COT3100
Summer’2001
Recitation on functions (Solutions)
06/21/2001
1. Given
g
={(1,
b
), (2,
c
), (3,
a
)}, a function from
X
={1, 2, 3} to
Y
={
a, b, c, d
}, and
f
=
{(
a
,
x
), (
b
,
x
), (
c
,
z
), (
d
,
w
)}, a function from
Y
to
Z
={
w
,
x
,
y
,
z
}
a)
write
f
ο
g
as a set of ordered pairs and draw the arrow diagram of
f
ο
g
;
f
ο
g
= {(1,
x
), (2,
z
), (3,
x
)}
b)
Determine if
f
ο
g
is surjective or injective.
f
ο
g
is
neither surjective nor injective.
2. Let
A
=R

{

1}, where R is the set of real numbers, and define the function
f
:
A
→
R by
the formula
1
2
)
(
+
=
x
x
x
f
. Prove that
f
is injective but not surjective.
1)
To prove that
f
(
x
) is injective, it is sufficient to show that if
f
(
x
1
)=
f
(
x
2
),
then
x
1
=
x
2
. So, assume
f
(
x
1
)=
f
(
x
2
). Then we have
2
x
1
/(
x
1
+1) = 2
x
2
/(
x
2
+1).
Using the fact that denominators are not zero (
x
1
,
x
2
≠

1) we get by
algebra :
2
x
1
⋅
(
x
2
+1) = 2
x
2
⋅
(
x
1
+1),
or:
2
x
1
⋅
x
2
+ 2
x
1
= 2
x
2
⋅
x
1
+.2
x
2
x
1
=
x
2
2) To prove that
f
(
x
) is not surjective, it is sufficient to find the value
y
∈
R
, that
has no preimage in
A
. To find such
y
let
y
= 2
x
/(
x
+1) and try to solve for
x
as a
function of
y
. We have by algebra:
y
(
x
+1)=2
x
x
(2

y
)=
y
x
=
y
/(2

y
).
From here we see, that if we take
y
= 2, then there is no real
x
such that
f
(
x
) =2.
3.
a) Let
f
be a function,
f
:
S
→
S
, where 
S
 =
n
. Prove that
f
is injective iff it is
1
1
•
2
•
3
•
a
y
z