Real Numbers form Vector Space

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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 $\left({\R, +, \times}\right)$.

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

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

where:

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

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

$\mathbf a + \mathbf b = \left({a + b}\right)$
$\lambda \cdot \mathbf a = \left({\lambda \times a}\right)$

where:

$\mathbf a, \mathbf b \in \R$
$\lambda \in \R$
$\mathbf a = \left({a}\right)$
$\mathbf b = \left({b}\right)$

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

$\blacksquare$


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