III. Existence of a zero? Well, we need an element (
p, q
) such that (
p, q
) +
(
x, y
) = (
x, y
) for all pairs (
x, y
).
x
+
p
=
x
and
yq
=
y
for all
x, y
∈
R
.
Choosing the zero to be the element (0
,
1) is the only possible choice,
and when you make that choice, Axiom III holds.
IV. Existence of an additive inverse?
This is where it goes wrong for
the first time.
Our vectors are allowed to have a zero in the second
position e.g. (
x,
0), but there is no vector you can add to that (using
the definition of addition in this part) to get the zero vector, which
has to be (0
,
1).
V. This axiom is OK. Scalar multiplication by 1 does not change any
vector.
VI. Associativity of scalar multiplication is obviously OK. Observe:
c
1
(
c
2
(
a
1
, a
2
))
=
c
1
(
c
2
a
1
, a
2
)
=
(
c
1
(
c
2
a
1
)
, a
2
)
=
((
c
1
c
2
)
a
1
, a
2
)
=
(
c
1
c
2
)(
a
1
, a
2
)
.
VII. Distribution in the vector position. This axiom holds. Observe:
x
((
a
1
, a
2
) + (
b
1
, b
2
))
=
x
(
a
1
+
b
1
, a
2
b
2
)
=
(
x
(
a
1
+
b
1
)
, a
2
b
2
)
=
(
xa
1
+
xb
1
, a
2
b
2
)
=
(
xa
1
, a
2
) + (
xb
1
, b
2
)
=
x
(
a
1
, a
2
) +
x
(
b
1
, b
2
)
.
VIII. Distribution in the scalar position.
This axiom fails
. To show that
an axiom fails we need to find a particular example where the equation
is false. A simple one is the case where the two scalars are
x
=
y
= 1
and the vector
v
= (1
,
2). Then the left hand side of the axiom becomes
(1 + 1)(1
,
2) = 2(1
,
2) = (2
,
2).
And the right hand side becomes
1(1
,
2) + 1(1
,
2) = (1
,
2) + (1
,
2) = (2
,
2
.
2), which is different.
2