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

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

Let $+$ denote addition on $\R$.

Then $\left({\R, +}\right)$ 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 $\left({\R, +}\right)$.

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

Let $+$ denote addition on $\R$.

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

Then:
 * $\left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] + \left[{\!\left[{\left\langle{0}\right\rangle}\right]\!}\right] = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] = \left[{\!\left[{\left\langle{0}\right\rangle}\right]\!}\right] + \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right]$

Hence, $\left[{\!\left[{\left\langle{0}\right\rangle}\right]\!}\right]$ is the identity of $\left({\R, +}\right)$.

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

Let $+$ denote addition on $\R$.

Define:
 * $0^* = \left\{{q \in \Q: q < 0}\right\}$

We show that $0^*$ is the identity of $\left({\R, +}\right)$.

First, we show that $0^* \in \R$.

Let $q \in 0^*$.

If $p \in \Q$ and $p < q$, then $p \in 0^*$.

We have that $\left({\Q, +, \times, \le}\right)$ is a totally ordered field.

By Properties of Totally Ordered Field, we have that $1 + 1 > 1 > 0$, and so $\left({1 + 1}\right)^{-1} > 0$.

Therefore, $q < \left({1 + 1}\right)^{-1} \times q$.

By Properties of Totally Ordered Field, we have that $\left({1 + 1}\right)^{-1} \times q \in 0^*$.

Let $\alpha \in \R$.

We only show that $\alpha + 0^* = \alpha$; the equality $0^* + \alpha = \alpha$ holds similarly.

Suppose that $p \in \alpha$.

If $q \in 0^*$, then $p + q < p$, and so $p + q \in \alpha$.

Hence, $\alpha + 0^* \subseteq \alpha$.

Now, let $p \in \alpha$.

By the definition of a Dedekind cut, we can choose $q \in \alpha$ such that $p < q$.

We have that $-q + p < 0$.

Therefore, $p = q + \left({-q + p}\right) \in \alpha + 0^*$.

Hence, $\alpha \subseteq \alpha + 0^*$.

By Equality of Sets, we have $\alpha = \alpha + 0^*$.

Proof 4
Let $\R$ denote the set of real numbers, as defined as the Dedekind completion of the rational numbers.

Let $+$ denote addition on $\R$.

By definition, $\left({\R, +}\right)$ is a group.

By the group axioms, $\left({\R, +}\right)$ has an identity.