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Thread: prove in algebra

  1. #1
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    prove in algebra

    a/a-b = c/c-d
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  2. #2
    Super Member Bacterius's Avatar
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    Er ... and what are you trying to prove exactly ? Please add more detail ...
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  3. #3
    Member rowe's Avatar
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    Quote Originally Posted by the undertaker View Post
    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 $\displaystyle a, b, c, d \in R$, 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.
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  4. #4
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    Quote Originally Posted by rowe View Post
    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 $\displaystyle \frac{a}{a-b}\cdot(a-b)\cdot(c-d) = \frac{c}{c-d}\cdot(a-b)\cdot(c-d)$
    You assume here that $\displaystyle \frac{a}{a-b}=\frac{c}{c-d}$ which is actually what you're trying to prove...

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

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

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

    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.
    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
    $\displaystyle a=1, ~ b=2, ~ c=2, ~ d=1$ ?
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  5. #5
    Member rowe's Avatar
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    Yeah, that was a pretty bad effort. Thanks for your pointers
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