Question

Is $\dfrac{9}{8}$ the multiplicative inverse of $ - 1\dfrac{1}{8}$? Why or why not?

Answer 1

Hint: First we’ll understand the concept of mixed fraction, improper fraction, and how they are transformed, using this we’ll simplify the given term and then learning about multiplicative inverse that multiplicative inverse of a number ‘n ’ is be given by $\dfrac{1}{n}$.
Keeping these properties in mind we’ll approach our answer.

Complete step by step Answer:

Given: To find, Is $\dfrac{9}{8}$ the multiplicative inverse of$ - 1\dfrac{1}{8}$
A mixed fraction is a whole number plus a fractional part. An improper fraction is a fraction where the numerator is larger than the denominator and if the numerator is smaller than the denominator then it is called a proper fraction, we can only interchange the improper fraction to mixed fraction and vice versa.
If we have a mixed fraction $a\dfrac{b}{c}$then its form in improper fraction will be $\dfrac{{ac + b}}{c}$
Therefore, $ - 1\dfrac{1}{8} = - \dfrac{{1(8) + 1}}{8}$
$ = - \dfrac{9}{8}$
Now, we know that if we have a number ‘n ’ then the multiplicative inverse of ‘n’ will be given by $\dfrac{1}{n}$ .
Or we can say that the product of any number and its multiplicative inverse is equal to 1.
i.e. $n \times \dfrac{1}{n} = 1$
Therefore, if $\dfrac{9}{8}$is a multiplicative inverse of $ - 1\dfrac{1}{8}$then their product must be 1
i.e. \[ - 1\dfrac{1}{8} \times \dfrac{9}{8} = - \dfrac{9}{8} \times \dfrac{9}{8}\]but \[ - \dfrac{9}{8} \times \dfrac{9}{8} \ne 1\]
Therefore, $\dfrac{9}{8}$is not the multiplicative inverse of $ - 1\dfrac{1}{8}$

Note: Some of the students misinterpret this mixed fraction and just do the simple multiplication of the whole number and the fraction, which is wrong and will lead us to a wrong solution to the question. Some students write this mixed fraction $a\dfrac{b}{c}$as $a \times \dfrac{b}{c}$which is wrong and $a\dfrac{b}{c} = \dfrac{{ac + b}}{c} \ne a \times \dfrac{b}{c}$.