Verify commutativity of addition of rational numbers pairs of rational numbers -4 and $$\dfrac{4}{-7}$$

Asked by Sakshi | 1 year ago |  57

##### Solution :-

Firstly we need to convert the denominators to positive numbers.

$$\dfrac{ 4}{-7} = \dfrac{(4 × -1)}{ (-7 × -1)} = \dfrac{-4}{7}$$

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

$$\dfrac{ -4}{1} and \dfrac{-4}{7}$$ as

$$\dfrac{ -4}{1}+ \dfrac{-4}{7}$$ and $$\dfrac{-4}{7}+ \dfrac{ -4}{1}$$

The denominators are 1 and 7

By taking LCM for 1 and 7 is 7

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

Now, $$\dfrac{-4}{1} = \dfrac{(-4 × 7)}{ (1×7)} = \dfrac{-28}{7}$$

$$\dfrac{-4}{7} = \dfrac{(-4 ×1) }{ (7 ×1)} =\dfrac{ -4}{7}$$

Since the denominators are same we can add them directly

$$\dfrac{ -28}{7} +\dfrac{ -4}{7} = \dfrac{(-28 + (-4))}{7}$$

$$= \dfrac{(-28-4)}{7} = \dfrac{-32}{7}$$

$$\dfrac{ -4}{7} +\dfrac{ -4}{1}$$

The denominators are 7 and 1

By taking LCM for 7 and 1 is 7

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

Now, $$\dfrac{ -4}{7}= \dfrac{(-4 ×1) }{ (7 ×1) }= \dfrac{ -4}{7}$$

$$\dfrac{ -4}{1} =\dfrac{ (-4 × 7) }{ (1×7)} =\dfrac{ -28}{7}$$

Since the denominators are same we can add them directly

$$\dfrac{ -4}{7} + \dfrac{ -28}{7} = \dfrac{(-4 + (-28))}{7}$$

$$= \dfrac{(-4-28)}{7} = \dfrac{-32}{7}$$

$$\dfrac{ -4}{1} +\dfrac{ -4}{7} = \dfrac{-4}{7} + \dfrac{-4}{1}$$ is satisfied.

Answered by Sakshi | 1 year ago

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