For each set of rational numbers, given below, verify the associative property of addition of rational numbers:

$\dfrac{1}{2} , \dfrac{2}{3}$ and $\dfrac{-1}{6}$

To prove:

$\Big(\dfrac{1}{2} + \dfrac{2}{3}\Big) + \dfrac{-1}{6} = \dfrac{1}{2} + \Big(\dfrac{2}{3} + \dfrac{-1}{6}\Big)$

Taking LHS:

$\Big(\dfrac{1}{2} + \dfrac{2}{3}\Big) + \dfrac{-1}{6} \\[1em]$ LCM of 2 and 3 is 2 x 3 = 6

$= \Big(\dfrac{1 \times 3}{2 \times 3} + \dfrac{2 \times 2}{3 \times 2}\Big) + \dfrac{-1}{6} \\[1em] = \Big(\dfrac{3}{6} + \dfrac{4}{6}\Big) + \dfrac{-1}{6} \\[1em] = \Big(\dfrac{3 + 4}{6}\Big) + \dfrac{-1}{6} \\[1em] = \dfrac{7}{6}+ \dfrac{-1}{6} \\[1em] = \dfrac{7-1}{6} \\[1em] = \dfrac{6}{6} \\[1em] = 1$

Taking RHS: $\dfrac{1}{2} + \Big(\dfrac{2}{3} + \dfrac{-1}{6}\Big)$

LCM of 3 and 6 is 2 x 3 = 6 $\dfrac{1}{2} + \Big(\dfrac{2 \times 2}{3 \times 2} + \dfrac{-1 \times 1}{6 \times 1}\Big) \\[1em] = \dfrac{1}{2} + \Big(\dfrac{4}{6} + \dfrac{-1}{6}\Big) \\[1em] = \dfrac{1}{2} + \Big( \dfrac{4 +(-1)}{6}\Big) \\[1em] = \dfrac{1}{2} + \dfrac{3}{6} \\[1em]$

LCM of 2 and 6 is 2 x 3 = 6

$= \dfrac{1 \times 3}{2 \times 3} + \dfrac{3 \times 1}{6 \times 1} \\[1em] = \dfrac{3}{6} + \dfrac{3}{6} \\[1em] = \dfrac{3 + 3}{6} \\[1em] = \dfrac{6}{6} \\[1em] = 1$

∴ LHS = RHS

$\Big(\dfrac{1}{2} + \dfrac{2}{3}\Big) + \dfrac{-1}{6} = \dfrac{1}{2} + \Big(\dfrac{2}{3} + \dfrac{-1}{6}\Big)$

So, the associative property for the addition of the rational number is verified.