38
Solutions
to
Exercises
Not antisymmetric: (2,1)
En
because
2
1
=
1
$
3
and
(1,
2)
En
because
1
2
=
1
$
3,
but
1"#
2.
Not transitive: (5,3)
E
n
because
5 
3
=
2
$
3
and
(3,1)
E
n
because
3 
1
=
2
$
3,
but
(5,1)
sf
n
because
5 
1
=
41.
3.
(t)
Reflexive:
For
any
(a,
b)
E
A,
a

a
=
b

b;
thus,
«a,
b),
(a, b))
En.
Symmetric:
If
«a,
b),
(c,
d))
En,
then
ac
=
bd,
so
ca
=
db
and, hence,
«c,
d), (a,
b))
E
n.
Notantisymmetric:
«5,2),
(15, 12))
En
because
515
=
212
and similarly,
«15,12), (5,2))
E
n;
however,
(15,12)
"#
(5,2).
If
«a,
b),
(c,d))
E
nand
«c,d),
(e,f))
E
n
then
a

c
=
b

d
and c 
e
=
d

I.
Thus,
a

e
=
(a

c) + (c 
e)
=
(b

d)
+
(d

f)
=
b

I
and so
«a,
b), (e,
f))
E
n.
(g)
Not reflexive:
If
n
E
N,
then
n
"#
n
is not true.
Symmetric:
Ifnl
"#
n2,
then
n2
"#
nl.
Not antisymmetric: 1
"#
2
and
2"# 1
so both
(1,2)
and
(2,
1)
are in
n,
yet
1
"#
2.
Not transitive: 1
"#
2,
2
"#
1,
but
1
=
1.
(h)
Not reflexive: (2,2)
sf
n
because
2
+
2
"#
10.
Symmetric:
If
(x,
y)
E
n,
then
x
+
y
=
10,
so
y
+
x
=
10,
and hence,
(y,
x)
En.
Not antisymmetric:
(6,
4)
E
n
because
6
+
4
=
10
and similarly,
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 Summer '10
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 Graph Theory, Transitive relation, Symmetric relation, nd

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