Question

Find the multiplicative inverse of $\dfrac{4}{7}\times \dfrac{21}{28}$.\[\]

Answer 1

Hint: We know that the multiplicative inverse of a fraction $\dfrac{a}{b}$ where $a,b\ne 0$ are real number is given by $\dfrac{b}{a}$. We simplify the given expression to the form $\dfrac{a}{b}$ by dividing the terms in numerator and denominator by a common factor and then find the multiplicative inverse.\[\]

Complete step by step answer:
The given numerical expression from the question is
\[\dfrac{4}{7}\times \dfrac{21}{28}\]

We see that the given expression has two fractions multiplied with each other, The first fraction is $\dfrac{4}{7}$ and the second fraction is$\dfrac{21}{28}$. We are asked to the multiplicative inverse of their product.\[\]

 The fractions belong to the set of rational number .We know that multiplicative identity of rational number set is 1 as we can multiply any rational number with 1 and get the same number. The multiplicative inverse of a rational number say $\dfrac{a}{b}$ is a rational number with whom by multiplying we get the multiplicative identity 1. Let $x$ be the multiplicative inverse of$\dfrac{a}{b}$. So we have
\[\begin{align}
  & \dfrac{a}{b}\times x=1 \\
 & \Rightarrow ax=b\left( \text{cross-multiplication} \right) \\
 & \Rightarrow x=\dfrac{b}{a}(\text{by dividing }a\text{ both side}) \\
\end{align}\]
So the multiplicative inverse of $\dfrac{a}{b}$ is$\dfrac{b}{a}$. Let us simplify the given expression to a form of$\dfrac{a}{b}$.by dividing the terms in numerator and denominator by a common factor. We have,
\[\dfrac{4}{7}\times \dfrac{21}{28}\]
We divide the number 4 in the numerator in the first fraction and the number 28 in the denominator of second fraction by 4 and get,
\[\dfrac{\dfrac{4}{4}}{7}\times \dfrac{21}{\dfrac{28}{4}}=\dfrac{1}{7}\times \dfrac{21}{7}\]
We divide the number 7 in the numerator and the number 28 in the denominator of second fraction to have,
\[\dfrac{\dfrac{4}{4}}{7}\times \dfrac{21}{\dfrac{28}{4}}=\dfrac{1}{7}\times \dfrac{21}{7}=\dfrac{1}{7}\times \dfrac{\dfrac{21}{7}}{\dfrac{7}{7}}=\dfrac{1}{7}\times \dfrac{3}{1}\]
We multiply the fractions numerator by numerator and denominator by denominator. We get
\[\dfrac{\dfrac{4}{4}}{7}\times \dfrac{21}{\dfrac{28}{4}}=\dfrac{1}{7}\times \dfrac{21}{7}=\dfrac{1}{7}\times \dfrac{\dfrac{21}{7}}{\dfrac{7}{7}}=\dfrac{1}{7}\times \dfrac{3}{1}=\dfrac{3}{7}\]
The obtained fraction $\dfrac{3}{7}$ is in the form of $\dfrac{a}{b}$ where $a=3,b=7.$ Its multiplicative inverse is
\[\dfrac{b}{a}=\dfrac{7}{3}\]

Note:
We can alternatively solve by first finding the multiplicative inverse of both the fractions and then multiplying them. We note that multiplicative inverse does not exist for 0 as we cannot divide by 0. The other name of the multiplicative inverse is reciprocal. Similarly additive inverse of $\dfrac{a}{b}$ is $-\dfrac{a}{b}.$