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64
(c)
(f
0
g)1
= {(I, 4), (2,3), (3,2),
(4, 1), (5,
5n
g1
0
f
1
=
{(I, 4), (2,3), (3,2), (4, 1), (5,
5n
=
(f
0
g)I.
f
1
0
g1
=
{(I, 5), (2,3), (3,2), (4,4), (5,
In
=I
(f
0
g)I.
Solutions to Exercises
20.
(a)[BB]ForxEB,(fog)(x)=fC:~I)= 2~
=
2t
)=22
x
=x.
xI
 2
2x
 2
xI
2(~)
2x
2x
(b) [BB] For
x
E
A, (g
0
f)(x)
=
g(x:'2)
=
~1
=
x
_
(x
_
2) =
2"
=
x
and so, by
x2
Proposition 3.2.7,
f
and
9
are inverses.
21. (a) Suppose 0 E
A
and let a
=
f(O).
Then 0
=
fl(a)
=
fta)
which is not possible for any real
number
f(a).
(b) Since
f
and
f4
each have domain
A,
we have only to prove that
f4(a)
=
a for all a
E
A.
Let
then a be some element of
A.
Let al
=
f(a),
a2
=
f(al)
=
f2(a),
a3
=
f(a2)
=
f3(a)
and
a4
=
f(a3)
=
f4(a).
We must show that
a4=
a. From al
=
f(a),
we have a
=
f
1
(al) =
f(~J)
and so a2
=
f(al)
=
~.
From a3
=
f(a2),
we obtain a2
=
f
1
(a3)
=
f(~3)
and so
a4
=
f(a3)
=
.!.
.
= _1_
==
a as desired.
a2
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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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