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

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 $\dfrac{1+2}{2+4}=0.5$.

• The answers below answer your question completely, but as a side note, this phenomenon is called Componendo-Dividendo. – Shraddheya Shendre Apr 25 '17 at 13:58
• Perhaps it is worth mentioning that if you have two fractions $\frac ac$ and $\frac bd$, then the fraction $\frac{a+b}{c+d}$ is called mediant. – Martin Sleziak Apr 26 '17 at 3:28
• $\frac{a+b}{c+d}$ is between $\frac{a}{c}$ and $\frac{b}{d}$ so when the latter two are the same, so is the third – Henry Apr 26 '17 at 7:35
• @Henry Don't get your dirty calculus in our clean algebra! ;) – Yakk Apr 26 '17 at 15:22

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 $1+\lambda\not=0$.

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 $a=1,b=2,k=2$

• Such a difference in up-votes for being a mere 25 seconds behind Michael! – Paul Sinclair Apr 25 '17 at 23:17
• @PaulSinclair I know, I even upvoted this one too! – Michael Burr Apr 26 '17 at 0:38
• -1 needs more $\lambda$ – Aza Apr 26 '17 at 7:48
• This answer lacks a condition on k+1 – Stephan Apr 26 '17 at 10:13

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}

We know $ad=bc$, so $ab+bc=ab+ad$. If you factor out this equation you get $b(a+c)=a(b+d)$ and then you get $\frac{a}{b}=\frac{a+c}{b+d}$. Similarly, you can prove the other equality.

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:

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