If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$?

Question

Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.

Answer 1

Sketch: If you have $\frac{p}{q}$ and $\frac{\lambda p}{\lambda q}$, then $$ \frac{p+\lambda p}{q+\lambda q}=\frac{(1+\lambda)p}{(1+\lambda)q}=\frac{p}{q} $$ provided $1+\lambda\not=0$.

Answer 2

Consider $\frac{a}{b}=\frac{ka}{kb}$ Then, $$\frac{a+ka}{b+kb}=\frac{(k+1)a}{(k+1)b}=\frac{a}{b}$$ which is exactly what you noticed, but with $a=1,b=2,k=2$

Answer 3

Or, with less variables, you can treat your fraction as reducible to some number $a$ (e.g. a decimal), which you can write as $\frac{a}{1}$. Then:

$\frac{a+a}{1+1} = \frac{2a}{2} = \frac{a}{1} = a$

Answer 4

An alternative solution, not to disparage the other answers.

$$\begin{aligned} \frac{a}{b}=\frac{c}{d}\quad&\Rightarrow\quad\frac{ad}{b}=c&\text{solve for $c$}\\ \frac{a+c}{b+d}&=\frac{a+\left(\frac{ad}{b}\right)}{b+d}&\text{substitute $c$}\\ &=a\cdot\frac{1+\left(\frac{d}{b}\right)}{b+d}&\text{factor $a$ from numerator}\\ &=a\cdot\frac{b+d}{b(b+d)}&\text{multiply by $\frac{b}{b}$}\\ &=\frac{a}{b}\quad\blacksquare&\text{cancel $(b+d)$}\\ \end{aligned}$$