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66
Solutions to Exercises
(c) To see that
t
is onto, let
y
E R. Then we have
y
= LyJ
+
b
for 0
$;
b
<
1.
If
b
= 0, then
t(y)
=
y.
Otherwise,O
<
1 
b
<
1. So, with
x
= LyJ
+
1
+
(1 
b),
we have LxJ = LyJ
+
1. Thus,
t(x)
= LxJ 
(1 
b)
= (LyJ
+
1) 
(1
b)
= LyJ
+
b
=
y.
In any case, we have an
x
such that
t(x)
=
y.
{
X
ifxEZ
(d)
rl:R4Risgivenbyrl(x)=
n+1+(1b)
ifx=n+b,nEZ,0<b<1
Exercises
3.3
1. [BB] Ask everyone to find a seat.
2. [BB] The two lists 1
2
,2
2
,3
2
,4
2
,
... and 1,2,3,4, .
.. obviously have the same length;
a
2
f+
a
is a
onetoone correspondence between the set of perfect squares and N.
3. (a)
x
+t
14,
Y
+t
3,
{a, b,
c}
+t
t.
(b) The function
f:
2Z 4 17Z defined by
f(2k)
=
17k
for2k E 2Z.
(c) [BB] The function
f:
N x N 4
C
defined by
f(m, n)
= m
+
ni
for all m,
n
E N.
(d) The function
f:
N 4
Q
defined by
f(n)
=
n/2.
4.
fI(z)
=
{2Z
if
z
>
0
1 2z
if
z
$;
0
5. (a)
If
g(ml' nl)
=
g(m2' n2),
then (ml'
f(nI})
=
(m2' f(n2)),
so ml = m2 and
f(nl)
=
f(n2).
Since
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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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