Four exponentials conjecture

In algebra, componendo and dividendo (or componendo et dividendo) is a method of simplification based on fractions provided that they are in proportion. It states that

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

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 $x_0$ and $y_0$ are nonzero and can be expressed in terms of the values of

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

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 C and D

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

$=>\frac{2\sqrt{3}}{2x}=\frac{3}{1}$

$=>\frac{\sqrt{3}}{x}=3$

$=>x=\frac{1}{\sqrt{3}}$

See also

• Reduction (mathematics)
• Fraction (mathematics)

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