MATH 135
Fall 2005
Assignment 1 Solutions
HandIn Problems
Exercise 112:
Write down the truth tables for each expression. NOT
P
=
⇒
(
Q
⇐⇒
R
).
Solution:
P
Q
R
NOT
P
Q
⇐⇒
R
NOT
P
=
⇒
(
Q
⇐⇒
R
)
T
T
T
F
T
T
T
T
F
F
F
T
T
F
T
F
F
T
T
F
F
F
T
T
F
T
T
T
T
T
F
T
F
T
F
F
F
F
T
T
F
F
F
F
F
T
T
T
Exercise 122:
Show that the statements NOT (
P
OR
Q
) and (NOT
P
) AND (NOT
Q
) have the same truth tables and give an
example of the equivalence of these statements in everyday language.
Solution:
P
Q
P
OR
Q
NOT (P OR Q)
T
T
T
F
T
F
T
F
F
T
T
F
F
F
F
T
P
Q
NOT
P
NOT
Q
(NOT
P
) AND (NOT
Q
)
T
T
F
F
F
T
F
F
T
F
F
T
T
F
F
F
F
T
T
T
The ﬁnal columns of each table are the same, so the two statements have the same truth tables.
This equivalence can be illustrated in everyday language. Consider the statement “I do not want pizza or doughnuts”.
This means that “I do not want pizza” and “I do not want doughnuts”.
Exercise 134:
Express each statement as a logical expression using quantiﬁers. State the universe of discourse.
There is no smallest positive real number.
Solution:
The universe of discourse is the set of all positive real numbers. The statement “there is no smallest positive real
number” is equivalent to “for every positive real number, there is a smaller positive real number” which is equivalent
to
∀
r
∃
x,
(
x < r
)
.
1
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View Full DocumentExercise 138:
Express each statement as a logical expression using quantiﬁers. State the universe of discourse.
For every real number
y
, there is a real number
x
such that
x
3
+
x
=
y
.
Solution:
If we assume that the universe of discourse is the set of real numbers, we can express the statement as
∀
y
∃
x,
(
x
3
+
x
=
y
)
.
Exercise 142:
Negate each expression, and simplify your answer.
∀
x,
((
P
(
x
) AND
Q
(
x
)) =
⇒
R
(
x
)).
Solution:
Using Example 1.23, NOT (
A
=
⇒
B
) is equivalent to
A
AND (NOT
B
), we have
NOT
∀
x,
[(
P
(
x
) AND
Q
(
x
)) =
⇒
R
(
x
)]
∃
x,
NOT [(
P
(
x
) AND
Q
(
x
)) =
⇒
R
(
x
)]
∃
x,
[(
P
(
x
) AND
Q
(
x
)) AND NOT
R
(
x
)]
Exercise 144:
Negate each expression, and simplify your answer.
∃
x
∀
y,
(
P
(
x
) AND
Q
(
y
)).
Solution:
NOT
∃
x
∀
y,
(
P
(
x
) AND
Q
(
y
))
∀
x
NOT
∀
y,
(
P
(
x
) AND
Q
(
y
))
∀
x
∃
y,
NOT (
P
(
x
) AND
Q
(
y
))
∀
x
∃
y,
(NOT
P
(
x
)) OR (NOT
Q
(
y
))
since NOT(
A
AND
B
) is equivalent to (NOT
A
) OR (NOT
B
).
Exercise 146:
If the universe of discourse is the real numbers, what does each statement mean in English? Are they true or false?
∃
x
∃
y,
(
x
≥
y
)
.
Solution:
“For some real number there is a real number that is less than or equal to it.”
This statement is true since for instance there is a real number (say, 0) less than 1 (here,
x
= 1 and
y
= 0).
(In fact, this statement is true for all real numbers
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 Spring '08
 Vanderburgh
 Logic, Logical connective, exclusive or

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