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Basic rule is that: a/b=c/d then, a+b/a-b = c+d/c-dbut suppose

  1. Jan 18, 2012 #1
    Basic rule is that:

    a/b=c/d then, a+b/a-b = c+d/c-d

    but suppose if we apply "componendo dividendo" just to the RHS TWICE, we get the original number... consider the example : 16/4 (which we know is equal to 4 or rather 4/1)
    now applying componendo dividendo just once to 16/4 ,
    we get 20/12 , then again applying componendo dividendo to
    20/12 , we get 32/8 ,which is equal to 4/1 or 4.
    but i know this is not even componendo -dividendo theorem,
    but when we apply it twice to the RHS v get back the RHS......
    this was quite useful when solving a trigonometry problem....but according to the teachers there is no such theory.......so, not very useful.
    so, the question is ,what is it that you find wrong with this "theory" i used.(if any,specify)......??
  2. jcsd
  3. Jan 18, 2012 #2
    Re: Componendo-dividendo

    Whatever you said in your post is complete correct. What exactly did you want to ask??
  4. Jan 18, 2012 #3


    Staff: Mentor

    Re: Componendo-dividendo

    If the second equation is
    [tex]\frac{a+b}{a-b} = \frac{c+d}{c-d}[/tex]
    then what you have written is incorrect. What you wrote is the same as a + (b/a) - b = c + (d/c) - d.

    When you write fractions with numerators or denominators with multiple terms, you need to used parentheses around the entire numerator or denominator, like so:
    (a+b)/(a-b) = (c+d)/(c-d)

    (Or learn to write then using LaTeX...)
  5. Jan 19, 2012 #4
    Re: Componendo-dividendo

    micromass, thakyou for replying.My question is whether you can point out any mistake in it.

    and Mark44
    Sorry for not putting it in the parenthesis.What i actually meant to post was
    (a+b)/(a-b) = (c+d)/(c-d).
  6. Jan 19, 2012 #5


    User Avatar
    Science Advisor

    Re: Componendo-dividendo

    What statement was proved?
  7. Jan 20, 2012 #6
    Re: Componendo-dividendo

    the above statement ofcourse.
  8. Jan 28, 2012 #7
    Re: Componendo-dividendo

    nice theory
  9. Oct 27, 2013 #8
    Your teachers are doing you a disservice by stating that there is " no such theory" when what they really mean is that there is no such commonly known useful theory. However, your proposition is correct, and provable.
    For conciseness, let there be a Componendo et Dividendo operator, which we shall show as CeD{}, such that CeD{a/b} = (a+b)/(a-b)
    Then the original theorem says, in our nomenclature, if a/b = c/d then Ced{a/b} = Ced{c/d}

    What you are calculating then is Ced{Ced{a/b}}
    Expanding, Ced{Ced{a/b}} = Ced{(a+b)/(a-b)} = ((a+b)+(a-b))/((a+b)-(a-b)) = 2a/2b = a/b. QED

    Therefore the CeD of a CeD will always return the original ratio (specifically, to double the original expression). Thus it could well be called the Alicia theorem, a provable theorem. But it is unlikely to become a widely known one, since, other than its curiosity value, it does not appear to have any wide applicability as it does nothing to simplify the original expression.
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