74
Solutions
to
Exercises
17.
Let
A
=
{a1,a2,
...
,an}
and
B
=
{b1,b2,
...
,b
m
}.
Since
A
n
B
=
0,
ai
=f
bj
for any
i,j.
Let
C
=
{I,
2,
...
,
n
+
m}.
If
we
can
establish a onetoone correspondence
between
A
U
B
and
C,
then
we will have
IA
U
BI
=
ICI
=
n
+
m.
Define
f:
AU
B
+
C
by
f(aI)
=
1,
f(a2)
=
2,
...
,
f(a
n)
=
n, f(b1)
=
n
+
1,
f(b2)
=
n
+
2,
...
,
f(b
m )
=
n
+
m.
Since
f(x)
=
f(y)
+
x
=
y,
f
is
onetoone. Clearly
f
is onto, so
it
is a onetoone correspondence.
18.
The
function
f:
(2,2)
+
(1,9)
defined
by
f(x)
=
2x
+
5
is a bijection.
1
1
19. Define
f:
(1,3)
+
R
by
f(x)
=


2
and suppose
f(X1)
=
f(X2).
xI
1
1
1
1
1
1
Then



=


,
so

=
,
Xl

1
=
X2

1
and
Xl
=
X2.
Therefore,
f
Xl

1
2
X2

1
2
Xl

1
X2

1
.
G·
(0)
I
2y
+
3
1S
onetoone.
lVen
y
E
,00,
et
X
=
2y
+
1·
We leave
it
to the student
to
verify that
f(x)
=
y.
Also
1
<
~::~
<
3.
Finally,
rng
f
=
(0,00),
so
f
is a onetoone correspondence between
(1,3)
an.d
(0,00).
20. (a)
Assume
to
the contrary that
S
is countable.
In
that case,
we
can
list the elements
of
S
as
a1, a2,
a3,
....
Writing each
of
these using their decimal expansions,
we
have
a1
=
O.an
a12a13a14
..
.
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 Summer '10
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 Graph Theory, Trigraph, Decimal, onetoone correspondence, Finite set, Howard Staunton, 0.999...

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