Real Numbers form Algebra

Theorem
The set of real numbers $\R$ forms an algebra over the Field of Real Numbers.

This algebra is:
 * $(1): \quad$ An associative algebra.
 * $(2): \quad$ A commutative algebra.
 * $(3): \quad$ A normed division algebra.
 * $(4): \quad$ A nicely normed $*$-algebra whose $*$ operator is the identity mapping.
 * $(5): \quad$ A real $*$-algebra.

Construction of Algebra
We have that $\left({\R, +, \times}\right)$ is a field.

Let this be expressed as $\left({\R, +_\R, \times_\R}\right)$ in order to call attention to the precise scope of the operators.

From Real Numbers form Vector Space, we have that $\left({\R^1, +, \cdot}\right)_\R$ is a vector space, where:
 * the field is $\left({\R, +_R, \times_R}\right)$
 * the abelian group is $\left({\R, +_G}\right)$ where $+_G$ is real addition.

In Real Numbers form Vector Space, it is established that elements of $\left({\R^1, +, \cdot}\right)_\R$ are in fact just real numbers.

So, let $\times$ be the binary operation on $\left({\R^1, +, \cdot}\right)_\R$ defined as:
 * $\forall x, y \in \left({\R^1, +, \cdot}\right)_\R: x \times y = x \times_R y$

where $\times_R$ is real multiplication.

Proof of an Algebra
We need to show that $\times$ as defined on $\left({\R^1, +, \cdot}\right)_\R$ as:
 * $\forall x, y \in \left({\R^1, +, \cdot}\right)_\R = x \times_R y$

is bilinear.

That is: $\forall a, b \in \R, x, y \in \R^1$:
 * $\left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right) \times z = \left({a \cdot \left({x \times z}\right)}\right) + \left({b \cdot \left({y \times z}\right)}\right)$
 * $z \times \left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right) = \left({a \cdot \left({z \times x}\right)}\right) + \left({b \cdot \left({z \times y}\right)}\right)$

So:

Similarly:

So the set of real numbers forms an algebra $\left({\R, \times}\right)$.

Proof of Associativity
Elements of $\left({\R, \times}\right)$ are merely real numbers, and $\times$ is just real multiplication.

Associativity of $\times$ therefore follows directly from Real Multiplication is Associative.

Proof of Commutativity
Elements of $\left({\R, \times}\right)$ are merely real numbers, and $\times$ is just real multiplication.

Associativity of $\times$ therefore follows directly from Real Multiplication is Commutative.

Proof of Normed Division Algebra
Elements of $\left({\R, \times}\right)$ are merely real numbers, and $\times$ is just real multiplication.

So from Real Multiplication Identity is One, $\left({\R, \times}\right)$ has a unit, which is $1$.

So $\left({\R, \times}\right)$ is a unitary algebra.

From Inverses for Real Multiplication, every element of $\left({\R, \times}\right)$ except $0$ has a multiplicative inverse.

So $\left({\R, \times}\right)$ is a division algebra and hence a unitary division algebra.

We define a norm on $\left({\R, \times}\right)$ by:
 * $\forall a \in \R: \left \Vert {a} \right \Vert = \left \vert {a} \right \vert = \sqrt {a^2}$

That is, by the absolute value of $a$.

This is a norm because:


 * $(1): \quad \left \Vert x \right \Vert = 0 \iff x = \mathbf 0$
 * $(2): \quad \left \Vert \lambda x \right \Vert = \left \vert \lambda \right \vert \left \vert x \right \vert = \left \vert \lambda \right \vert \left \Vert x \right \Vert$
 * $(3): \quad \left \Vert x - y \right \Vert \le \left \Vert x - z \right \Vert + \left \Vert z - y \right \Vert$ which follows from Real Number Line is Metric Space.

It also follows that:
 * $\left \Vert x \times y \right \Vert = \left \vert x \times y \right \vert = \left \vert x \right \vert \times \left \vert y \right \vert = \left \Vert x \right \Vert \times \left \Vert y \right \Vert$

and so $\left({\R, \times}\right)$ is a normed division algebra.

Proof of Nicely Normed $*$-Algebra
We define the conjugation $*$ by making it the identity mapping on $\R$.

That is:
 * $\forall a \in \R: a^* = a$

We have that:

demonstrating that $*$ is indeed a conjugation.

Then we have that:

Similarly for $a^* + a$.

Trivially, $a^* + a$ and $a \times a^*$ are both real.

So $\left({\R, \times}\right)$ is a nicely normed $*$-algebra.

Proof of Real $*$-Algebra
By definition of $*$ we have that $\forall a \in \R: a^* = a$.

Hence, trivially, $\forall a \in \R: a^* \in \R$.

That is, $\left({\R, \times}\right)$ is a real $*$-algebra.