MATH 135, Final Exam Solutions
Page 2 of 10
Combining these inequalities, we get 50
≤
n
≤
51.
If
n
= 50, (
x, y
) = (1
,
15).
If
n
= 51, (
x, y
) = (9
,
2).
Therefore, the nonnegative solutions are (1
,
15) and (9
,
2).
2.
(a) Determine the truth table for the statement (
P
=
⇒
Q
) OR (
Q
=
⇒
R
).
[4]
Solution:
P
Q
R
(
P
=
⇒
Q
)
(
Q
=
⇒
R
)
(
P
=
⇒
Q
) OR (
Q
=
⇒
R
)
T
T
T
T
T
T
T
T
F
T
F
T
T
F
T
F
T
T
T
F
F
F
T
T
F
T
T
T
T
T
F
T
F
T
F
T
F
F
T
T
T
T
F
F
F
T
T
T
(b) Find a statement which does not contain the connective NOT and is equivalent to the
[3]
statement
NOT(
∃
x
∀
y, x
≤
y
)
No justification is required.
Solution:
Using the negation rules, we obtain
∀
x
∃
y, x > y
(c) Give examples of statements
P
(
x
) and
Q
(
x
), each with universe of discourse equal to the
[2]
set of real numbers
R
, which demonstrate that the following statements are not equivalent:
∃
x,
(
P
(
x
) AND
Q
(
x
))
and
(
∃
x, P
(
x
)) AND (
∃
x, Q
(
x
))
Give a brief explanation.
Solution:
For
∃
x,
(
P
(
x
) AND
Q
(
x
)) to be true, there needs to be an
x
for which both
P
(
x
) and
Q
(
x
) are true.
For (
∃
x, P
(
x
)) AND (
∃
x, Q
(
x
)) to be true, there needs to be an
x
for which
P
(
x
) and
there needs to be a (possibly different)
x
for which
Q
(
x
) is true.
If we choose
P
(
x
) = “
x >
0” and
Q
(
x
) = “
x <
0”, then the first statement is false, since
there is no
x
for which both
P
(
x
) and
Q
(
x
) are true.
However, the second statement is true, since there is an
x
(for example,
x
= 1) for which
P
(
x
) is true, and there is an
x
(for example,
x
=

1) for which
Q
(
x
) is true.
Therefore, the statements are not equivalent.