Arrange the following rational numbers in ascending order.

$\frac{3}{8}, \frac{4}{12}, \frac{- 7}{16}, \frac{- 2}{3}$ .

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Solution

LCM of denominators

$8, 12, 16, a n d 3 = 2 \times 2 \times 2 \times 2 \times 3 = 48.$

$T h e n \frac{3}{8} = \frac{3 \times 6}{8 \times 6} = \frac{18}{48} : \frac{4}{12} = \frac{4 \times 4}{12 \times 4} = \frac{16}{48};$

$\frac{- 7}{16} = \frac{- 7 \times 3}{16 \times 3} = \frac{- 21}{48}; \frac{- 2}{3} = \frac{- 2 \times 16}{3 \times 16} = \frac{- 32}{48}$

$T h e e q u i v a l e n t r a t i o n a l n u m b e r s a r e \frac{18}{48}, \frac{16}{48}, \frac{- 21}{48} a n d \frac{- 32}{48}$

Therefore, the smallest rational number is $\frac{- 32}{48},$ then comes, $\frac{- 21}{48},$ then comes $\frac{16}{48}$ , and the greatest rational number is $\frac{18}{48}$ . Hence, their ascending order is $\frac{- 2}{3}, \frac{- 7}{16}, \frac{4}{12}, \frac{3}{8}$ .

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