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Basic inequality question

  1. Jun 6, 2009 #1

    Tried to solve first problem in the book "Firfty Challenging Problems in Probability" and solved it although very ugly.

    Then I check the answers and see the author use the following inequality:

    [tex]r > \frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b[/tex]

    Now correct me if I am wrong, but this implies that


    \frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b


    Right? Well, this does not seem right, does it?
  2. jcsd
  3. Jun 6, 2009 #2
    Given: [tex] \frac{1}{\sqrt{2}-1}b=(\sqrt{2} + 1)b [/tex]

    Multiply by [tex]\frac{\sqrt{2}+1}{\sqrt{2}+1}[/tex]

    [tex] \frac{1}{\sqrt{2}-1}*\frac{\sqrt{2}+1}{\sqrt{2}+1}b=(\sqrt{2} + 1)b [/tex]

    [tex] \frac{\sqrt{2}+1}{2-1}b=(\sqrt{2} + 1)b [/tex]

    [tex] \frac{\sqrt{2}+1}{1}b=(\sqrt{2} + 1)b [/tex]
  4. Jun 6, 2009 #3
    sweet mother of god. thanks.

    It is always the 1 that is omitted that confuses me. Plus I used sqrt(9) to ease the calculation.
  5. Jun 6, 2009 #4
    Sure, I just hope you understood the steps I took.

    I'll lay it out in English just in case.

    Basically you multiply top and bottom by the same thing (aka 1), then when you multiply the denominator, you'll remember that (x + A) (x - A) = x^2 - A^2. So for this we get 2 - 1 = 1 :)
  6. Jun 6, 2009 #5
    yes I understood from the first reply, it's just so simple, hence my amazement.
  7. Jun 6, 2009 #6
    Ah no worries. I wouldn't have thought to do this step either if this was me several years ago.

    Once you see it the first time, it sticks to you. Once you practice it on a couple of problems it becomes natural. So whenever I see square roots like that in the denominator, I automatically turn on simplification mode (since I've seen it so many times).

    Such is knowledge and life.
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