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

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    $\begingroup$ The answers below answer your question completely, but as a side note, this phenomenon is called Componendo-Dividendo. $\endgroup$ Apr 25, 2017 at 13:58
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    $\begingroup$ 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. $\endgroup$ Apr 26, 2017 at 3:28
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    $\begingroup$ $\frac{a+b}{c+d}$ is between $\frac{a}{c}$ and $\frac{b}{d}$ so when the latter two are the same, so is the third $\endgroup$
    – Henry
    Apr 26, 2017 at 7:35
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    $\begingroup$ @Henry Don't get your dirty calculus in our clean algebra! ;) $\endgroup$
    – Yakk
    Apr 26, 2017 at 15:22
  • $\begingroup$ @Henry squeeze theorem isn't it? $\endgroup$ Mar 26, 2023 at 10:40

5 Answers 5

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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 $1+\lambda\not=0$.

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

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    $\begingroup$ Such a difference in up-votes for being a mere 25 seconds behind Michael! $\endgroup$ Apr 25, 2017 at 23:17
  • $\begingroup$ @PaulSinclair I know, I even upvoted this one too! $\endgroup$ Apr 26, 2017 at 0:38
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    $\begingroup$ -1 needs more $\lambda$ $\endgroup$
    – user66698
    Apr 26, 2017 at 7:48
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    $\begingroup$ This answer lacks a condition on k+1 $\endgroup$
    – Stephan
    Apr 26, 2017 at 10:13
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An alternative solution, not to disparage the other answers.

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

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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.

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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:

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

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