# 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
7 years, 4 months ago

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

- 7 years, 1 month ago

See the statements contained in the first box.

Staff - 7 years, 1 month ago

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

- 5 years, 4 months ago

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

Staff - 5 years, 4 months ago

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?

- 3 years, 6 months ago

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).

Staff - 3 years, 6 months ago

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.

- 3 years, 6 months ago

That's a great writeup. I like how you highlighted Method 1, which "by right" should be how people understand this identity, and "by left" it is placed in this wiki because of the form it took.

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

Staff - 3 years, 6 months ago