Verify commutativity of addition of rational numbers pairs of rational numbers $$\dfrac{-3}{5}$$ and $$\dfrac{-2}{-15}$$

Asked by Sakshi | 1 year ago |  54

##### Solution :-

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

$$\dfrac{ -3}{5} and \dfrac{-2}{-15}$$ as

$$\dfrac{ -3}{5} + \dfrac{-2}{-15}$$ and $$\dfrac{-2}{-15}+ \dfrac{ -3}{5}$$

$$\dfrac{-2}{-15} = \dfrac{2}{15}$$

The denominators are 5 and 15

By taking LCM for 5 and 15 is 15

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

Now, $$\dfrac{ -3}{5} = \dfrac{(-3 × 3) }{ (5×3) }= \dfrac{-9}{15}$$

$$\dfrac{2}{15} =\dfrac{ (2 ×1) }{ (15 ×1)} = \dfrac{2}{15}$$

Since the denominators are same we can add them directly

$$\dfrac{-9}{15}+\dfrac{2}{15}=\dfrac{ (-9 + 2)}{15} =\dfrac{ -7}{15}$$

$$\dfrac{-2}{-15}+ \dfrac{ -3}{5}$$

The denominators are 15 and 5

By taking LCM for 15 and 5 is 15

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

Now, $$\dfrac{2}{15}= \dfrac{(2 ×1) }{ (15 ×1)} = \dfrac{2}{15}$$

$$\dfrac{ -3}{5} =\dfrac{ (-3 × 3) }{ (5×3) }= \dfrac{-9}{15}$$

Since the denominators are same we can add them directly

$$\dfrac{2}{15}+ \dfrac{-9}{15} = \dfrac{(2 + (-9))}{15 }=\dfrac{ (2-9)}{15} = \dfrac{-7}{15}$$

$$\dfrac{-3}{5 } + \dfrac{-2}{-15 } = \dfrac{-2}{-15 } + \dfrac{-3}{5 }$$ is satisfied.

Answered by Aaryan | 1 year ago

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