symbols, we need to determine whether or not P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ R.
They are not equivalent. See the counterexample below:
P
F
1
Q
F
2
R
T
3
P ∧ (Q ∨ R)
F
4
(P ∧ Q) ∨ R
T
5
(4) The cat is gray, or it has stripes and speckles; and the cat is gray or
it has stripes, and the cat is gray or it has speckles.
Solution: Two statements are eqivalent by rule 7 in Fact 1.3.2. (Distributive
Law).
Eercise 1.3.8. State the inverse, converse and contrapositive of each of the
follow- ing statements.
(1) If its Tuesday, it must be Belgium.
Contrapositive: If it is not Belgium, it must not be Tuesday.
Converse: If it is Belgium, it must be Tuesday.
Inverse: If its not Tuesday, it must not be Belgium.
(3) Good fences make good neighbors.
Contrapositive: Bad neighbors make bad fences.
Converse: Good neighbors make good fences .
Inverse: Bad fences make bad neighbors.
(5) If you like him, you should give him a hug.
Contrapositive: If you don’t give him a hug, you don’t like him.
Converse: If you give him a hug, you like him.
Inverse: If you don’t like him, you should not give him a hug.
Eercise 1.3.10. Negate each of the following statements.
(1) e5 > 0: e5 ≤ 0
(5) w − 3 > 0 implies w2 + 9 > 6w:
It is not the case that w − 3 > 0 implies w2 + 9 > 6w.
Or: There are some values of w such that w − 3 > 0 and w2 + 9 ≤ 6w.
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