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Proof of neighbor fractions

  1. Jan 4, 2014 #1
    proof.jpg


    *If I turn out to have a wrong answer, please no hints or showing an valid proof. I want to do it on my own !
    ad/bd-bc/bd=+-1/bd is neighbor fraction

    Now, reduce the common numbers :

    a/b-c/d=+-1/bd

    We must now prove that the left hand side has irreductible fractions. Lets see what would happen if these fraction were reductible.

    let a=z*y , b=z*l , c=p*m d=p*n

    z*y/z*l-p*m/p*n equality to be determined +-1/(z*l)*(p*n)

    Reduce:

    *ln (y/l-m/n) equality to be determined (+-1/(z*l)*(p*n))*ln

    yn-lm equality to be determined +-1/z*p

    yn-lm not= +-1/z*p

    We have considered the initial fractions to be reductible and have arrived at a false result. An integer cannot be equal to (+-1/z*p)

    So, the initial fractions must be irreductible.

    *Of course, I'm considering the variables to represents integers.
     
  2. jcsd
  3. Jan 4, 2014 #2

    mfb

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    I see a shorter way, but your approach is possible, with two caveats:

    Currently, you are showing that at least one of the fractions has to be irreducible. Do you want to show that both have to be irreducible?

    It can, in a special case (that does not change the main result, but you have to consider this case).
     
  4. Jan 4, 2014 #3
    Wait, what do you mean I considered only one case ? I tried to consider two factorizable fractions at the sametime...

    Also, for the special case you were talking about, I did notice that there was an exception but I rejected it, didn't think it would have great consequence. Is it the following? :

    x=1/x
    If we were to obtain a result where x=-1, or 1 (I don't know how we could obtain such result) then this would create an equality, if such result was possible. But this would be the only exception.
     
  5. Jan 4, 2014 #4

    mfb

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    You found a contradiction for the assumption that both fractions are not irreducible at the same time, indeed. This does not rule out the case of 1 irreducible and one reducible fraction.

    z and p equal to +-1 is indeed this exception, but this exception corresponds to irreducible fractions, so it's fine (as long as you consider it in this way).
     
  6. Jan 4, 2014 #5
    Ah ok, then I'll post here my new finding when I create the other part of the proof.

    I just didn't follow the following :
    but this exception corresponds to irreducible fractions, so it's fine (as long as you consider it in this way)

    Could you rephrase ??? (I'm tired, so maybe I'm not reading correctly lol.) From what I've understood what you meant is that this exception could only be achieved if were able to have irreductible fractions be reductible (Which I proved in part wasn't possible and a few numerical examples follow what I'm trying to prove.) Is that what you meant ??

    Thank you!
     
  7. Jan 8, 2014 #6
    Ok, sorry for being late to respond :


    a/b-c/d=+-1/bd

    xi/xj-c/d (to be determined) +-1/((xj)*d)

    (i/j-c/d(to be determined)+-1/xj*d) * jd

    id-jc not equal to +-1/x

    except for 1 and -1

    The same is done for the second fractions on the left side.
     
  8. Jan 8, 2014 #7

    mfb

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    Right.
     
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