Which of the following expressions shows that rational numbers are associative under multiplication?

1. $\dfrac{4}{5} \times \left(\dfrac{-6}{7} \times \dfrac{8}{9}\right) = \left(\dfrac{4}{5} \times \dfrac{-6}{7}\right) \times \dfrac{8}{9}$

2. $\dfrac{4}{5} \times \left(\dfrac{-6}{7} \times \dfrac{8}{9}\right) = \dfrac{4}{5} \times \left(\dfrac{8}{9} \times \dfrac{-6}{7}\right)$

3. $\dfrac{4}{5} \times \left(\dfrac{6}{7} \times \dfrac{8}{9}\right) = \left(\dfrac{8}{9} \times \dfrac{4}{5}\right) \times \dfrac{-6}{7}$

4. $\left(\dfrac{4}{5} \times \dfrac{-6}{7}\right) \times \dfrac{8}{9} = \left(\dfrac{-6}{7} \times \dfrac{4}{5}\right) \times \dfrac{8}{9}$

If $\dfrac{a}{b}, \dfrac{c}{d} \text{ and } \dfrac{e}{f}$ are any three rational numbers, then the associative law of multiplication states that $\dfrac{a}{b} \times \left(\dfrac{c}{d} \times \dfrac{e}{f}\right) = \left(\dfrac{a}{b} \times \dfrac{c}{d}\right) \times \dfrac{e}{f}$

In Option 1, only the grouping of the same three rational numbers, taken in the same order, has been changed.

In Options 2 and 4, the order of the rational numbers has been changed, which shows the commutative law.

In Option 3, the order of the rational numbers has been changed and $\dfrac{6}{7}$ on the left side has been replaced by $\dfrac{-6}{7}$ on the right side, so the two sides are not equal at all.

Hence, Option 1 is the correct option.