Solutions to Homework 4  Math 2000
(# 2.18) Consider the open sentences
P
(
n
) : 5
n
+ 3 is prime.
and
Q
(
n
) : 7
n
+ 1 is
prime over the domain
N
.
(a). State
P
(
n
) =
⇒
Q
(
n
) in words.
(b). State
P
(2) =
⇒
Q
(2) in words. Is this statement true or false?
(c). State
P
(6) =
⇒
Q
(6) in words. Is this statement true or false?
Solution
. (a). If 5
n
+ 1 is prime, then 7
n
+ 1 is prime.
(b). If 5(2) + 1 is prime, then 7(2) + 1 is prime. Since 5(2) + 1 = 11,
P
(2) is true and
7(2) + 1 = 15 thus
Q
(2) is false. Therefore
P
(2) =
⇒
Q
(2) is false.
(c). If 5(6) + 1 is prime, then 7(6) + 1 is prime. Since 5(6) + 1 = 31,
P
(5) is true and
7(6) + 1 = 43 thus
Q
(7) is true. Therefore
P
(7) =
⇒
Q
(7) is true.
(# 2.20) In each of the following, two open sentences
P
(
x
) and
Q
(
x
) over a domain
S
are given. Determine all
x
∈
S
for which
P
(
x
) =
⇒
Q
(
x
) is a true statement.
(a).
P
(
x
) :
x

3 = 4;
Q
(
x
) :
x
≥
8;
S
=
R
.
(b).
P
(
x
) :
x
2
≥
1;
Q
(
x
) :
x
≥
1;
S
=
R
.
(c).
P
(
x
) :
x
2
≥
1;
Q
(
x
) :
x
≥
1;
S
=
N
.
(d).
P
(
x
) :
x
∈
[

1
,
2];
Q
(
x
) :
x
2
≤
2;
S
= [

1
,
1].
Solution
. (a).
P
(
x
) is true if
x
= 7 and
P
(
x
) is false if
x
= 7.
Q
(
x
) is true if
x
≥
8 and
Q
(
x
) is false if
x <
8. We know that
P
(
x
) =
⇒
Q
(
x
) is false only if
P
(
x
) is true and
Q
(
x
) is false. This only occurs if
x
= 7. Therefore
P
(
x
) =
⇒
Q
(
x
) is true for all other
x
. That is, it is true if
x
= 7.
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 Spring '09
 dd
 Math, Logic, Logical connective, If and only if, Logical equality, open sentences

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