# Componendo and Dividendo

**Componendo and dividendo** is a theorem on proportions that allows for a quick way to perform calculations and reduce the amount of expansions needed.

It is particularly useful when dealing with equations involving fractions or rational functions in mathematical Olympiads, especially when you see fractions of the following form:

\[\begin{array} &\frac{ a}{b}, &\frac{a+b}{a-b}, &\frac{ a + kb } { a - k b }.\end{array} \]

#### Contents

## Componendo and Dividendo Theorems

Here are the rules based on componendo and dividendo.

If \(a, b, c\) and \(d\) are numbers such that \(b\) and \(d\) are non-zero and \( \frac{a}{b} = \frac{c}{d} \), then the following holds:

- Componendo: \(\dfrac{ a+b}{b} = \dfrac{ c+d}{d}.\)
- Dividendo: \(\dfrac{ a-b}{b} = \dfrac{ c-d} {d}.\)
- For \(k \neq \dfrac{a}{b}, \dfrac{ a+kb}{a-kb} = \dfrac{ c+kd}{c-kd}.\)
- For \( k \neq \dfrac{-b}{d}, \dfrac{ a}{b} = \dfrac{ a + kc } { b + kd }.\)

## Proof of Componendo and Dividendo Theorems

We can prove the 4 componendo and dividendo statements as follows:

- \( \dfrac{ a+b}{b} = \dfrac{ \frac{a}{b} + 1} {1} = \dfrac{ \frac{c}{d} + 1} {1} = \dfrac{ c+d}{d}. \\ \)
- \(\dfrac{ a-b}{b} = \dfrac{ \frac{a}{b} - 1} {1} = \dfrac{ \frac{c}{d} - 1} {1} = \dfrac{ c-d}{d}. \\ \)
- \(\dfrac{ a+kb}{a-kb} = \dfrac{ \frac{a}{b} + k } { \frac{a}{b} - k} = \dfrac{ \frac{c}{d} + k } { \frac{ c}{d} -k} = \dfrac{ c+kd} { c-kd}. \\ \)
- \(\dfrac{ a + kc} { b+ kd} = \dfrac{ a}{b} \times \dfrac{ 1 + k \frac{c}{a} } { 1 + k \frac{d}{b} } = \dfrac{ a}{b}. \ _\square\)

The converse of componendo and dividendo also holds, and we can prove it by applying dividendo and componendo, respectively.

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 \( \frac{ a}{b} = \frac{c}{d} \).

We apply componendo and dividendo with \(k=1 \) \(\big(\)which is valid since \( \frac{a+b}{a-b} \neq 1\big)\) and get that

\[ \frac{ 2a } { 2b} = \frac{ (a+b) + (a-b) } { (a+b) - (a-b) } = \frac{ (c+d) + (c-d) } { (c+d) - (c-d) } = \frac{ 2c} { 2d}. \ _\square\]

## Problem Solving with Componendo and Dividendo

The power of these equations is best demonstrated by working through several examples.

If \(\frac{a}{b}=\frac{16}{3}\), what is \(\frac{a+b}{a-b}?\)

By using equation \(3\) above with \(k=1\), we know that if \(\frac{a}{b}=\frac{c}{d}\), then \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\). Thus,

\[\dfrac{a+b}{a-b}=\dfrac{16+3}{16-3}=\dfrac{19}{13}. \ _\square\]

Without using the theorem, one approach would be to substitute \(a = \frac{16}{3} b \) to obtain that the expression is equal to \( \frac{ \frac{16}{3} b + b } { \frac{16}{3} b - b } \). This can be eventually simplified to the answer of \( \frac{19}{13} \), but will involve more steps.

If \(\frac{a+b}{a-b}=\frac{19}{13}\), what is \(\frac{a}{b}?\)

We will apply equation 3 again to obtain that

\[ \frac{ (a+b) + (a-b) } { ( a+b) - (a -b) } = \frac{ 19 + 13 } { 19 - 13 }. \]

Simplifying both sides, we obtain \( \frac{2a}{2b} = \frac{ 32} { 6} \implies \frac{a}{b} = \frac{ 16}{3} \). \( _ \square\)

Observe that this answer matches the condition given in the previous example. This is not surprising, as the statements are converses of each other. In fact, the converse of componendo can be proved by applying the theorem again as done above.

If \(\frac{a}{b} = \frac{c}{d},\) prove that \(\frac{2a - 9b}{2a + 9b} = \frac{2c - 9d}{2c + 9d}\).

