Question

Write the additive inverse of each of the following.
(i) $ \dfrac{2}{8} $ (ii) $ \dfrac{-5}{9} $ (iii) $ \dfrac{-6}{-5} $ (iv) $ \dfrac{2}{-9} $ (v) $ \dfrac{19}{-6} $

Answer 1

Hint: For answering this question we need to write the additive inverse of the given numbers. In mathematics, the additive inverse of a number $ a $ is the number which when added to $ a $, yields zero. This number can be given as $ -a $. For a real number, the additive inverse of a number has opposite sign with respect to the sign of the given number and same magnitude.

Complete step by step answer:
(i) For the first part of the question, we need to find the additive inverse of $ \dfrac{2}{8} $ .
Let the Additive inverse of the given number be $ x $.
From the basic concept, we know the additive inverse of a given number is the number which when added to the given number gives zero.
Hence we can say that $ \dfrac{2}{8}+x=0 $
By performing calculations we will have
 $ x=-\dfrac{2}{8} $ is the additive inverse of the given number.
(ii) For the first part of the question, we need to find the additive inverse of $ \dfrac{-5}{9} $ .
Let the Additive inverse of the given number by $ x $.
From the basic concept, we know the additive inverse of a given number is the number which when added to the given number gives zero.
Hence we can say that $ \dfrac{-5}{9}+x=0 $
By performing calculations we will have
 $ x=\dfrac{5}{9} $ is the additive inverse of the given number.
(iii) For the first part of the question, we need to find the additive inverse of $ \dfrac{-6}{-5} $.
Let the Additive inverse of the given number by $ x $.
From the basic concept, we know the additive inverse of a given number is the number which when added to the given number gives zero.
Hence we can say that $ \dfrac{-6}{-5}+x=0 $
By performing calculations we will have
 $ x=\dfrac{-6}{5} $ is the additive inverse of the given number.
(iv) For the first part of the question, we need to find the additive inverse of $ \dfrac{2}{-9} $.
Let the Additive inverse of the given number by $ x $.
From the basic concept, we know the additive inverse of a given number is the number which when added to the given number gives zero.
Hence we can say that $ \dfrac{2}{-9}+x=0 $
By performing calculations we will have
 $ x=\dfrac{2}{9} $ is the additive inverse of the given number.
(v) For the first part of the question, we need to find the additive inverse of $ \dfrac{19}{-6} $.
Let the Additive inverse of the given number be $ x $.
From the basic concept, we know the additive inverse of a given number is the number which when added to the given number gives zero.
Hence we can say that $ \dfrac{19}{-6}+x=0 $
By performing calculations we will have
 $ x=\dfrac{19}{6} $ is the additive inverse of the given number.

Note:
We should be careful while calculating the additive inverse of the given number. Similar to the additive inverse we have multiplicative inverse also which is defined as the number which gives 1 when multiplied by the given number. The additive identity is 0 and the multiplicative identity is 1. The definition of inverse of a number for an operation is the number that results in the respective identity when the operation is performed between both the numbers.