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

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:

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

Similarly:

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

\((1):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a^*}\right)^*\)	\(=\)	\(\displaystyle a^*\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $*$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $*$

\((2):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a + b}\right)^*\)	\(=\)	\(\displaystyle a + b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $*$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b + a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as Real Addition is Commutative
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b^* + a^*\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

demonstrating that $*$ is indeed a conjugation.

Then we have that:

\((3):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a + a^*\)	\(=\)	\(\displaystyle a + a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $*$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\in\)	\(\displaystyle \R\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

\((4):\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a \times a^*\)	\(=\)	\(\displaystyle a \times a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of $*$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(>\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Even Powers are Positive

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 $*$ 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.

$\blacksquare$