Proof that multiplying by the scalar 1 does not change the vector in a normed vector space.

Proof that multiplying by the scalar 1 does not change the vector in a
normed vector space.

I'm beginning a self-study of functional analysis, and I seem to have come
to a halt trying to solve the first problem in the first problem set, and
was wondering if someone could give me a pointer.
The problem asks to show that, while the property $1\mathbf{a}=\mathbf{a}$
must be included in the axioms of a vector space, it is a property that
can be derived from the axioms of a normed vector space.
The axioms for a vector space $V$ over the field $\mathbf{R}$ given in the
book are as follow:
$V$ is a group.
For each $\alpha\in\mathbf{R}$ and $\mathbf{a}\in V$, $\alpha\mathbf{a}\in V$
$\alpha(\mathbf{a}+\mathbf{b})=\alpha\mathbf{a}+\alpha\mathbf{b}$
$(\alpha+\beta)\mathbf{a}=\alpha\mathbf{a}+\beta\mathbf{a}$
$\alpha(\beta\mathbf{a})=(\alpha\beta)\mathbf{a}$
$1\mathbf{a}=\mathbf{a}$
The axioms for a normed vector space are 1--5 plus the additional axiom
that there is a map $\Vert\cdot\Vert:V\to\mathbf{R}$ that satisfies all
the properties of a norm.
The problem formally stated now is to show that 1--5 do not imply 6, but
1--5 together with the norm do imply 6.
This problem seems similar to the question posed here, and from the
discussion of that question it seems an easy example of a vector space
satisfying 1--5 but not 6 would be one where scalar multiplication always
yields the zero vector. This cannot be the case in a normed vector space,
because for $\mathbf{a}\neq0$ and $\alpha\neq0$ we know
$\Vert\mathbf{a}\Vert\neq0$, and a norm satisfies
$\Vert\alpha\mathbf{a}\Vert=\vert\alpha\vert\Vert\mathbf{a}\Vert\neq0$,
but we also have $\Vert\alpha\mathbf{a}\Vert=\Vert\mathbf{0}\Vert=0$ from
our definition of scalar multiplication, which is a contradiction.
It is trivial that $1\mathbf{a}=\mathbf{a}$ if every vector in $V$ can be
written as a scalar multiple of a vector in $V$, since then
$\mathbf{a}=\alpha\mathbf{b}=(1\alpha)\mathbf{b}=1(\alpha\mathbf{b})=1\mathbf{a}$,
but while the norm has ruled out the strange case of scalar multiplication
that means only $\mathbf{0}$ can be written as a scalar multiple of a
vector, I haven't been able to convince myself that other cases where
there exist vectors that are not scalar multiples of a vector have been
ruled out.
Can anyone offer a pointer for how to show every vector in a normed vector
space can be written as a scalar multiple of a vector in the space, or
perhaps point me to a more natural way of approaching this problem?
Thanks.