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

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

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 $x_{0}$ 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 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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