Rational Numbers under Multiplication do not form Group

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Theorem

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


Proof

Aiming for a contradiction, suppose that $\left({\Q, \times}\right)$ 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, $\left({\Q, \times}\right)$ is the trivial group.

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

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

$\blacksquare$


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