Real Numbers under Multiplication do not form Group

Theorem
The algebraic structure $\left({\R, \times}\right)$ consisting of the set of real numbers $\R$ under multiplication $\times$ is not a group.

Proof
Aiming for a contradiction, suppose that $\left({\R, \times}\right)$ is a group.

By the definition of the number $0 \in \R$:
 * $\forall x \in \R: x \times 0 = 0 = 0 \times x$

Thus $0$ is a zero in the abstract algebraic sense.

From Group with Zero Element is Trivial, $\left({\R, \times}\right)$ is the trivial group.

But $\R$ contains other elements besides $0$.

From this contradiction it follows that $\left({\R, \times}\right)$ is not a group.

Also see

 * Non-Zero Real Numbers under Multiplication form Abelian Group