Verify commutativity of addition of rational numbers pairs of rational numbers \( \dfrac{2}{-7}\) and \( \dfrac{12}{-35}\)

Asked by Sakshi | 1 year ago |  51

1 Answer

Solution :-

Firstly we need to convert the denominators to positive numbers.

\(\dfrac{ 2}{-7} =\dfrac{ (2 × -1)}{ (-7 × -1)} = \dfrac{ 2}{-7}\)

\(\dfrac{ 12}{-35} = \dfrac{(12 × -1)}{ (-35 × -1) }= \dfrac{ -12}{35}\)

By using the commutativity law, the addition of rational numbers is commutative.

\(\dfrac{ a}{b} + \dfrac{c}{d} =\dfrac{c}{d} + \dfrac{ a}{b}\)

In order to verify the above property let us consider the given fraction

\(\dfrac{ -2}{7} and -\dfrac{12}{35}\) as

\( \dfrac{ -2}{7} + -\dfrac{12}{35}\)and \(\dfrac{-12}{35}+ \dfrac{ -2}{7} \)

The denominators are 7 and 35

By taking LCM for 7 and 35 is 35

We rewrite the given fraction in order to get the same denominator

Now, \( \dfrac{ -2}{7} = \dfrac{(-2 × 5) }{ (7 ×5) }= \dfrac{-10}{35}\)

\(\dfrac{ -12}{35} = \dfrac{(-12 ×1) } {(35 ×1) }= \dfrac{ -12}{35} \)

Since the denominators are same we can add them directly

\(\dfrac{ -10}{35} +\dfrac{ (-12)}{35} = \dfrac{(-10 + (-12))}{35 }\)

\( = \dfrac{(-10-12)}{35} = \dfrac{-22}{35}\)

\( \dfrac{-12}{35 }+ \dfrac{-2}{7}\)

The denominators are 35 and 7

By taking LCM for 35 and 7 is 35

We rewrite the given fraction in order to get the same denominator

Now, \(\dfrac{ -12}{35} =\dfrac{ (-12 ×1) }{ (35 ×1)} = \dfrac{-12}{35}\)

\(\dfrac{ -2}{7} =\dfrac{ (-2 × 5) }{ (7 ×5)} = \dfrac{-10}{35}\)

Since the denominators are same we can add them directly

\( \dfrac{-12}{35} + \dfrac{-10}{35} = \dfrac{(-12 + (-10))}{35} \)

\( = \dfrac{(-12-10)}{35} = \dfrac{-22}{35}\)

\(\dfrac{ -2}{7} +\dfrac{ -12}{35} = \dfrac{-12}{35} +\dfrac{ -2}{7}\) is satisfied.

Answered by Sakshi | 1 year ago

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