We have

\[\begin{align} \dfrac{a}{b} & = \dfrac{c}{d}\\ \\ \dfrac{2a}{9b} & = \dfrac{2c}{9d} &&\qquad (\text{mutliplying both sides by 2 and dividing both sides by 9})\\ \\ \dfrac{2a + 9b}{2a - 9b} & = \dfrac{2c + 9d}{2c - 9d} &&\qquad (\text{by componendo and divdendo})\\\\ \dfrac{2a - 9b}{2a + 9b} & = \dfrac{2c - 9d}{2c + 9d}.\ _\square &&\qquad (\text{by invertendo}) \end{align}\]

If \(\frac{7a - 3b}{7c - 3d} = \frac{7a + 3b}{7c + 3d},\) prove that \(\frac{a}{b} = \frac{c}{d}\).

We have

\[\begin{align} \dfrac{7a - 3b}{7c - 3d} & = \dfrac{7a + 3b}{7c + 3d}\\ \\ \dfrac{7a + 3b}{7a - 3b} & = \dfrac{7c + 3d}{7c - 3d}&\qquad (\text{by alternendo})\\ \\ \dfrac{7a + 3b + 7a - 3b}{7a + 3b - 7a + 3b} & = \dfrac{7c + 3d + 7c - 3d} {7c + 3d -7c + 3d}\\ \\ \dfrac{14a}{6b} & = \dfrac{14c}{6d} \\\\ \dfrac{a}{b} &= \dfrac{c}{d}.\ _\square \end{align}\]

If \(x = \frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}\), prove that \(3bx^2 - 2ax + 3b = 0\).

We have

\[\begin{align} \dfrac{x}{1} & = \dfrac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}}\\ \\ \dfrac{x+1}{x-1} & = \dfrac{\sqrt{a + 3b} + \sqrt{a - 3b} + \sqrt{a + 3b} - \sqrt{a - 3b}}{\sqrt{a + 3b} + \sqrt{a - 3b} - \sqrt{a + 3b} + \sqrt{a - 3b}}\\ \\ & = \dfrac{2\sqrt{a + 3b}}{2\sqrt{a - 3b}}\\ \\ & = \dfrac{\sqrt{a + 3b}}{\sqrt{a - 3b}}\\ \\ {\left(\dfrac{x +1}{x - 1}\right)}^2 & = {\left(\dfrac{\sqrt{a + 3b}}{\sqrt{a - 3b}}\right)}^2\\ \\ \dfrac{{(x+1)}^2}{{(x-1)}^2} & = \dfrac{a + 3b}{a - 3b}\\ \\ \dfrac{{(x+1)}^2 + {(x-1)}^2}{{(x+1)}^2 - {(x-1)}^2} & = \dfrac{a + 3b + a - 3b}{a + 3b - a + 3b}\\ \\ \dfrac{2(x^2 + 1)}{4x} & = \dfrac{2a}{6b}\\ \\ \dfrac{(x^2 + 1)}{2x} & = \dfrac{a}{3b}\\ \\ 3bx^2 + 3b & = 2ax\\ 3bx^2 - 2ax + 3b & = 0.\ _\square \end{align}\]

If \(\frac{a^3 + 3ab^2}{3a^2b + b^3} = \frac{x^3 + 3xy^2}{3x^2y + y^3}\), prove that \(\frac{x}{a} = \frac{y}{b}\).

We have

\[\begin{align} \dfrac{a^3 + 3ab^2}{3a^2b + b^3} & = \dfrac{x^3 + 3xy^2}{3x^2y + y^3}\\ \\ \dfrac{a^3 + 3ab^2 + 3a^2b + b^3}{a^3 + 3ab^2 - 3a^2b - b^3} & = \dfrac{x^3 + 3xy^2 + 3x^2y + y^3}{x^3 + 3xy^2- 3x^2y- y^3}&&\qquad (\text{by componendo and dividendo})\\ \\ \dfrac{a^3 + b^3 + 3ab(a + b)}{a^3 - b^3 - 3ab(a - b)} & = \dfrac{x^3 + y^3 + 3xy(x + y)}{x^3 - y^3 - 3xy(x - y)}\\ \\ \dfrac{{(a + b)}^3}{{(a - b)}^3} & = \dfrac{{(x + y)}^3}{{(x - y)}^3}\\ \\ \dfrac{(a + b)}{(a - b)} & = \dfrac{(x + y)}{(x - y)}&&\qquad (\text{taking cube root on both sides})\\ \\ \dfrac{(a + b + a - b)}{(a + b - a + b)} & = \dfrac{(x + y + x - y)}{(x + y - x + y)}\\ \\ \dfrac{2a}{2b} & = \dfrac{2x}{2y}\\ \\ \dfrac{a}{b} & = \dfrac{x}{y}\\ \\ \dfrac{x}{a} & = \dfrac{y}{b}.\ _\square &&\qquad (\text{by alternendo}) \end{align}\]

