# prove in algebra

• Dec 1st 2009, 10:57 PM
the undertaker
prove in algebra
a/a-b = c/c-d
• Dec 1st 2009, 10:59 PM
Bacterius
Er ... and what are you trying to prove exactly ? Please add more detail ...
• Dec 2nd 2009, 07:54 AM
rowe
Quote:

Originally Posted by the undertaker
a/a-b = c/c-d

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

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

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

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

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

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

$1=1$, QED.
• Dec 2nd 2009, 08:22 AM
Defunkt
Quote:

Originally Posted by rowe
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 $a, b, c, d \in R$, and $a \neq b$ and $c \neq d$:

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

You assume here that $\frac{a}{a-b}=\frac{c}{c-d}$ which is actually what you're trying to prove...

Quote:

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

$\frac{c-d}{c} = \frac{a-b}{a}$
For this step to be correct, you must assume that a,c are non-zero (if they are, the equation holds trivially, although I don't really know what we're trying to "prove" here...).

Quote:

If $x = \frac{1}{x}$ then $x^2 = 1$, and thus $x = 1$.
What about $x=-1$?

Quote:

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

$1=1$, QED.
You basically said, "assume the equation holds, then it holds". This is obviously not a valid proof!

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
$a=1, ~ b=2, ~ c=2, ~ d=1$ ?
• Dec 2nd 2009, 08:40 AM
rowe