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52
Solutions to Review Exercises
(b) We have
0'R7
and
7'R0,
yet 0
=f
7. The relation is not antisymmetric, so it is not a partial order.
17. (+) Assume
x
f'V
a.
Let
t
E
x.
Then
t
f'V
x
so
t
rv
a
by transitivity. Thus
tEa.
This proves
x
~
a.
Similarly,
a
~
x,
so
X
=
a.
(f)
Assume
x
=
a.
Since
x
E
x,
by symmetry,
x
E
a.
Thus
x
f'V
a.
18. Since
a
E
b,
a
f'V
b
and hence
a
=
b
by Proposition 2.4.3. Similarly
dEb
implies
d
=
b,
hence
a
=
d.
Now
d
¢:
c
implies
d =f
c,
so
end
=
0
by Proposition 2.4.4. Since
a
=
d,
so also
a
n c
= 0.
19. (a) We must show that
~
is reflexive, anti symmetric and transitive on
A.
The relation is reflexive: For
any
(a, b)
E
A, (a, b)
f'V
(a, b)
because
a::; a
and
b
::;
b.
It
is antisymmetric:
If
(a, b), (c,
d)
E
A,
with
(a, b)
f'V
(c,
d)
and
(c,
d)
f'V
(a, b),
then
a
::;
c,
d::;
b, c
::;
a
and
b
::;
d.
So
a
=
c, b
=
d,
hence,
(a, b)
=
(c,
d).
It
is transitive: If
(a, b), (c,
d),
(e,
f)
E
A
with
(a, b)
f'V
(c,
d)
and
(c,
d)
f'V
(e,
f), then
a
::;
c,
d
::;
b,
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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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