MODEL ANSWERS TO THE FIRST HOMEWORK
1. Chapter 1,
§
1: 1. Suppose that
a
and
b
are elements of
S
. By rule
(1)
a
b
=
a.
But by rule (2),
a
b
=
b
a.
Applying rule (1) we get
a
b
=
b
a
=
b
.
Thus
a
=
a
b
=
b
. As
a
and
b
are arbitrary,
S
can have at most one
element.
1. Chapter 1
§
1: 2. (a) Suppose that
a
and
b
are two integers and that
a
b
=
b
a
.
Now
a
b
=
a

b
and
b
a
=
b

a
so that then
a

b
=
b

a
. Applying
the standard rules of arithmetic, we get 2
a
= 2
b
and so
a
=
b
.
(b) Suppose that
a
,
b
and
c
are integers. Then
a
(
b
c
) =
a
(
b

c
) =
a

(
b

c
) =
a
+
c

b.
On the other hand
(
a
b
)
c
= (
a

b
)
c
= (
a

b
)

c
=
a

(
b
+
c
)
.
Thus equality holds i
ff
a
+
c

b
=
a

(
b
+
c
), that is cancelling
c
=

c
so that
c
= 0. Thus
is not associative. For example,
0
(0
1) = 1
but
(0
0)
1 =

1
.
(c) Let
a
be an integer. Then
a
0 =
a

0 =
a.
(d) Let
a
be an integer. Then
a
a
=
a

a
= 0
.
2. Chapter 2,
§
1: (a). No, by the question above, this rule of multipli
cation is not associative.
1
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(b) No this is not a group.
We consider the three axioms.
Suppose
that
a
,
b
and
c
are three integers. Then
a
(
b
c
) =
a
(
b
+
c
+
bc
)
=
a
+ (
b
+
c
+
bc
) +
a
(
b
+
c
+
bc
)
=
a
+
b
+
c
+
bc
+
ab
+
ac
+
abc.
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 Spring '08
 LONG
 Math, Algebra, Multiplication, Identity element

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