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Fractions with roots: voodoo

  1. Mar 30, 2008 #1
    Starting with simple fractions, it's known that:

    [tex]{{{a \over b}} \over {{c \over d}}} = {{ad} \over {bc}}[/tex]

    So when b == d:

    [tex]{{{a \over b}} \over {{c \over b}}} = {a \over c}[/tex]

    But what if in the case of:

    [tex]{{{{1 + \sqrt 2 } \over {\sqrt 2 }}} \over {{{1 - \sqrt 2 } \over {\sqrt 2 }}}}[/tex]

    Following the above (for abc), it should result in:

    [tex]{{1 + \sqrt 2 } \over {1 - \sqrt 2 }}[/tex]

    But if you follow the abcd version:

    [tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]

    ...which is not equal to the first expression.

    Why?
     
  2. jcsd
  3. Mar 30, 2008 #2
    Are you sure that you did the calculations right, at frist place?????
    Check your work, you did not distrubute correctly the [tex]\sqrt{2}[/tex]

    remember [tex]\sqrt{2}*\sqrt{2}=2 \ \ \ and \ \ \ not \ \ \ \sqrt{2}[/tex]
     
    Last edited: Mar 30, 2008
  4. Mar 30, 2008 #3
    Yes.

    [tex]\displaylines{
    \sqrt 2 \left( {1 + \sqrt 2 } \right) = \left( {\sqrt 2 \times 1} \right) + \left( {\sqrt 2 \times \sqrt 2 } \right) \cr
    = \sqrt 2 + 2 \cr}[/tex]
     
  5. Mar 30, 2008 #4

    Check the denominator, too.
     
  6. Mar 30, 2008 #5
    You got the terms reversed in the demoninator: [tex]{{\left( {1 + \sqrt 2 } \right)\sqrt 2 } \over {\sqrt 2 \left( {1 - \sqrt 2 } \right)}} = {{2 + \sqrt 2 } \over {2 - \sqrt 2 }}[/tex]

    Denominator should be [tex]\sqrt{2}-2[/tex]
     
    Last edited: Mar 30, 2008
  7. Mar 30, 2008 #6
    Ah, that solves it then. Thanks.

    As an aside, how can I make this:

    [tex]{{dy} \over {dx}} = {{2y - x^2 } \over {y^2 - 2x}}[/tex]

    into this:

    [tex]{{dy} \over {dx}} = {{x^2 -2y } \over {2x - y^2}}[/tex]

    They're meant to be identical, but I can't think how.

    Thanks!
     
  8. Mar 30, 2008 #7
    multiply the top and the bottom by -1.

    You can do this because in essence it is multiplying the entire expression by 1.
     
  9. Mar 30, 2008 #8
    Are you working on diff. eq. or someone just gave u that expression. Don't tell me that you are!
     
  10. Mar 30, 2008 #9
    It's part of a larger exercise on a differential equation. Just a simple example for an implicit function.
     
  11. Mar 30, 2008 #10
    This was someting that i really was affraid i would hear. Damn.!!
     
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