1.
A
=
{
1
,
3
,
5
,
7
,
9
,
11
}
,B
=
{
3
,
6
,
9
,
12
,
15
,
18
}
,A
∩
B
=
{
3
,
9
}
,A

B
=
{
1
,
5
,
7
,
11
}
.
2. (1)
A
∩
B
∩
C
=
{
4
}
.
(2)
A

B
=
{
1
,
3
}
.
(3) (
A
∪
B
)
C
=
{
5
,
7
,
9
,
10
}
.
(4)
A
C
∪
B
C
∪
C
C
=
{
1
,
2
,
3
,
5
,
6
,
7
,
8
,
9
,
10
}
. It is easy to see that
{
4
}
C
=
{
1
,
2
,
3
,
5
,
6
,
7
,
8
,
9
,
10
}
.
4.
6. Solve
(
y

2 =
x

1
2
x
+ 1 =
y
+ 2
⇒
(
y
=
x
+ 1
2
x
+ 1 =
y
+ 2
⇒
2
x
+1 =
x
+1+2
⇒
x
= 2 and
y
= 3.
8.
R
=
{
(1
,
3)
,
(1
,
5)
,
(2
,
3)
,
(2
,
5)
,
(3
,
5)
,
(4
,
5)
}
, so dom
R
=
{
1
,
2
,
3
,
4
}
and range
R
=
{
3
,
5
}
.
10. (a)
Reﬂexive:
(1
,
1)
,
(2
,
2)
,
(3
,
3)
,
(4
,
4)
,
(5
,
5)
∈
R
.
(b)
Symmetric:
(1
,
2)
∈
R
⇒
(2
,
1)
∈
R
, (3
,
4)
∈
R
⇒
(4
,
3)
∈
R
. No other pairs.
(c)
Transitive:
(1
,
2)
,
(2
,
1)
∈
R
⇒
(1
,
1)
∈
R
; (3
,
4)
,
(4
,
3)
∈
R
⇒
(3
,
3)
∈
R
; (2
,
1)
,
(1
,
2)
∈
R
⇒
(2
,
2)
∈
R
; (4
,
3)
,
(3
,
4)
∈
R
⇒
(4
,
4)
∈
R
. No other pairs.
Then 1
∼
2
,
3
∼
4
,
5 is alone, i.e.
E
=
{
1
,
2
} ∪ {
3
,
4
} ∪ {
5
}
.
12.
g
(

2) =

7
,g
(

1) = 0
,g
(0) = 1
,g
(1) = 2
,g
(2) = 9. We can verify that for all
x
1
,x
2
∈
X
,
x
1
6
=
x
2
will imply
g
(
x
1
)
6
=
g
(
x
2
), so
g
is onetoone, and im(
g
) =
{
7
,
0
,
1
,
2
,
9
}
.
By solving
y
=
x
3
+ 1, we have
y

1 =
x
3
, i.e.
x
=
3
√
y

1. Then
g

1
(
x
) =
3
√
x

1 (if
we assume that the independent variable is always
x
).
14. Since
f
(
a
) =
b,f
(
b
) =
d,f
(
c
) =
a
, and
f
(
d
) =
c
,
f
maps diﬀerent elements of
A
to
diﬀerent elements in
A
, so
f
is onetoone. Meanwhile, im(
f
) =
{
b,d,a,c
}
=
A
, so
f
is
onto. Its inverse function is:
f

1
(
b
) =
a,f

1
(
d
) =
b,f

1
(
a
) =
c
, and
f

1
(
c
) =
d
, i.e.
f
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 Spring '11
 astina
 Math, Linear Algebra, Sets, Dot Product, Trigraph, Cauchy–Schwarz inequality

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