Real Vector Space is Vector Space

Theorem
Let $\R$ be the set of real numbers.

Then the real vector space $\R^n$ is a vector space.

Construction of Real Vector Space
From the definition, a vector space is a unitary module whose scalar ring is a division ring.

In order to call attention to the precise scope of the operators, let real addition and real multiplication be expressed as $+_\R$ and $\times_\R$ respectively.

Then we can express the field of real numbers] as $\left({\R, +_\R, \times_\R}\right)$.

By definition of a field, $\left({\R, +_\R, \times_\R}\right)$ is also a division ring.

From Real Numbers under Addition form Abelian Group, $\left({\R, +}\right)$ is a group.

Again, in order to call attention to the precise scope of the operator, let real addition be expressed on $\left({\R, +}\right)$ as $+_G$.

That is, the group under consideration is $\left({\R, +_G}\right)$.

Consider the cartesian product:
 * $\displaystyle \R^n = \prod_{i \mathop = 1}^n \left({\R, +_G}\right) = \underbrace{ \left({\R, +_G}\right) \times \cdots \times \left({\R, +_G}\right) }_{n \ \text{copies}}$

Let:
 * $\mathbf a = \left({a_1, a_2, \ldots, a_n}\right)$
 * $\mathbf b = \left({b_1, b_2, \ldots, b_n}\right)$

be arbitrary elements of $\R^n$.

Let $\lambda$ be an arbitrary element of $\R$.

Let $+$ be the binary operation defined on $\R^n$ as:
 * $\mathbf a + \mathbf b = \left({a_1 +_G b_1, a_2 +_G b_2, \ldots, a_n +_G b_n}\right)$

Also let $\cdot$ be the binary operation defined on $\R \times \R^n$ as:
 * $\lambda \cdot \mathbf a = \left({\lambda \times_\R a_1, \lambda \times_\R a_2, \ldots, \lambda \times_\R a_n}\right)$

In this context, $\lambda \times_\R a_j$ is defined as real multiplication, as is appropriate (both $\lambda$ and $a_j$ are real numbers).

With this set of definitions, the structure $\left({\R^n, +, \cdot}\right)$ is a vector space, as is shown in Proof of Real Vector Space below.

Proof of Real Vector Space
In order to show that $\left({\R^n, +, \cdot}\right)$ is a vector space, we need to show that:

$\forall \mathbf x, \mathbf y \in \R^n, \forall \lambda, \mu \in \R$:
 * $(1): \quad \lambda \cdot \left({\mathbf x + \mathbf y}\right) = \left({\lambda \cdot \mathbf x}\right) + \left({\lambda \cdot \mathbf y}\right)$


 * $(2): \quad \left({\lambda +_R \mu}\right) \cdot x = \left({\lambda \cdot \mathbf x}\right) + \left({\mu \cdot \mathbf x}\right)$


 * $(3): \quad \left({\lambda \times_R \mu}\right) \cdot x = \lambda \cdot \left({\mu \cdot \mathbf x}\right)$


 * $(4): \quad \forall \mathbf x \in \R^n: 1 \cdot \mathbf x = \mathbf x$.

where $1$ in this context means the real number one.

From External Direct Product of Groups is Group, we have that $\left({\R^n, +}\right)$ is a group in its own right.

Let:
 * $\mathbf x = \left({x_1, x_2, \ldots, x_n}\right)$
 * $\mathbf y = \left({y_1, y_2, \ldots, y_n}\right)$

Checking the criteria in turn:

$(1): \quad \lambda \cdot \left({\mathbf x + \mathbf y}\right) = \left({\lambda \cdot \mathbf x}\right) + \left({\lambda \cdot \mathbf y}\right)$:

So $(1)$ has been shown to hold.

$(2): \quad \left({\lambda +_R \mu}\right) \cdot x = \left({\lambda \cdot \mathbf x}\right) + \left({\mu \cdot \mathbf x}\right)$:

So $(2)$ has been shown to hold.

$(3): \quad \left({\lambda \times_R \mu}\right) \cdot x = \lambda \cdot \left({\mu \cdot \mathbf x}\right)$:

So $(3)$ has been shown to hold.

$(4): \quad \forall \mathbf x \in \R^n: 1 \cdot \mathbf x = \mathbf x$:

So $(4)$ has been shown to hold.

So the $\R$-module $\R^n$ is a vector space, as we were to prove.

Warning
Notice that the factors of $\R^n$ are considered to be elements of a group, where $+_\R$ is the only operator defined. That is, in this particular context, $\times_G$ is not defined and can not be used.

This stands to reason, as in the context of a vector space, there is no unique and canonical definition for multiplication of vectors. Several are defined (for example, dot product and cross product), but these are not canonical.

Also see
It follows directly, by setting $n=1$, that the $\R$-module $\R$ itself can also be regarded as a vector space.

This is expanded upon in Real Numbers form Vector Space.