$\dfrac{a}{b} = \dfrac{c}{d} \,\,\,$ $\Leftrightarrow \,\,\,$ $\dfrac{a+b}{a-b} = \dfrac{c+d}{c-d}$

$a$, $b$, $c$ and $d$ are four literals but $b$ and $d$ are non-zero literal numbers. Assume, the ratio of $a$ to $b$ is equal to the ratio of $c$ to $d$.

$\dfrac{a}{b} = \dfrac{c}{d}$

Add one both sides of the left and right hand side fractions.

$\implies \dfrac{a}{b} + 1 = \dfrac{c}{d} + 1$

$\implies \dfrac{a+b}{b} = \dfrac{c+d}{d}$

Subtract one both sides of the left and right hand side fractions.

$\implies \dfrac{a}{b} -1 = \dfrac{c}{d} -1$

$\implies \dfrac{a-b}{b} = \dfrac{c-d}{d}$

Divide the equation obtained in step 1 by the equation obtained in step 2.

$\implies \dfrac{\Bigg[\dfrac{a+b}{b}\Bigg]}{\Bigg[\dfrac{a-b}{b}\Bigg]}$ $=$ $\dfrac{\Bigg[\dfrac{c+d}{d}\Bigg]}{\Bigg[\dfrac{c-d}{d}\Bigg]}$

$\implies \Bigg[\dfrac{a+b}{b}\Bigg] \times \Bigg[\dfrac{b}{a-b}\Bigg]$ $=$ $\Bigg[\dfrac{c+d}{d}\Bigg] \times \Bigg[\dfrac{d}{c-d}\Bigg]$

$\implies \Bigg[\dfrac{a+b}{a-b}\Bigg] \times \Bigg[\dfrac{b}{b}\Bigg]$ $=$ $\Bigg[\dfrac{c+d}{c-d}\Bigg] \times \Bigg[\dfrac{d}{d}\Bigg]$

$\implies \require{cancel} \Bigg[\dfrac{a+b}{a-b}\Bigg] \times \Bigg[\dfrac{\cancel{b}}{\cancel{b}}\Bigg]$ $=$ $\Bigg[\dfrac{c+d}{c-d}\Bigg] \times \Bigg[\dfrac{\cancel{d}}{\cancel{d}}\Bigg]$

$\implies \Bigg[\dfrac{a+b}{a-b}\Bigg] \times 1$ $=$ $\Bigg[\dfrac{c+d}{c-d}\Bigg] \times 1$

$\therefore \,\,\,\,\,\, \dfrac{a+b}{a-b} = \dfrac{c+d}{c-d}$

Therefore, it is proved that if ratio of $a$ to $b$ is equal to the ratio of $c$ to $d$, then the ratio of $a+b$ to $a-b$ is equal to the ratio of $c+d$ to $c-d$. This property is called the componendo and dividendo rule.

Let us test this theorem by verifying it in number system. Take two equal fractions to understand this property.

$\dfrac{1}{2} = 0.5$ and $\dfrac{2}{4} = 0.5$

$\therefore \,\,\,\,\,\, \dfrac{1}{2} = \dfrac{2}{4}$

Now, verify the Componendo and Dividendo law.

$\dfrac{1+2}{1-2} = \dfrac{3}{-1} = -3$

$\require{cancel} \dfrac{2+4}{2-4} = \dfrac{\cancel{6}}{\cancel{-2}} = \dfrac{3}{-1} = -3$

$\therefore \,\,\,\,\,\, \dfrac{1+2}{1-2} = \dfrac{2+4}{2-4}$

This example has verified the Componendo and Dividendo identity. Hence, if the ratio of any two numbers is equal to ratio of another two numbers, then the ratios of sum of numerator and denominator to difference of numerator and denominator of both rational numbers are equal.

Latest Math Topics

Jul 24, 2024

Dec 13, 2023

Jul 20, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved