a/a-b = c/c-d

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- Dec 1st 2009, 10:57 PMthe undertakerprove in algebra
a/a-b = c/c-d

- Dec 1st 2009, 10:59 PMBacterius
Er ... and what are you trying to prove exactly ? Please add more detail ...

- Dec 2nd 2009, 07:54 AMrowe
Here is an attempted proof, although it is probably wrong, I thought I'd have a go. It relies on some unwritten obvious proof, but you can prove that yourself.

For $\displaystyle a, b, c, d \in R$, and $\displaystyle a \neq b$ and $\displaystyle c \neq d$:

Then $\displaystyle \frac{a}{a-b}\cdot(a-b)\cdot(c-d) = \frac{c}{c-d}\cdot(a-b)\cdot(c-d)$

$\displaystyle a(c-d) = c(a-b)$

$\displaystyle \frac{c-d}{c} = \frac{a-b}{a}$

If $\displaystyle x = \frac{1}{x}$ then $\displaystyle x^2 = 1$, and thus $\displaystyle x = 1$.

Therefore, as shown above, $\displaystyle \frac{a}{a-b} = \frac{a-b}{a} = 1$, as does $\displaystyle \frac{c}{c-d} = \frac{c-d}{c} = 1$.

$\displaystyle 1=1$, QED. - Dec 2nd 2009, 08:22 AMDefunkt
You assume here that $\displaystyle \frac{a}{a-b}=\frac{c}{c-d}$ which is actually what you're trying to prove...

Quote:

$\displaystyle a(c-d) = c(a-b)$

$\displaystyle \frac{c-d}{c} = \frac{a-b}{a}$

Quote:

If $\displaystyle x = \frac{1}{x}$ then $\displaystyle x^2 = 1$, and thus $\displaystyle x = 1$.

Quote:

Therefore, as shown above, $\displaystyle \frac{a}{a-b} = \frac{a-b}{a} = 1$, as does $\displaystyle \frac{c}{c-d} = \frac{c-d}{c} = 1$.

$\displaystyle 1=1$, QED.

Not to mention that it is quite obvious at a glance that this equation will definitely not hold for any four real numbers... what if we choose

$\displaystyle a=1, ~ b=2, ~ c=2, ~ d=1$ ? - Dec 2nd 2009, 08:40 AMrowe
Yeah, that was a pretty bad effort. Thanks for your pointers