Componendo and dividendo

Componendo and dividendo is a method of simplification in mathematics.

According to it,

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

Proof

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

Comment on the proof

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

$$\frac{x_0 + \ldots + x_n}{y_0 + \ldots +y_n}$$ 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:

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

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.