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
From Real Numbers form Field, $\struct {\R, +, \times}$ is a field.

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

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

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

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

where $\times_\R$ is real multiplication.

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

is bilinear.

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

So:

Similarly:

So the set of real numbers forms an algebra $\struct {\R, \times}$.

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

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

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

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

So $\struct {\R, \times}$ is a unitary algebra.

From Inverse for Real Multiplication, every element of $\struct {\R, \times}$ except $0$ has a multiplicative inverse.

So $\struct {\R, \times}$ is a division algebra and hence a unitary division algebra.

We define a norm on $\struct {\R, \times}$ by:
 * $\forall a \in \R: \norm a = \size a = \sqrt {a^2}$

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

This is a norm because:


 * $(1): \quad \norm x = 0 \iff x = \mathbf 0$
 * $(2): \quad \norm {\lambda x} = \size \lambda \size x = \size \lambda \norm x$
 * $(3): \quad \norm {x - y} \le \norm {x - z} + \norm {z - y}$ which follows from Real Number Line is Metric Space.

It also follows that:
 * $\norm {x \times y} = \size {x \times y} = \size x \times \size y = \norm x \times \norm y$

and so $\struct {\R, \times}$ 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 $\struct {\R, \times}$ is a nicely normed $*$-algebra.

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

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

That is, $\struct {\R, \times}$ is a real $*$-algebra.