Homework 9 Solutions
1. (
α
0
+
α
1
i
+
α
2
j
+
α
3
k
)
2
= (
α
2
0

α
2
1

α
2
2

α
2
3
) + 2
α
0
α
1
i
+ 2
α
0
α
2
j
+ 2
α
0
α
3
k
Hence, (
α
0
+
α
1
i
+
α
2
j
+
α
3
k
)
2
=

1
⇔
α
2
0

α
2
1

α
2
2

α
2
3
=

1 and
α
0
α
1
=
α
0
α
2
=
α
0
α
3
= 0
⇔
α
2
1
+
α
2
2
+
α
2
3
= 1 (and
α
0
= 0).
Thus there are infinitely many solutions (1 corresponding to each point on the unit sphere).
2. Call the proposed field
R
, to simplify matters.
•
additive identity =
0
0
0
0
∈
R
•
commutative under addition since
R
is.
•
closed under addition since
a
+
c
b
+
d

b

d
a
+
c
∈
R
•
addition is associative  clear
•
multiplicative identity =
1
0
0
1
∈
R
•
for
A
∈
R
,
det
(
A
) =
a
2
+
b
2
= 0
⇔
a
=
b
= 0 thus every element has an inverse except
the additive identity (as desired).
•
a
b

b
a
c
d

d
c
=
ac

bd
ad
+
bc

ad

bc
ac

bd
=
c
d

d
c
a
b

b
a
So commuta
tive under multiplication.
•
multiplication is associative  clear
•
a
b

b
a
c
d

d
c
+
e
f

f
e
=
ac
+
ae

bd

bf
ad
+
af
+
bc
+
be

ad

af

bc

be
ac
+
ae

bd

bf
=
ac

bd
ad
+
bc

ad

bc
ac

bd
+
ae

bf
af
+
be

af

be
ae

bf
=
a
b

b
a
c
d

d
c
+
a
b

b
a
e
f

f
e
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 Spring '11
 Johnson
 Algebra, Addition, CD AB, bd ef ab

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