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 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 \(\dfrac{5 }{ 2} and \dfrac{– 11 }{ 12}\)