$For each pair of rational numbers, given below, verify that the multiplicatin is commutative : (i) \frac{- 1}{5} and \frac{2}{9} (i i) \frac{5}{- 3} and \frac{13}{- 11} (i i i) 3 and \frac{- 8}{9} (i v) 0 and \frac{- 12}{17}$

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Solution

$(i) \frac{- 1}{5} and \frac{2}{9} \frac{- 1}{5} \times \frac{2}{9} = \frac{(- 1) \times 2}{9 \times 5} = \frac{- 2}{45} And, \frac{2}{9} \times (\frac{- 1}{5}) = \frac{2 \times (- 1)}{9 \times 5} = \frac{- 2}{45} ∴ \frac{- 1}{5} \times \frac{2}{9} = \frac{2}{9} \times \frac{- 1}{5} (i i) \frac{5}{- 3} and \frac{13}{- 11} \frac{5}{- 3} \times \frac{13}{- 11} = \frac{5 \times 13}{(- 3) \times (- 11)} = \frac{65}{33} And, \frac{13}{- 11} \times \frac{5}{- 3} = \frac{13 \times 5}{(- 3) \times (- 11)} = \frac{65}{33} ∴ \frac{5}{- 3} \times \frac{13}{- 11} = \frac{13}{- 11} \times \frac{5}{- 3} (i i i) 3 and \frac{- 8}{9} \frac{3}{1} \times \frac{- 8}{9} = \frac{1 \times (- 8)}{1 \times 3} = \frac{- 8}{3} And, \frac{- 8}{9} \times \frac{3}{1} = \frac{(- 8) \times 1}{3 \times 1} = \frac{- 8}{3} ∴ 3 \times \frac{- 8}{9} = \frac{8}{9} \times 3 (i v) 0 and \frac{- 12}{17} = 0 \times \frac{- 12}{17} = \frac{0 \times (- 12)}{1 \times 17} = 0 And \frac{- 12}{17} \times 0 = \frac{(- 12) \times 0}{17 \times 1} = 0 ∴ 0 \times \frac{(- 12)}{17} = \frac{(- 12)}{17} \times 0$

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