What quantity must be added to each term of the ratio \(\frac{a + 2b}{a - 2b}\), to make it equal to \(\frac{{(a + b)}^2}{{(a - b)}^2}?\)

Let the quantity be \(x\). Then

\[\begin{align} \dfrac{a + 2b + x}{a - 2b + x} & = \dfrac{{(a + b)}^2}{{(a - b)}^2}\\ \\ \dfrac{a + 2b + x + a - 2b + x}{a + 2b + x - a + 2b - x} & = \dfrac{{(a + b)}^2 + {(a - b)}^2}{{(a + b)}^2 - {(a - b)}^2}\\ \\ \dfrac{2(a + x)}{4b} & = \dfrac{2(a^2 + b^2)}{4ab}\\ \\ \dfrac{(a + x)}{2b} & = \dfrac{(a^2 + b^2)}{2ab}\\ \\ a + x & = a + \dfrac{b^2}{a}\\ \\ x &= \dfrac{b^2}{a}.\ _\square \end{align}\]

If \(3x - 5y = 2x + y\), find the value of \(\frac{3x + 5y}{3x - 5y}\).

We have

\[\begin{align} 3x - 5y & = 2x + y\\ 3x - 2x & = y + 5y\\ x & = 6y\\ \dfrac{x}{y} & = \dfrac{6}{1}\\ \\ \dfrac{3x}{5y} & = \dfrac{3 × 6}{5 × 1}&&\qquad (\text{mutliplying both sides by 3 and dividing both sides by 5})\\ \\ & = \dfrac{18}{5}\\ \\ \dfrac{3x + 5y}{3x - 5y} & = \dfrac{18+5}{18-5}&&\qquad (\text{by componendo and dividendo})\\ \\ & = \dfrac{23}{13}.\ _\square \end{align}\]

Solve for \(x\): \( \frac{ x + \sqrt{5} } { x - \sqrt{ 5 } } = 2. \)

Applying equation 3 above with \( k = 1 \), we obtain

\[ \frac{ \left( x + \sqrt{5} \right) + \left( x - \sqrt{ 5 } \right) } { \left( x + \sqrt{ 5 } \right) - \left( x - \sqrt{5} \right) } = \frac{ 2+1}{2-1}. \]

This simplifies to

\[ \frac{ 2x} { 2 \sqrt{5} } = 3 \Rightarrow x = 3 \sqrt{5}. \ _\square \]

If \(\frac{a}{b} = \frac{c}{d},\) show that \(\frac{a + mb}{b + na} = \frac{c + md}{d + nc}\) for \( n \neq \frac{-b}{a}.\)

Method 1:Direct ProofWe have \[\dfrac{a + mb}{b + na} = \dfrac{b}{a} \times \dfrac{\frac{a}{b} + m} {\frac{b}{a} + n} = \dfrac{d}{c} \times \dfrac{\frac{c}{d} + m} {\frac{d}{c} + n} = \dfrac{c + md}{d + nc}. \\ \]

Method 2:Using Theorem 4We have \[\begin{align} \dfrac{a}{b} & = \dfrac{c}{d}\\ \\ \dfrac{a}{c} & = \dfrac{b}{d} &&\qquad (\text{by invertendo})\\ \\ \dfrac{a + mb}{c + md} & = \dfrac{b + na}{d + nc} &&\qquad (\text{by theorem 4})\\ \\ \dfrac{a + mb}{b + na} & = \dfrac{c + md}{d + nc}. &&\qquad (\text{by invertendo again})\\ \\ \end{align}\]

Note that Method 2 does require the additional condition \( m \neq \frac{-c}{d}\), though if \( m = \frac{-c}{d}\) the statement is trivially true. \(_\square\)

Thus far, the examples have only involved linear terms for simplicity. The variables \(a\) and \(b\) are allowed to be polynomials or even exponential functions. Consider the following example, where the fraction isn't initially in the form of \( \frac{ a+b}{a-b} :\)

Solve for \(x\): \(\frac{ x^3+1} { x+ 1} = \frac{ x^3-1} { x-1} .\)

For the fractions to make sense, we must have \( x \neq 1, -1\).

