# Real Numbers form Vector Space

## Theorem

The set of real numbers $\R$, with the operations of addition and multiplication, forms a vector space.

## Proof

Let the field of real numbers be denoted $\struct {\R, +, \times}$.

From Real Vector Space is Vector Space, we have that $\struct {\R^n, +, \cdot}$ is a vector space, where:

$\mathbf a + \mathbf b = \tuple {a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n}$
$\lambda \cdot \mathbf a = \tuple {\lambda \times a_1, \lambda \times a_2, \ldots, \lambda \times a_n}$

where:

$\mathbf a, \mathbf b \in \R^n$
$\lambda \in \R$
$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$

When $n = 1$, the vector space degenerates to:

$\mathbf a + \mathbf b = \tuple {a + b}$
$\lambda \cdot \mathbf a = \tuple {\lambda \times a}$

where:

$\mathbf a, \mathbf b \in \R$
$\lambda \in \R$
$\mathbf a = \tuple a$
$\mathbf b = \tuple b$

Thus it can be seen that the vector space $\struct {\R^1, +, \cdot}$ is identical with the field of real numbers denoted by $\struct {\R, +, \times}$.

$\blacksquare$