MATH 135 Assignment 7, Winter 2010
Due 8:30am on Wednesday March 10, 2010 in the drop boxes opposite the Tutorial Centre, MC 4066.
Question 1. Missing proofs in Section 5.1
[20 marks]
Consider the set
S
=
Z
×
(
Z
{
0
}
) =
{
(
a, b
)

a, b
∈
Z
, b
±
= 0
}
and a relation
∼
deﬁned on
S
as (
a, b
)
∼
(
c, d
)
iﬀ
ad
=
bc
.
(
i
) [10 marks]
Show that
∼
is an equivalence relation on
S
.
(
ii
) [10 marks]
Let
a
b
denote the equivalence class of
∼
containing (
a, b
).
Addition and multiplication are deﬁned on the equivalence classes as:
a
b
+
c
d
=
ad
+
bc
bd
,
a
b
·
c
d
=
ac
bd
.
Show that they are well deﬁned. In other words:
if
(
a, b
)
∼
(
a
±
, b
±
) and (
c, d
)
∼
(
c
±
, d
±
),
then (
ad
+
bc, bd
)
∼
(
a
±
d
±
+
b
±
c
±
, b
±
d
±
) and (
ac, bd
)
∼
(
a
±
c
±
, b
±
d
±
).
Answer:
(
i
)
Since
∀
(
a, b
)
∈
S
, it holds
ab
=
ba
so that (
a, b
)
∼
(
a, b
), therefore
∼
is reﬂexive.
Suppose (
a, b
)
∼
(
c, d
). Then,
ad
=
bc
, and this can be rewritten as
cb
=
da
. So (
c, d
)
∼
(
a, b
). Since this
holds for any (
a, b
), (
c, d
) in
S
,
∼
is symmetric.
Suppose (
a, b
)
∼
(
c, d
) and (
c, d
)
∼
(
e, f
). Then, (i)
ad
=
bc
and (ii)
cf
=
de
. Multiplying (i) and (ii) gives
adcf
=
bcde
. Since
d
±
= 0, we have (iii)
acf
=
bce
.
Now, we treat the cases
c
= 0 and
c
±
= 0 separately.
•
If
c
= 0, then, by (i),(ii) and that
d
±
= 0, we conclude
a
=
e
= 0. Thus,
af
=
be
= 0.
•
If
c
±
= 0, we conclude from (iii) that
af
=
be
.
In either case,
af
=
be
so (
a, b
)
∼
(
e, f
).
Since this holds for any (
a, b
), (
c, d
), and (
e, f
) in
S
,
∼
is transitive.
Answer: