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Homework Help: Simplifying rational numbers help

  1. Dec 22, 2009 #1
    Express
    [tex]\frac{1+\sqrt{2}}{3-\sqrt{2}}[/tex] as [tex]a+b\sqrt{2}[/tex] where a and b are rational numbers.

    I started by
    [tex]\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{3+\sqrt{2}}{3+\sqrt{2}}[/tex]

    But, I obtain

    [tex]\frac{5}{7}-\frac{4}{7}\sqrt{2}[/tex]

    I believe that, here, a and b are rational, but is there a more tidy version? I tried playing with the square root so that it is a multiple of 2, so that I could 'split' it into two square roots, on of the square root of two so that I could use it to cancel out the square root of two on the bottom, then I could also remove the other square root if it was a rational square root. For example, I tried:

    [tex]\frac{1+\sqrt{2}}{3-\sqrt{2}} * \frac{x+\sqrt{8}}{x+\sqrt{8}}[/tex]
    because
    [tex]\sqrt{8}= \sqrt{2*4} = \sqrt{2} * \sqrt{4} = 2\sqrt{2}[/tex]

    I haven't specified x, since its just an example. I actually tried making a relationship between x and the square root I introduced, since I could represent it algebraically as
    [tex]\sqrt{g}[/tex]
    where g is a multiple of two, would give
    [tex]\sqrt{2}*\sqrt{\frac{1}{2}g}[/tex].

    I'm guessing its more straight forward than this.
    Thanks in advance.
     
    Last edited: Dec 22, 2009
  2. jcsd
  3. Dec 22, 2009 #2

    Mark44

    Staff: Mentor

    Re: Simplifying

    The sign in the middle is wrong -- should be +.
    I don't believe there is. Assuming you make the correction I mentioned, you will have written the original expression in the form a + b*sqrt(2), where a and be are rational. You can't get any tidier than that.
     
  4. Dec 22, 2009 #3

    Mentallic

    User Avatar
    Homework Helper

    Re: Simplifying

    Quite ambitious, but no. Since the original expression is irrational, if you're going to simplify it, it will always still be irrational. That is why the [itex]\sqrt{2}[/itex] must be there. You can of course "simplify" it into many different ways, but there wil always be an irrational portion of the expression.
     
  5. Dec 23, 2009 #4
    Re: Simplifying

    Many thanks Mark44 and Mentallic.
     
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