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Square roots and fractions

  1. Mar 17, 2008 #1
    How do you get 3/sqrt2 from 3sqrt2/2?
     
  2. jcsd
  3. Mar 17, 2008 #2
    Ask yourself what's (sqrt x)^2?
     
  4. Mar 17, 2008 #3

    Math Is Hard

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    hint: sqrt2/sqrt2 = 1
     
  5. Mar 17, 2008 #4
    sorry dont follow
     
  6. Mar 17, 2008 #5

    Integral

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    Try going the other way.
    Start with
    [tex] \frac 3 {\sqrt 2} [/tex]

    to get

    [tex] \frac {3 \sqrt 2} 2 [/tex]
     
  7. Mar 17, 2008 #6
    If you're going from 3sqrt2/2 to 3/sqrt2 you're basically de-rationalizing the denominator.
     
  8. Mar 18, 2008 #7

    HallsofIvy

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    By definition [itex](\sqrt{2})^2= \sqrt{2}\sqrt{2}= 2[/itex]. That means that
    [tex]\frac{\sqrt{2}}{2}= \frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}[/tex]
     
  9. Mar 21, 2008 #8

    Air

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    [itex]\frac{3}{\sqrt 2} . \frac{\sqrt 2}{\sqrt 2} = \frac{3 \sqrt 2}{2}[/itex].

    This is rationalising the denominator as you cannot have a surd on the denominator. The question that you are asking is how you get the reverse, de-rationalising the denominator. Try opposite steps to the one I shown.
     
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