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38
Solutions to Exercises
Not antisymmetric: (2,1)
En because 2
1
=
1 $ 3 and (1, 2)
En because
12
=
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)
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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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