# Componendo and Dividendo: Proof & Example

Componendo and Dividendo is one basic property about proportions that provide a quick way to perform calculations and reduce the number of required extensions.

This is especially useful when working with equations involving fractions or rational functions in math competitions, especially when you’re looking at fractions. i.e \frac{a+b}{a-b}=\frac{c+d}{c-d}

Proof of Componendo and Dividendo \frac{a+b}{a-b}=\frac{c+d}{c-d}:

Add one both sides of the left and right-hand side fractions.
\begin{array}{l} \Rightarrow \frac{a}{b}+1=\frac{c}{d}+1 \\ \Rightarrow \frac{a+b}{b}=\frac{c+d}{d} \end{array} Let this equation be (A)
Step: 2
Subtract one both sices of the left and right hand side fractions:
\begin{array}{l} \Rightarrow \frac{a}{b}-1=\frac{c}{d}-1 \\ \Rightarrow \frac{a-b}{b}=\frac{c-d}{d} \end{array} Let this equation be (B)
Now divide equation divide A by B:
\begin{array}{l} \Rightarrow \frac{\left[\frac{a+b}{b}\right]}{\left[\frac{a-b}{b}\right]}=\frac{\left[\frac{c+d}{d}\right]}{\left[\frac{c-d}{d}\right]}\\ \Rightarrow\left[\frac{a+b}{b}\right] \times\left[\frac{b}{a-b}\right]=\left[\frac{c+d}{d}\right] \times\left[\frac{d}{c-d}\right]\\ \Rightarrow\left[\frac{a+b}{a-b}\right] \times\left[\frac{b}{b}\right]=\left[\frac{c+d}{c-d}\right] \times\left[\frac{d}{d}\right]\\ \Rightarrow\left[\frac{a+b}{a-b}\right] \times\left[\frac{\not b}{\not b}\right]=\left[\frac{c+d}{c-d}\right] \times\left[\frac{\not d}{\not d}\right]\\ \Rightarrow\left[\frac{a+b}{a-b}\right] \times 1=\left[\frac{c+d}{c-d}\right] \times 1\\ \therefore \quad \frac{a+b}{a-b}=\frac{c+d}{c-d} \end{array}

#### Definitions Related To Componendo and Dividendo

If  a / b=c / d

• Invertendo: b / a=d / c
• Alternendo : a / c=b / d
• Componendo: (a+b), b=(c+d) / d
• Dividendo : (a-b) / b=(c-d) / d
• Componendo – Dividendo: (a+b) /(a-b)=(c+d)/(c-d)

Componendo and Dividendo Example:

If \frac{a}{\delta}=\frac{c}{d}, prove that \frac{3 a-8 b}{2 a+8 b}=\frac{2 c-8 d}{3 c-8 d} Method 1. We have
\begin{array}{l} \frac{a}{b}=\frac{c}{d} \\ \frac{3 a}{8 b}=\frac{3 c}{8 d} \\ \frac{3 a+9 b}{2 a-8 b}=\frac{3 c+8 d}{3 c-8 d} \\ \frac{3 a-9 b}{2 a+8 b}=\frac{3 c-8 d}{3 c+8 d} \end{array}

Method 2. Alternatively, we can make use of the property that \frac{a+k h}{a-k b}=\frac{c+k d}{c-k d} If \frac{a}{b}=\frac{c}{d} Then,
\frac{a+\left(\frac{8}{3}\right) b}{a-\left(\frac{8}{3}\right) b}=\frac{c+\left(\frac{8}{3}\right) d}{c-\left(\frac{8}{3}\right) d} Multiplying both the numerator and the denominator by 2 on both sides:
\begin{array}{l} \frac{3 a+8 b}{3 a-8 b}=\frac{3 c+8 d}{2 c-8 d} \\ \frac{3 a-8 b}{3 a+5 b}=\frac{3 ec-8 d}{2 c+8 d} \end{array} (applying invertendo)

In this article, we have seen Componendo and Dividendo proof and example. Hope you like it, and please share with your maths nerds and helped them out to clear their doubts.