Non-Zero Real Numbers under Multiplication form Abelian Group

Theorem

Let $\R_{\ne 0}$ be the set of real numbers without zero:

$\R_{\ne 0} = \R \setminus \set 0$

The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.

Proof 1

Taking the group axioms in turn:

G0: Closure

From Non-Zero Real Numbers Closed under Multiplication: Proof 2, $\R_{\ne 0}$ is closed under multiplication.

Note that proof 2 needs to be used specifically here, as proof 1 rests on this result.

$\Box$

G1: Associativity

$\Box$

G2: Identity

The identity element of real number multiplication is the real number $1$:

$\exists 1 \in \R: \forall a \in \R_{\ne 0}: a \times 1 = a = 1 \times a$

$\Box$

G3: Inverses

Each element $x$ of the set of non-zero real numbers $\R_{\ne 0}$ has an inverse element $\dfrac 1 x$ under the operation of real number multiplication:

$\forall x \in \R_{\ne 0}: \exists \dfrac 1 x \in \R_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$

$\Box$

C: Commutativity

$\Box$

Infinite

$\blacksquare$

Proof 2

Because Real Numbers under Multiplication form Monoid and the non-zero numbers are exactly the invertible elements of real multiplication (Inverses for Real Multiplication) then the non zero reals under multiplication form a group by Invertible Elements of Monoid form Subgroup of Cancellable Elements.

The group must also be Abelian because Real Multiplication is Commutative and Subset Product within Commutative Structure is Commutative

$\blacksquare$

Proof 3

From Non-Zero Real Numbers under Multiplication form Group, $\struct {\R_{\ne 0}, \times}$ forms a group.

$\Box$

From Real Multiplication is Commutative it follows that $\struct {\R_{\ne 0}, \times}$ is abelian.

$\Box$

From Real Numbers are Uncountably Infinite it follows that $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.

$\blacksquare$