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Easy question about the root (of a real number)

  1. Oct 23, 2009 #1
    [SOLVED] Easy question about the root (of a real number)

    Hi, i'm a bit embarresed to ask this but does anybody know how to get this:
    [tex]\sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1[/tex]

    ?
     
    Last edited: Oct 23, 2009
  2. jcsd
  3. Oct 23, 2009 #2
    [tex]\sqrt{3 - 2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}[/tex]

    what is the relationship between [tex]\left(2-2\sqrt{2}+1\right)[/tex] and [tex]\left(\sqrt{2} - 1\right)[/tex]??
     
  4. Oct 23, 2009 #3
    I must say that this one does require a bit of clever thinking. I can add that the best way to "see" the answer is to take your equation

    [tex]\sqrt{3-2\sqrt{2}}=\sqrt{2}-1[/tex]

    and solve it as is (this will ultimately lead you to what S_David is pointing out).
     
  5. Oct 23, 2009 #4
    OK! Got it !
    Sorry for wasting your time and thank you !! :)

    It's true that it is one you have to 'see' !
    And if you 'see it' it's easy, but if you don't ...

    But thanks a lot you guys, got it now!
     
    Last edited: Oct 23, 2009
  6. Oct 23, 2009 #5
    Definitely didn't waste my time. You forced me to think, which is always good! :biggrin:
     
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