Math 25: Advanced Calculus
UC Davis, Spring 2011
Math 25 — Solutions to Homework Assignment #3
Solutions
1. Let
F
be a field.
(a) Fix some
a, b, c
∈
F
.
By commutativity of multiplication (ax
iom A1), distributivity of multiplication over addition (axiom
AM1), and commutativity of multiplication again (A1), respec
tively,
c
(
a
+
b
) = (
a
+
b
)
c
=
ac
+
bc
=
ca
+
cb
.
(b) Fix some
a, b
∈
F
such that
ab
= 0 and assume (without loss
of generality)
a
6
= 0.
Then by the existence of a multiplicative
inverse (axiom M4)
a
has an inverse element
a

1
such that
a

1
a
=
1. By a claim proved in class,
x
·
0 = 0 for any
x
∈
F
. Applying
this with
x
=
a

1
and then using associativity (M2) and the
defining property (M3) of 1 gives that 0 =
a

1
·
0 =
a

1
(
ab
) =
(
a

1
a
)
b
= 1
·
b
=
b
. Thus,
b
= 0.
(c) Fix some
a, b, c
∈
F
, with
ac
=
bc
and
c
6
= 0. By the existence of
multiplicative inverse, substitution, and assicativity of multipli
cation,
ac
=
bc
=
⇒
(
ac
)
c

1
= (
bc
)
c

1
=
⇒
a
(
cc

1
) =
b
(
cc

1
) =
⇒
a
1 =
b
1 =
⇒
a
=
b
.
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 Winter '11
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 Differential Equations, Calculus, Addition, Equations, Multiplication, Supremum, Order theory, Max E

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