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
\(\dfrac{ 4}{1} \;and\; \dfrac{-3}{5}\) as
\( \dfrac{ 4}{1} \;and\; \dfrac{-3}{5}\)and \(\dfrac{-3}{5}+ \dfrac{ 4}{1}\)
The denominators are 1 and 5
By taking LCM for 1 and 5 is 5
We rewrite the given fraction in order to get the same denominator
Now, \(\dfrac{ 4}{1} = \dfrac{(4 × 5) }{ (1×5)} =\dfrac{ 20}{5}\)
\(\dfrac{ -3}{5} =\dfrac{ (-3 ×1) }{ (5 ×1)} = \dfrac{-3}{5}\)
Since the denominators are same we can add them directly
\( \dfrac{20}{5} + \dfrac{-3}{5} = \dfrac{(20 + (-3))}{5}\)
\( = \dfrac{(20-3)}{5} = \dfrac{17}{5}\)
\(\dfrac{ -3}{5} + \dfrac{4}{1}\)
The denominators are 5 and 1
By taking LCM for 5 and 1 is 5
We rewrite the given fraction in order to get the same denominator
Now, \(\dfrac{ -3}{5} =\dfrac{ (-3 ×1) }{ (5 ×1)} = \dfrac{ -3}{5} \)
\(\dfrac{ 4}{1} =\dfrac{ (4 × 5) }{ (1×5) }= \dfrac{20}{5}\)
Since the denominators are same we can add them directly
\(\dfrac{ -3}{5} + \dfrac{20}{5} = \dfrac{(-3 + 20)}{5} = \dfrac{17}{5}\)
\( \dfrac{4}{1} +\dfrac{ -3}{5} = \dfrac{ -3}{5} + \dfrac{4}{1}\) is satisfied.
Answered by Sakshi | 1 year agoBy what number should 1365 be divided to get 31 as quotient and 32 as remainder?
Which of the following statement is true / false?
(i) \(\dfrac{ 2 }{ 3} – \dfrac{4 }{ 5}\) is not a rational number.
(ii) \( \dfrac{ -5 }{ 7}\) is the additive inverse of \( \dfrac{ 5 }{ 7}\)
(iii) 0 is the additive inverse of its own.
(iv) Commutative property holds for subtraction of rational numbers.
(v) Associative property does not hold for subtraction of rational numbers.
(vi) 0 is the identity element for subtraction of rational numbers.
If x = \( \dfrac{4 }{ 9}\), y =\( \dfrac{-7 }{ 12}\) and z = \( \dfrac{-2 }{ 3}\), then verify that x – (y – z) ≠ (x – y) – z
If x = \(\dfrac{ – 4 }{ 7}\) and y = \( \dfrac{2 }{ 5}\), then verify that x – y ≠ y – x
Subtract the sum of \(\dfrac{ – 5 }{ 7} and\dfrac{ – 8 }{ 3}\) from the sum of \(\dfrac{5 }{ 2} and \dfrac{– 11 }{ 12}\)