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Math Help - Proof that sum is rational number

  1. #1
    Senior Member OReilly's Avatar
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    Proof that sum is rational number

    I have to prove that
    \frac{1}{{\sqrt 1  + \sqrt 2 }} + \frac{1}{{\sqrt 2  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99}  + \sqrt {100} }}
    is rational number.

    I have one idea, I hope it's not crazy!

    I have spoted that the sum of first three fractions are equal to 1:

    \frac{1}{{\sqrt 1  + \sqrt 2 }} + \frac{1}{{\sqrt 2  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 4 }}=1

    I have then discovered that next five fractions are also equal to 1.
    \frac{1}{{\sqrt 4  + \sqrt 5 }} + \frac{1}{{\sqrt 5  + \sqrt 6 }} + \frac{1}{{\sqrt 6  + \sqrt 7 }} + \frac{1}{{\sqrt 7  + \sqrt 8 }} + \frac{1}{{\sqrt 8  + \sqrt 9 }} = 1.

    Continuing, we have that every odd number of fractions are equal to 1.
    Considering that there are 99 fractions in that sum we get that because of 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99 whole sum is equal to 9: \frac{1}{{\sqrt 1  + \sqrt 2 }} + \frac{1}{{\sqrt 2  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99}  + \sqrt {100} }}=9


    I don't know how to prove that but it seems to me that there is a rule in that sum which I have described.
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  2. #2
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    Quote Originally Posted by OReilly
    I have to prove that
    \frac{1}{{\sqrt 1  + \sqrt 2 }} + \frac{1}{{\sqrt 2  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99}  + \sqrt {100} }}
    is rational number.

    I have one idea, I hope it's not crazy!

    I have spoted that the sum of first three fractions are equal to 1:

    \frac{1}{{\sqrt 1  + \sqrt 2 }} + \frac{1}{{\sqrt 2  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 4 }}=1

    I have then discovered that next five fractions are also equal to 1.
    \frac{1}{{\sqrt 4  + \sqrt 5 }} + \frac{1}{{\sqrt 5  + \sqrt 6 }} + \frac{1}{{\sqrt 6  + \sqrt 7 }} + \frac{1}{{\sqrt 7  + \sqrt 8 }} + \frac{1}{{\sqrt 8  + \sqrt 9 }} = 1.

    Continuing, we have that every odd number of fractions are equal to 1.
    Considering that there are 99 fractions in that sum we get that because of 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99 whole sum is equal to 9: \frac{1}{{\sqrt 1  + \sqrt 2 }} + \frac{1}{{\sqrt 2  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 4 }} + ... + \frac{1}{{\sqrt {99}  + \sqrt {100} }}=9


    I don't know how to prove that but it seems to me that there is a rule in that sum which I have described.
    One word,
    rationalize.
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by ThePerfectHacker
    One word,
    rationalize.
    Bit more words?
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  4. #4
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    Quote Originally Posted by OReilly
    Bit more words?
    Okay whenever you have,
    \frac{1}{\sqrt{x}\pm \sqrt{y}} you can change the fraction, called "rationalizing". This is done by mutiplying the numerator and denominator by \sqrt{x}\mp \sqrt{y}. This changes the denominator into whole number because you use the difference of two squares.
    Thus,
    \frac{1}{\sqrt{x}\pm \sqrt{y}}\cdot \frac{\sqrt{x}\mp \sqrt{y}}{\sqrt{x}\mp \sqrt{y}}=\frac{\sqrt{x}\mp \sqrt{y}}{x-y}
    ------------
    Hence you have,
    \frac{1}{\sqrt{1}+ \sqrt{2} }+\frac{1}{\sqrt{2}+ \sqrt{3} }+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{ \sqrt{99}+\sqrt{100}}
    Upon rationalizing,
    \frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{99}-\sqrt{100}}{-1}
    Simplyfy,
    -1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{99}+\sqrt{100}
    Each one cancels each other out (such a series is called telescoping), thus,
    -1+\sqrt{100}=-1+10=9
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  5. #5
    Senior Member OReilly's Avatar
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    Thanks!
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  6. #6
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    Welcomed.
    That happens to be a nice problem which I really like. Where did you get it from?
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  7. #7
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    Quote Originally Posted by ThePerfectHacker
    Welcomed.
    That happens to be a nice problem which I really like. Where did you get it from?
    From high school book. Author has also wrote book with variety of problems which i find most interesting.
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