Real Numbers form Algebra

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

\(\displaystyle \left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right) \times z\) \(=\) \(\displaystyle \left({\left({a \times_\R x}\right) + \left({b \times_\R y}\right)}\right) \times_\R z\) $\cdot$ and $\times$ are Real Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({a \times_\R x}\right) \times_\R z}\right) + \left({\left({b \times_\R y}\right) \times_\R z}\right)\) Real Multiplication Distributes over Addition
\(\displaystyle \) \(=\) \(\displaystyle \left({a \times_\R \left({x \times_\R z}\right)}\right) + \left({b \times_\R \left({y \times_\R z}\right)}\right)\) Real Multiplication is Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({a \cdot \left({x \times z}\right)}\right) + \left({b \cdot \left({y \times z}\right)}\right)\) $\cdot$ and $\times$ are Real Multiplication


Similarly:

\(\displaystyle z \times \left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right)\) \(=\) \(\displaystyle z \times_\R \left({\left({a \times_\R x}\right) + \left({b \times_\R y}\right)}\right)\) $\cdot$ and $\times$ are Real Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({z \times_\R \left({a \times_\R x}\right)}\right) + \left({z \times_\R \left({b \times_\R y}\right)}\right)\) Real Multiplication Distributes over Addition
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({z \times_\R a}\right) \times_\R x}\right) + \left({\left({z \times_\R b}\right) \times_\R y}\right)\) Real Multiplication is Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({a \times_\R z}\right) \times_\R x}\right) + \left({\left({b \times_\R z}\right) \times_\R y}\right)\) Real Multiplication is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left({a \times_\R \left({z \times_\R x}\right)}\right) + \left({b \times_\R \left({z \times_\R y}\right)}\right)\) Real Multiplication is Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({a \cdot \left({z \times x}\right)}\right) + \left({b \cdot \left({z \times y}\right)}\right)\) $\cdot$ and $\times$ are Real Multiplication

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

$\Box$


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.

$\Box$


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.

$\Box$


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.

$\Box$


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:

\((1):\quad\) \(\displaystyle \left({a^*}\right)^*\) \(=\) \(\displaystyle a^*\) Definition of $*$
\(\displaystyle \) \(=\) \(\displaystyle a\) Definition of $*$
\((2):\quad\) \(\displaystyle \left({a + b}\right)^*\) \(=\) \(\displaystyle a + b\) Definition of $*$
\(\displaystyle \) \(=\) \(\displaystyle b + a\) Real Addition is Commutative
\(\displaystyle \) \(=\) \(\displaystyle b^* + a^*\)

demonstrating that $*$ is indeed a conjugation.


Then we have that:

\((3):\quad\) \(\displaystyle a + a^*\) \(=\) \(\displaystyle a + a\) Definition of $*$
\(\displaystyle \) \(\in\) \(\displaystyle \R\)
\((4):\quad\) \(\displaystyle a \times a^*\) \(=\) \(\displaystyle a \times a\) Definition of $*$
\(\displaystyle \) \(=\) \(\displaystyle a^2\)
\(\displaystyle \) \(>\) \(\displaystyle 0\) Square of Real Number is Non-Negative

Similarly for $a^* + a$.

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

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

$\Box$


Proof of Real $*$-Algebra

By definition of $*$:

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

Hence, trivially:

$\forall a \in \R: a^* \in \R$

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

$\blacksquare$