# Rational Numbers under Multiplication do not form Group

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## Theorem

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

## Proof

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

By the definition of the number $0 \in \Q$:

- $\forall x \in \Q: x \times 0 = 0 = 0 \times x$

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

From Group with Zero Element is Trivial, $\struct {\Q, \times}$ is the trivial group.

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

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

$\blacksquare$

## Also see

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$