Math Doubts

Componendo and Dividendo Rule

Formula

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

Proof

$a$, $b$, $c$ 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 $d$.

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

Step: 1

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

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

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

Step: 2

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

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

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

Step: 3

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.

Verification

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.

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