Write four more rational numbers of the following patterns:
$\frac{2}{- 3}, \frac{- 2}{3}, \frac{4}{- 6}, \frac{- 6}{9} . . . . . .$

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Solution

Given,
$\frac{2}{- 3}, \frac{- 2}{3}, \frac{4}{- 6}, \frac{- 6}{9} . . . . . .$
We have,
$\frac{2}{- 3} = \frac{2 \times 1}{- 3 \times 1}, \frac{- 2}{3} = \frac{2 \times (- 1)}{- 3 \times (- 1)}, \frac{4}{- 6} = \frac{2 \times 2}{- 3 \times 2}, \frac{- 6}{9} = \frac{2 \times (- 3)}{- 3 \times (- 3)}$
Thus the following pattern is observed as
$\frac{2 \times 1}{- 3 \times 1}, \frac{2 \times (- 1)}{- 3 \times (- 1)}, \frac{2 \times 2}{- 3 \times 2}, \frac{2 \times (- 3)}{- 3 \times (- 3)}$
Therefore following the same pattern next four rational numbers will be
$\frac{2 \times 4}{- 3 \times 4}, \frac{2 \times (- 5)}{- 3 \times (- 5)}, \frac{2 \times 6}{- 3 \times 6}, \frac{2 \times (- 7)}{- 3 \times (- 7)}$
or, $\frac{8}{- 12}, \frac{- 10}{15}, \frac{12}{- 18}, \frac{- 14}{21}$
Hence, the required four rational numbers are
$\frac{8}{- 12}, \frac{- 10}{15}, \frac{12}{- 18} a n d \frac{- 14}{21}$

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