**Problem 1.** Prove that
for every
.

*Solution.* By definition we have
. But
. So by
uniqueness of the additive inverse (1.27), we conclude that
.

**Problem 2.** Suppose
,
, and
. Prove that
or
.

*Solution.* Suppose that
. Then
admits a
multiplicative inverse
,
and it follows that
by (1.31).

**Problem 3.** Suppose
. Explain why
there exists a unique
such that
.

*Proof.* Let
.
This vector exists because
admits an additive
inverse in
and
is closed under
addition and scalar multiplication. It then follows that
which establishes the existence of such an
. Now, suppose that
for
. Adding
on the left and
cancelling yields
. Scaling these
vector by
yields
, so this vector
is unique.

**Problem 4.** The empty set is not a vector space. The
empty set fails to satisfy only one of the requirements listed in the
definition of a vector space (1.20). Which one?

*Solution.* The empty set contains no elements, and hence it
fails to admit an additive identity element. So a vector space
, as a set, must be
nonempty.

**Problem 5.** Show that in the definition of a vector
space (1.20) , the additive inverse condition can be replaced with the
condition that
for all
. Here the
on the left side is
the number
, and the
on the right side is
the additive identity of
.

*Solution.* Given the vector-space axioms, one has
for every
by (1.30).
Conversely, suppose that the additive-inverse axiom were replaced with
the condition that
for every
. Fix
. Then
. But
by (1.32),
so one has
meaning that
admits an additive
inverse, as required.