Real Numbers under Multiplication do not form Group

Theorem

The algebraic structure $\struct {\R, \times}$ consisting of the set of real numbers $\R$ under multiplication $\times$ is not a group.

Proof

Aiming for a contradiction, suppose that $\struct {\R, \times}$ 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, $\struct {\R, \times}$ is the trivial group.

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

From this contradiction it follows that $\struct {\R, \times}$ is not a group.

$\blacksquare$

Also see

• Non-Zero Real Numbers under Multiplication form Abelian Group

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.5$. Examples of groups: Example $82$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
• 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): $\S 1$: Some examples of groups