User:Abcxyz/Sandbox/Real Numbers/Identity for Real Multiplication

Theorem
Let $\R$ denote the set of real numbers.

Let $\times$ denote multiplication on $\R$.

Then $\left({\R, \times}\right)$ has an identity.

Proof 1
Let $\left({\R, +, \times, \le}\right)$ denote the real numbers, as axiomatically defined as a Dedekind complete totally ordered field.

By the field axioms, there exists an identity of $\left({\R, \times}\right)$.

Proof 2
Let $\R$ denote the set of real numbers, as constructed from Cauchy sequences.

Let $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] \in \R$.

Then:
 * $\left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] \left[{\!\left[{\left\langle{1}\right\rangle}\right]\!}\right] = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] = \left[{\!\left[{\left\langle{1}\right\rangle}\right]\!}\right] \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right]$

Hence, $\left[{\!\left[{\left\langle{1}\right\rangle}\right]\!}\right]$ is the identity of $\left({\R, \times}\right)$.

Proof 3
Let $\R$ denote the set of real numbers, as constructed from Dedekind cuts.

Define:
 * $1^* = \left\{{q \in \Q: q < 1}\right\}$

We show that $1^*$ is the identity of $\left({\R, \times}\right)$.

First, we show that $1^* \in \R$.

We have that $0 < 1$, so $1^*$ is non-empty.

We have that $1 \notin 1^*$, so $1^* \ne \Q$.

Suppose that $q \in 1^*$.

If $p \in \Q$ and $p < q$, then $p \in 1^*$.

We have that $q < \frac 1 2 \left({q + 1}\right) \in 1^*$.

Let $\alpha \in \R$.

$1^*$ is Left Identity of $\left({\R, \times}\right)$
We wish to show that $1^* \alpha = \alpha$.

Case where $\alpha \ge 0^*$
Suppose that $q \in 1^* \alpha$.

If $q < 0$, then $q \in \alpha$.

If $q \ge 0$, then, by the definition of real multiplication, we can choose $r \in 1^*$, $r > 0$ and $s \in \alpha$ such that $q = rs$.

We have that $s \ge 0$.

Hence, $q = rs \le s \in \alpha$, so $q \in \alpha$.

Since $q$ was arbitrary, it follows that $1^* \alpha \subseteq \alpha$.

Suppose that $q' \in \alpha$.

If $q' \le 0$, then $q' \le \frac 1 2 q' \in 1^* \alpha$, so $q' \in 1^* \alpha$.

If $q' > 0$, then we can choose $r' > q'$ such that $r' \in \alpha$.

Hence, $q' = \left({\dfrac {q'} {r'}}\right) r' \in 1^* \alpha$.

Since $q'$ was arbitrary, it follows that $\alpha \subseteq 1^* \alpha$.

By the definition of set equality, we have that $1^* \alpha = \alpha$.

Case where $\alpha < 0^*$

 * $1^* \alpha = -\left({1^* \left({-\alpha}\right)}\right) = -\left({-\alpha}\right) = \alpha$

$1^*$ is Right Identity of $\left({\R, \times}\right)$
Since real multiplication is commutative, we have that:
 * $\alpha 1^* = 1^* \alpha = \alpha$