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Here is the given information for the following proof I'm about to begin:

Suppose a, b, c, d, e and f are non-zero elements of field such that $$ \frac{a}{b} = \frac {c}{d} = \frac{e}{f}$$

I'm suppose to show the following identity is true, whenever the denominator in question is non-zero:

$$\frac{a}{b} = \frac{(a+c+e)}{(b+d+f)}$$

Here my thinking process: In order for this identity to hold, $c + e$ and $d + f$ must add up to $0$, or so I believe. Otherwise, the equality wouldn't hold true at all. Another way of thinking about this is that perhaps the letters a, c,e are multiples of one another (same with b, d, f).

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    What happens when you cross multiply? – 2017-02-28
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    We have $ad = bc$? – 2017-02-28
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    You also have $af = be$ and $cf = de$. It then follows that $a(b+d+f) = ab + ad + af = ba + bc + be = b(a+ c+e)$. – 2017-02-28
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    Ahh now I see it. So just cross multiplication then. – 2017-02-28
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    Basically. The part "whenever the denominator $b+d+f$ in question is nonzero" must be a hypothesis, though. It is not necessarily given that $b+d+f \neq 0$. Take $\frac{2}{2}=\frac{-1}{-1}=\frac{-1}{-1}$, for instance. – 2017-02-28
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    Yes I was thinking of counter examples as well like yours for example. And yes, we were given that the denominators are *non-zero*. – 2017-02-28
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    Hint: prove first that $a/b=c/d \implies a/b=(a+c)/(b+d)$ when $b,d,b+d \ne 0\,$, then apply it twice. – 2017-03-01

3 Answers 3

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$$\begin{align} b\,x\, &=\, a\\ d\,x\, &=\, c\\ f\,x\, &=\, e \\ \smash[t]{\overset{\rm add}\Longrightarrow}\,\ (b\!+\!d\!+\!f)\,x\, &=\, a\!+\!c\!+\!e \end{align}$$

Said geometrically: the vectors $\,(a,b),(c,d),(e,f)\,$ have the same slope hence so does their sum, i.e. the solutions of $\, ay = bx\,$ are a submodule of $\,\Bbb Z^2,\,$ just like in the vectors space case.

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$$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}\rightarrow c=ax,d=bx,e=ay,f=by\\\frac{a+c+e}{b+d+f}=\frac{a+ax+ay}{b+bx+by}=\frac{a(1+x+y)}{b(1+x+y)}=\frac{a}{b}$$

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We have $ad=bc$, $af=be$ and $cf=de$. Now let's compute

$$a(b+d+f)-b(a+c+e)=ab+ad+af-ab-bc-be=0$$

Therefore whenever the denominators are non zero

$${a\over b}={a+c+e\over b+d+f}$$

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    Yes, this is what joeb and I were discussing up in chat. It's all clear. – 2017-02-28