Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when x = \( \dfrac{-2}{5}\), y = \( \dfrac{4}{3}\), z = \( \dfrac{-7}{10}\)

Asked by Sakshi | 1 year ago |  70

1 Answer

Solution :-

As the property states (x + y) + z = x + (y + z)

Use the values as such,

\( \dfrac{-2}{5} + \dfrac{4}{3} + \dfrac{-7}{10} =\dfrac{ -2}{5 }+ \dfrac{4}{3} + \dfrac{-7}{10}\)

Let us consider LHS \(( \dfrac{-2}{5} + \dfrac{4}{3} )+ (\dfrac{-7}{10})\)

Taking LCM for 5 and 3 is 15

\(\dfrac{ (-2× 3)}{(5×3)} + \dfrac{(4×5)}{(3×5)}\)

\( \dfrac{-6}{15} + \dfrac{20}{15}\)

Since the denominators are same we can add them directly,

\( \dfrac{-6}{15} + \dfrac{20}{15}= \dfrac{(-6+20)}{15} = \dfrac{14}{15}\)

\( \dfrac{14}{15}+\dfrac{-7}{10}\)

Taking LCM for 15 and 10 is 30

\(\dfrac{ (14×2)}{(15×2)} + \dfrac{(-7×3)}{(10×3)}\)

\(\dfrac{ 28}{30 }+ \dfrac{(-21)}{30}\)

Since the denominators are same we can add them directly,

\( \dfrac{(28+(-21)}{30} = \dfrac{(28-21)}{30} = \dfrac{7}{30}\)

Let us consider RHS \(\dfrac{ -2}{5} + \dfrac{4}{3} + \dfrac{-7}{10}\)

Taking LCM for 3 and 10 is 30

\( \dfrac{4}{3} + \dfrac{-7}{10} = \dfrac{(4×10)}{(3×10)} + \dfrac{(-7×3)}{(10×3)}\)

= \(\dfrac{ 40}{30 }+ \dfrac{(-21)}{30}\)

Since the denominators are same we can add them directly,

\( \dfrac{ 40}{30 }+ \dfrac{(-21)}{30}= \dfrac{(40-21)}{30} = \dfrac{19}{30}\)

\( \dfrac{-2}{5} + \dfrac{19}{30}\)

Taking LCM for 5 and 30 is 30

\( \dfrac{-2}{5} + \dfrac{19}{30}=\dfrac{(-2×6)}{(5×6) }+\dfrac{ (19×1)}{(30×1)}\)

=\(\dfrac{ -12}{30} + \dfrac{19}{30}\)

Since the denominators are same we can add them directly,

= \(\dfrac{ (-12 + 19)}{30} = \dfrac{7}{30}\)

LHS = RHS associativity of addition of rational numbers is verified.

Answered by Sakshi | 1 year ago

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