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96
Solutions to Exercises
Transitivity: Suppose
a, b,
c
E
A,
a
j
b
and
b
j
c; that is,
ab
==
a
2
(mod
p)
and
be
==
b
2
(mod
p).
We want to prove that
a
j
c, that is, that
ae
==
a
2
(mod
p)
or, equivalently, that
p
I
(ae

a
2
).
Now we have that
(ab)(be)
==
a
2
b
2
(mod
p)
and, hence, that
p
I
b
2
(ae

a
2
).
Again, by Proposition 4.3.7,
p
I
b
2
or
p
I
(ae a
2
).
The latter case is what we want. In the former
case, we must have
p
I
b,
hence, that
ab
==
0 (mod
p).
Therefore,
a
2
==
ab
==
0, so
a
==
0 and
again
ae

a
2
==
0, the desired result.
(b) This is a partial order.
Reflexivity: For any
a
E
A,
a
j
a
because
a(a)
==
a
2
(mod
pq).
Antisymmetry:
If
a,
b
E
A,
a
j
band
b
j
a,
then
ab
==
a
2
(mod
pq)
and
ba
==
b
2
(mod
pq).
Thus,
ab
==
a
2
(mod
p)
and
ba
==
b
2
(mod
p)
and, as in (a), we get
a
==
b
(mod
p).
Similarly,
we obtain
a
==
b
(mod
q).
So we have
a

b
divisible by both
p
and
q.
By Exercise 14 of
Section 4.3,
pq
I
(a

b),
giving
a
==
b
(mod
pq)
as desired.
Transitivity: Suppose
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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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