While the fractions are not in a form that allows immediate application, we can cross multiply the denominators to obtain

\[ \frac{ x^3+1}{x^3-1} = \frac{ x+1}{x-1}. \]

This is in a recognizable form, so we apply Componendo and Dividendo with \(k=1 \) \(\big(\)which is valid since \( \frac{x+1}{x-1} \neq 1\big). \) Then we get that

\[ \frac{ 2x^3}{2} = \frac{ 2x}{2} \implies x(x^2-1) = 0 .\]

However, since \( x \neq 1, -1 \), we have \(x=0 \) as the only solution. \(_\square\)

If \(n\) is an integer, how many complex solutions are there to

\[ \frac{ x^n+1} { x+1 } = \frac{ x^n -1 } { x-1}? \]

Assuming that \(n \ge 2\), we manipulate the equation to obtain

\[ \frac{x^n+1}{x^n-1} \;=\; \frac{x+1}{x-1} .\]

Thus, using Componendo and Dividendo with \(k=1,\) we deduce that

\[ x^n \; =\; \frac{2x^n}{2} \; =\; \frac{(x^n+1)+(x^n-1)}{(x^n+1)-(x^n-1)} \; = \; \frac{(x+1)+(x-1)}{(x+1)-(x-1)} \; = \; \frac{2x}{2} \; = \; x, \]

and thus \(x(x^{n-1}-1)=0\). We need to exclude both \(x=1\) and \(x=-1\) as solutions. Thus,

- there are \((n-1)\) roots when \(n\) is even, namely \(0\) and the \((n-2)\) complex \((n-1)^\text{th}\) roots of unity \((1\) omitted\();\)
- there are \((n-2)\) roots when \(n\) is odd, namely \(0\) and the \((n-3)\) complex \((n-1)^\text{th}\) roots of unity \((1\) and \(-1\) omitted\().\)
If \(n=1,\) then there are infinitely many roots.

If \(n=0,\) then there are no solutions.

If \(n < 0,\) then the same algebraic steps as above hold. But we have to exclude \(0\) as a root, so the roots are the \(|n-1|^\text{th}\) roots of unity, with \(1\) and \(-1\) excluded. \(_\square\)

If \(\tan{\theta} = -\frac{1}{4}\), find the value of \( \frac{\sin{\theta} + 2\cos{\theta}}{\cos{\theta} - 3\sin{\theta}}.\)

Using \(\frac{a + mb}{b + na} = \frac{c + md}{d + nc}\) (see the proof in an above example) and rewriting the given condition as \(\frac{\sin{\theta}}{\cos{\theta}} = \frac{-1}{4}\), we immediately get

\[ \dfrac{\sin{\theta} + 2\cos{\theta}}{\cos{\theta} - 3\sin{\theta}} = \dfrac{-1+2(4)}{4-3(-1)} = \dfrac{7}{7} =1.\]

Note that we could have also written the given condition as \(\frac{\sin{\theta}}{\cos{\theta}} = \frac{1}{-4}\), but this would have not changed the final answer. \(_\square\)

Solve the system of equations \(2x + 3y = 4\) and \(5x + 7y = 10\) for \(x\) and \(y\).

We are given two equations \(2x + 3y = 4\) and \(5x + 7y = 10\). Dividing these two equations gives

\[\dfrac{2x + 3y }{ 5x + 7y }= \dfrac{4}{10} =\dfrac 2 5 .\]

Now we will apply componendo and dividendo method:

\[\dfrac{2x + 3y }{ 5x+ 7y -2(2x + 3y )}= \dfrac{2}{5-2\times2} \implies \dfrac{2x + 3y }{ x + y }= \dfrac{2}{1}.\]

Again applying componendo and dividendo to the above equation, we have

\[ \dfrac{2x + 3y-2(x + y) }{ x + y }=\dfrac{2-2}{1} \implies \dfrac{y }{ x + y }= \dfrac{0}{1} .\]

On cross multiplying, we get \(y=0\). Putting \(y=0\) in any one of the two equations \(2x + 3y = 4\) and \(5x + 7y = 10\), we get \(x=2\). Hence \((x,y)=(2,0)\) is the solution to the given equations. \(_\square\)

Note:The equations can also be solved using elimination method and substitution method.

## See Also

**Cite as:**Componendo and Dividendo.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/componendo-and-dividendo/