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

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

List of most recently solved mathematics problems.

Jun 22, 2018

Integral Calculus

Evaluate $\displaystyle \int \dfrac{1+\cos{4x}}{\cot{x}-\tan{x}} dx$

Jun 21, 2018

Limit

Evaluate $\displaystyle \large \lim_{x \to \infty} \normalsize {\sqrt{x^2+x+1}-\sqrt{x^2+1}}$

Jun 20, 2018

Differentiation

Learn how to find derivative of $\sin{(x^2)}$ with respect to $x$.

Jun 19, 2018

Limit (Calculus)

Find $\displaystyle \large \lim_{x \to 0} \normalsize \dfrac{x-\sin{x}}{x^3}$

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.