Componendo and dividendo

In algebra, componendo and dividendo is a method of simplification. It states that

$\text{If } \frac{a}{b} = \frac{c}{d} \text{ and } a \neq b \text{, then } \frac{a+b}{a-b} = \frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{\frac{c}{d} + 1}{\frac{c}{d} - 1} = \frac{c+d}{c-d}.$

Comment on the proof

We can similarly deduce the much more general fact that the value of any fraction

$\frac{x_0 + \cdots + x_n}{y_0 + \cdots +y_n}$

in which x₀ and y₀ are nonzero can be expressed in terms of the values of

$\frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}, \frac{y_1}{y_0}, \ldots, \frac{y_n}{y_0}$

and the value of $\frac{x_0}{y_0}$, and so depends only on the values of those 2n + 1 fractions:

$\frac{x_0 + \cdots + x_n}{y_0 + \cdots +y_n} = \frac{x_0}{y_0} \left(\frac{1 + \frac{x_1}{x_0} + \cdots + \frac{x_n}{x_0}}{1 + \frac{y_1}{y_0} + \cdots + \frac{y_n}{y_0}}\right)$

The original result is essentially a special case of this fact, because

$\frac{x+y}{x-y} = \frac{x+y}{x+(-y)}$

can be regarded as a fraction of the above form.

This method is very useful to evaluate some simple and complex algebraic problems.

For instance :

(√3 + x)/(√3 − x) = 2

Find the value of x.

Solution :

Applying C & D

[({√3 + x} + {√3 − x})/({√3 + x} − {√3 − x} )] = (2 + 1)/(2 − 1)

=> 2√3/(2x) = 3/1

=> √3/x = 3

=> x = 1/√3