Verify commutativity of addition of rational numbers pairs of rational numbers -3/5 and -2/-15

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.