SOLUTIONS FOR HOMEWORK 2
1.51. (a) Yes
,
I
f
(
X
∪
Y
) =
I
f
(
X
)
∪
I
f
(
Y
). To establish this equality, note that
a
∈
I
f
(
X
∪
Y
) iff
f
(
a
)
∈
X
∪
Y
. The last equality holds when either
f
(
a
)
∈
X
, or
f
(
a
)
∈
Y
. By the definition of
I
f
,
f
(
a
)
∈
X
is equivalent to
a
∈
I
f
(
X
), and similarly,
f
(
a
)
∈
Y
is equivalent to
a
∈
I
f
(
Y
). Thus,
f
(
a
)
∈
X
∪
Y
iff
a
∈
I
f
(
X
)
∪
I
f
(
Y
).
(b) Yes
,
I
f
(
X
∩
Y
) =
I
f
(
X
)
∩
I
f
(
Y
). Indeed,
a
∈
I
f
(
X
∩
Y
) iff
f
(
a
)
∈
X
∩
Y
, that
is,
f
(
a
)
∈
X
, and
f
(
a
)
∈
Y
. As noted in (a),
f
(
a
)
∈
X
iff
a
∈
I
f
(
X
), and
f
(
a
)
∈
Y
iff
a
∈
I
f
(
Y
). Thus,
f
(
a
)
∈
X
∩
Y
iff
a
∈
I
f
(
X
)
∩
I
f
(
Y
).
1.55. Multiplication.
Show first that
x
·
x
= 0, or equivalently,
x
=
x

1
.
By
Definition 1.39,
x
has the opposite element
x

1
=
y
∈
F
s.t.
x
·
y
= 0. We have to
show that
y
=
x
, or in other words,
y
negationslash
= 0, and
y
negationslash
= 1. However, these cases are easy
to rule out:
x
·
0 = 0
negationslash
= 1, and
x
·
1 =
x
negationslash
= 1. The only remaining possibility is
y
=
x
.
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 Fall '09
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 Math, Addition, Multiplication, multiplication table, ﬁeld axioms

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