Pedoe's inequality

In algebra, componendo and dividendo (or componendo et dividendo) is a method of simplification of equations implying fractions. It states that[1] [2]

\( \text{If } \frac{a}{b} = \frac{c}{d}\text{, then: } \)

\( \frac{b}{a}=\frac{d}{c} \) (1)

\( \frac{a}{c}=\frac{b}{d} \) (2)

Adding 1 to both sides of the main equation gives:

\( \frac{a+b}{b}=\frac{c+d}{d} \) (3)

Subtracting 1 from both sides of the main equation gives:

\( \frac{a-b}{b}=\frac{c-d}{d}(4) \)

Dividing equation 3 by equation 4, we have:

\( \frac{a+b}{a-b}=\frac{c+d}{c-d} \) (5)

Relationship 1 above is called invertendo.

Relationship 2 above is called alternendo.

Relationship 3 above is called componendo.

Relationship 4 above is called dividendo.

Relationship 5 above is called componendo and dividendo.

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 x_0 and y_0 are nonzero and 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.
Example

This method can be used in various situations.

For instance :

\( \frac{\sqrt{3} + x}{\sqrt{3} - x} = 2 \)

Find the value of x.

Solution :

Applying componendo and dividendo,

\( \frac{(\sqrt{3} + x) + (\sqrt{3} - x)}{(\sqrt{3} + x) - (\sqrt{3} - x)} = \frac{2 + 1}{2 - 1}
=> \frac{2 \sqrt{3}}{2 x} = \frac{3}{1}
=> \frac{\sqrt{3}}{x} = 3
=> x = \frac{1}{\sqrt{3}} \)

References

Bhamra, Partial Differential Equations. PHI Learning Pvt. Ltd. ISBN 978-81-203-3917-0

http://www.qc.edu.hk/math/Junior%20Secondary/Componendo%20et%20Dividendo.htm