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COT3100.01, Fall 2000
S. Lang
Solution Key to Assignment #4 (Question 2 corrected)
10/31/2000
First, we
define
when two functions are considered equal (or identical).
Definition.
Two functions
f
:
A
→
B
and
g
:
A
→
B
are said
equal
, denoted
f
=
g
, if
f
(
x
) =
g
(
x
) for
every
x
∈
A
, where
A
is the common domain of the two functions.
For example, two functions
f
(
x
) = (
x
+ 1)
2
and
g
(
x
) =
x
2
+ 2
x
+ 1, both defined from
R
to
R
, where
R
denotes the set of real numbers, are equal because (
x
+ 1)
2
=
x
2
+ 2
x
+ 1 by algebra laws.
As a reminder, when there are two functions
f
:
A
→
B
and
g
:
B
→
C
, they can be composed to
form a function denoted
g
o
f
:
A
→
C
, with
g
precedes
f
in the composition notation
g
o
f
.
However, when
f
and
g
are considered as relations (because functions are special cases of
relations), the notation for composing would be
f
o
g
, which as a relation has the property
f
o
g
⊂
A
×
C
.
The context should make it clear which convention (either
g
o
f
or
f
o
g
) is used.
1.
(11 pts.) Consider a set
A
= {
a
,
b
,
c
}, a set
B
= {1, 2}, and a set
C
= {
u
,
v
,
w
}.
Define a
relation
R
⊂
A
×
B
with
R
= {(
a
, 1), (
b
, 1), (
c
, 2)}, a relation
S
⊂
B
×
C
with
S
= {(1,
u
), (1,
v
), (2,
w
)}, and a relation
T
⊂
A
×
C
as
T
= {(
a
,
v
), (
b
,
w
), (
c
,
u
)}.
Now answer each of the
following questions with a short explanation.
(a)
Is
R
a function?
If so, is it an injection?
R
is a function (because for each element of
A
, there is one and only one element of
B
which is related via the relation
R
).
R
is not an injection because both
a
and
b
are related
to the same image 1.
(b) Is
S
–1
a function?
If so, is it a surjection?
Since
S
–1
= {(
u
, 1), (
v
, 1), (
w
, 2)}
⊂
C
×
B
, the relation
S
–1
defines a function from
C
to
B
.
It is a surjection because each element of
B
is the image of some element of
C
.
(More
precisely, element 1 is the image of
u
(and of
v
), and element 2 is the image of
w
.)
(c)
Is
T
a function?
Is
T
–1
a function?
Since
T
–1
= {(
v
,
a
), (
w
,
b
), (
u
,
c
)}, both of the relations
T
and
T
–1
are functions.
(In fact,
they both are bijections and they are inverse functions of each other.)
(d) Determine the composed
relation
S
–1
R
–1
, that is, find all the pairs in this relation and
express the relation as a subset of
C
×
A
.
Since
S
–1
= {(
u
, 1), (
v
, 1), (
w
, 2)} and
R
–1
= {(1,
a
), (1,
b
), (2,
c
)}, the composed relation
S
–1
R
–1
= {(
u
,
a
), (
u
,
b
), (
v
,
a
), (
v
,
b
), (
w
,
c
)}
⊂
C
×
A
.
(e)
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This document was uploaded on 06/09/2011.
 Fall '09

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