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.

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## Comments

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TopNewestif a/b =c/d what will be the result by componendo dividendo

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See the statements contained in the first box.

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

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Check out the examples on the componendo and dividendo wiki page.

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I found out (somewhat accidentally) that

$\frac{a+mb}{b+na} = \frac{c+md}{d+nc}$

is also true. It's quite easy to prove; it can also be derived from the 4th case stated above. It seems different enough, though, to be worth a mention, yet I never see it anywhere. Or is it perhaps that there are many other such corollaries and only the most basic ones are usually listed?

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Ohhh, that's a really nice identity. It is a generalization of Worked Example 1. Can you add it to the Componendo and Dividendo wiki under Problem Solving?

Like you said, it is essentially / can be derived from the 4th case, where we have $\frac{ a + mb } { c + md} = \frac{a}{c} = \frac{b}{d} = \frac{ b+na}{d + nc }$ (as long as the denominators are non-zero).

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Have finally had a chance to add the identity to the Wiki; please take a look when you can and let me know if I need to make any changes. (I accidentally first added it to the Theorem section before I remembered that you said Problem Solving, I did move it to the correct section after that, hope it didn't cause any problems.) I also added an example applying it a little further down, in the section that introduced using C&D with non-linear terms. The Wiki mentioned the terms could be polynomial or exponential; my example uses trig functions, I believe it's still valid but please take a look and let me know if there are any issues. Thanks.

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Yes, your example with a trigo substitution is valid. Good one. Thanks!

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