Componendo et Dividendo

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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,ca, b, c and dd are numbers such that b,db, d are non-zero and ab=cd \frac{a}{b} = \frac{c}{d} , then

1. Componendo:a+bb=c+dd.2. Dividendo: abb=cdd. Componendo et Dividendo: 3. For kab,a+kbakb=c+kdckd.4. For kbd,ab=a+kcb+kd. \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

 1.a+bb=ab+11=cd+11=c+dd. 2.abb=ab11=cd11=cdd. 3.a+kbakb=ab+kabk=cd+kcdk=c+kdckd. 4.a+kcb+kd=ab×1+kca1+kdb=ab. \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,ca, b, c and dd are numbers such that b,d,ab,cdb, d, a-b, c-d are non-zero and a+bab=c+dcd \frac{ a+b}{a-b} = \frac{c+d} { c-d} , then ab=cd \frac{ a}{b} = \frac{c}{d} .

Solution: We apply Componendo et Dividendo with k=1k=1 (which is valid since a+bab1 \frac{a+b}{a-b} \neq 1 ), and get that 2a2b=(a+b)+(ab)(a+b)(ab)=(c+d)+(cd)(c+d)(cd)=2c2d. \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 xx: x3+1x+1=x31x1 \frac{ x^3+1} { x+ 1} = \frac{ x^3-1} { x-1} .

Solution: For the fractions to make sense, we must have x1,1 x \neq 1, -1.

Cross multiplying, we get x3+1x31=x+1x1. \frac{ x^3+1}{x^3-1} = \frac{ x+1}{x-1}.

Apply Componendo et Dividendo with k=1k=1 (which is valid since x+1x11 \frac{x+1}{x-1} \neq 1 ), we get that 2x32=2x2x(x21)=0 \frac{ 2x^3}{2} = \frac{ 2x}{2} \Rightarrow x(x^2-1) = 0 . However since x1,1 x \neq 1, -1 , we have x=0x=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
5 years, 7 months ago

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Comments

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

Shamoli Barua - 5 years, 5 months ago

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See the statements contained in the first box.

Calvin Lin Staff - 5 years, 5 months ago

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

Kyran Gaypinathan - 3 years, 8 months ago

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

Calvin Lin Staff - 3 years, 8 months ago

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I found out (somewhat accidentally) that

a+mbb+na=c+mdd+nc\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?

zico quintina - 1 year, 10 months ago

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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 a+mbc+md=ac=bd=b+nad+nc \frac{ a + mb } { c + md} = \frac{a}{c} = \frac{b}{d} = \frac{ b+na}{d + nc } (as long as the denominators are non-zero).

Calvin Lin Staff - 1 year, 10 months ago

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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.

zico quintina - 1 year, 10 months ago

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@Zico Quintina 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!

Calvin Lin Staff - 1 year, 10 months ago

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