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# This note has been used to help create the Componendo and Dividendo wiki

See the complete wiki page here.

The method of Componendo et Dividendo allows a quick way to do some calculations, and can simplify the amount of expansion needed.

If $$a, b, c$$ and $$d$$ are numbers such that $$b, d$$ are non-zero and $$\frac{a}{b} = \frac{c}{d}$$, then

$\begin{array} {l r l } \text{1. Componendo:} & \frac{ a+b}{b} & = \frac{ c+d}{d}. \\ \text{2. Dividendo: } & \frac{ a-b}{b} & = \frac{ c-d} {d}. \\ \text{ Componendo et Dividendo: } & \\ \text{3. For } k \neq \frac{a}{b},& \frac{ a+kb}{a-kb} & = \frac{ c+kd}{c-kd} .\\ \text{4. For } k \neq \frac{-b}{d}, & \frac{ a}{b} & = \frac{ a + kc } { b + kd }. \\ \end{array}$

This can be proven directly by observing that

$\begin{array} {l r l } \text{ 1.} \frac{ a+b}{b} = \frac{ \frac{a}{b} + 1} {1} = \frac{ \frac{c}{d} + 1} {1} = \frac{ c+d}{d} . \\ \text{ 2.} \frac{ a-b}{b} = \frac{ \frac{a}{b} - 1} {1} = \frac{ \frac{c}{d} - 1} {1} = \frac{ c-d}{d} . \\ \text{ 3.} \frac{ a+kb}{a-kb} = \frac{ \frac{a}{b} + k } { \frac{a}{b} - k} = \frac{ \frac{c}{d} + k } { \frac{ c}{d} -k} = \frac{ c+kd} { c-kd} . \\ \text{ 4.} \frac{ a + kc} { b+ kd} = \frac{ a}{b} \times \frac{ 1 + k \frac{c}{a} } { 1 + k \frac{d}{b} } = \frac{ a}{b} . \end{array}$

## Worked examples

### 1. Show the converse, namely that if $$a, b, c$$ and $$d$$ are numbers such that $$b, d, a-b, c-d$$ are non-zero and $$\frac{ a+b}{a-b} = \frac{c+d} { c-d}$$, then $$\frac{ a}{b} = \frac{c}{d}$$.

Solution: We apply Componendo et Dividendo with $$k=1$$ (which is valid since $$\frac{a+b}{a-b} \neq 1$$ ), and get that $\frac{ 2a } { 2b} = \frac{ (a+b) + (a-b) } { (a+b) - (a-b) } = \frac{ (c+d) + (c-d) } { (c+d) - (c-d) } = \frac{ 2c} { 2d}.$

Note: The converse of Componendo and Dividendo also holds, and we can prove it by applying Dividendo and Componendo respectively.

### 2. Solve for $$x$$: $$\frac{ x^3+1} { x+ 1} = \frac{ x^3-1} { x-1}$$.

Solution: For the fractions to make sense, we must have $$x \neq 1, -1$$.

Cross multiplying, we get $$\frac{ x^3+1}{x^3-1} = \frac{ x+1}{x-1}.$$

Apply Componendo et Dividendo with $$k=1$$ (which is valid since $$\frac{x+1}{x-1} \neq 1$$ ), we get that $$\frac{ 2x^3}{2} = \frac{ 2x}{2} \Rightarrow x(x^2-1) = 0$$. However since $$x \neq 1, -1$$, we have $$x=0$$ as the only solution.

Note: We also need to check the condition that the denominators are non-zero, but this is obvious.

Note by Calvin Lin
2 years, 12 months ago

## Comments

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if a/b =c/d what will be the result by componendo dividendo · 2 years, 9 months ago

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See the statements contained in the first box. Staff · 2 years, 9 months ago

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Sorry I'm slightly confused, could you clarify what Componendo and Dividendo integrate to? · 1 year ago

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Check out the examples on the componendo and dividendo wiki page. Staff · 1 year ago

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