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.
== Proof ==
=== 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.
Category:Real Numbers
Category:Algebras