=== 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: {{begin-eqn}} {{eqn | l = \left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right) \times z       | r = \left({\left({a \times_\R x}\right) + \left({b \times_\R y}\right)}\right) \times_\R z       | c = $\cdot$ and $\times$ are Real Multiplication }} {{eqn | r = \left({\left({a \times_\R x}\right) \times_\R z}\right) + \left({\left({b \times_\R y}\right) \times_\R z}\right)       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \left({a \times_\R \left({x \times_\R z}\right)}\right) + \left({b \times_\R \left({y \times_\R z}\right)}\right)       | c = Real Multiplication is Associative }} {{eqn | r = \left({a \cdot \left({x \times z}\right)}\right) + \left({b \cdot \left({y \times z}\right)}\right)       | c = $\cdot$ and $\times$ are Real Multiplication }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = z \times \left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right)       | r = z \times_\R \left({\left({a \times_\R x}\right) + \left({b \times_\R y}\right)}\right)       | c = $\cdot$ and $\times$ are Real Multiplication }} {{eqn | r = \left({z \times_\R \left({a \times_\R x}\right)}\right) + \left({z \times_\R \left({b \times_\R y}\right)}\right)       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \left({\left({z \times_\R a}\right) \times_\R x}\right) + \left({\left({z \times_\R b}\right) \times_\R y}\right)       | c = Real Multiplication is Associative }} {{eqn | r = \left({\left({a \times_\R z}\right) \times_\R x}\right) + \left({\left({b \times_\R z}\right) \times_\R y}\right)       | c = Real Multiplication is Commutative }} {{eqn | r = \left({a \times_\R \left({z \times_\R x}\right)}\right) + \left({b \times_\R \left({z \times_\R y}\right)}\right)       | c = Real Multiplication is Associative }} {{eqn | r = \left({a \cdot \left({z \times x}\right)}\right) + \left({b \cdot \left({z \times y}\right)}\right)       | c = $\cdot$ and $\times$ are Real Multiplication }} {{end-eqn}} So the set of real numbers forms an algebra $\left({\R, \times}\right)$. {{qed|lemma}} === 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. {{qed|lemma}} === 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. {{qed|lemma}} === 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. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | n = 1       | l = \left({a^*}\right)^*       | r = a^*       | c = Definition of $*$ }} {{eqn | r = a       | c = Definition of $*$ }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 2       | l = \left({a + b}\right)^*       | r = a + b       | c = Definition of $*$ }} {{eqn | r = b + a       | c = Real Addition is Commutative }} {{eqn | r = b^* + a^*       | c =  }} {{end-eqn}} demonstrating that $*$ is indeed a conjugation. Then we have that: {{begin-eqn}} {{eqn | n = 3       | l = a + a^*       | r = a + a       | c = Definition of $*$ }} {{eqn | o = \in       | r = \R       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 4       | l = a \times a^*       | r = a \times a       | c = Definition of $*$ }} {{eqn | r = a^2       | c =  }} {{eqn | o = >       | r = 0       | c = Square of Real Number is Non-Negative }} {{end-eqn}} Similarly for $a^* + a$. Trivially, $a^* + a$ and $a \times a^*$ are both real. So $\left({\R, \times}\right)$ is a nicely normed $*$-algebra. {{qed|lemma}} === 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. {{qed}} Category:Real Numbers Category:Algebras g52sstetggd37595ci3qfc76fbeeapi