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42
Solutions to Exercises
Transitive: If
a
f"V
b
and
b
f"V
c,
then
3a
+
b
::::;
4k
for some integer
k
and
3b
+
c
=
4£ for some
integeri. Since4(k+£)
=
(3a+b)+(3b+c)
=
(3a+c) +4b,
we see that
3a+c
=
4(k+£)
4b
is a multiple of 4 and, hence, that
a
f"V
c.
(b)
0
=
{x
E Z
I
x
f"V
O}
=
{x
I
3x
=
4k
for some integer
k}.
Now if
3x
=
4k, k
must be a multiple
of 3. So
3x
=
12£ for some £ E Z and
x
=
4£.
0
=
4Z.
(c) 2
=
{x
E Z
I
x
f"V
2}
=
{x
I
3x
+ 2
=
4kforsomeintegerk}
=
{x
I
3x
=
4k
2 for some integer
k}
Now if
3x
=
4k

2, then
3x
=
3k
+
k

2 and so
k

2 is a multiple
of 3. Therefore,
k
=
3£ + 2 for some integer £,
3x
=
4(3£ + 2)  2
=
12£ + 6 and
x
=
4£ + 2.
S02
=
4Z+2.
(d) The quotient set is {4Z, 4Z + 1, 4Z + 2, 4Z + 3}.
to. (a) Reflexive: For any
a
E Z,
a
f"V
a
because
3a
+
4a
=
7a
and
a
is an integer.
Symmetric: If
a, b
E Z and
a
f"V
b,
then
3a
+
4b
=
7n
for some integer
n.
Then
3b
+
4a
=
(7a
+
7b)

(3a
+
4b)
=
7a+ 7b
7n
=
7(a
+
b

n)
and
a
+
b n
is an integer. Thus
b
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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