Homework 10
Section 8.1
1.(M) 48a) Since R and S are both having all pairs like (x,x) where x is an element of set A the union will
also have it. Thus
R
∪
S
is also reﬂexive.
48b)
R
∩
S
will be reﬂexive as both R and S have pairs like (x,x).
48c)
R
⊕
S
will be irreﬂexive as it does not contain any of the (x,x) type pairs.
48d) Since both R and S have all (x,x) kind of pairs
R

S
would have none. So
R

S
would be irreﬂexive.
48e) Since both R and S have all (x,x) kind of pairs the
S
◦
R
would have them too and hence would be
reﬂexive.
2.(E) 54a)
R
2
is
R
◦
R
. So
R
2
=
{
(1,2), (1,3), (1,4), (1,1), (1,5), (2,1), (2,4), (2,5), (2,2), (3,1), (3,2), (3,3), (3,5), (3,4),
(4,3), (4,4), (4,1), (4,2), (5,1), (5,2), (5,3), (5,4), (5,5)
}
54b)
R
3
=
{
(1,2), (1,3), (1,4), (1,1), (1,5), (2,1), (2,3), (2,4), (2,5), (2,2), (3,1), (3,2), (3,3), (3,5), (3,4), (4,3), (4,4),
(4,1), (4,2), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5)
}
3.(C) 56) We prove this by induction. For
n
= 1
this is trivial as
R
is given as symmetric. Assume the result
to be true for
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 Spring '07
 YaoyunShi
 Binary relation, Transitive relation, Symmetric relation, R1 R2, diagonal entries

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