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
f"V
a.
Transitive: Suppose
a,
b,
c
E Z with
a
f"V
b
and
b
f"V
c.
Then
3a
+
4b
=
7n
and
3b
+
4c
=
7m
for some integers
n
and
m.
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 Summer '10
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 Graph Theory, Trigraph, Negative and nonnegative numbers, Transitive relation, nd

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