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Math Help - Who is bigger !!

  1. #1
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    Who is bigger !!

    Hello to all !

    I have a little problem !!

    Who is bigger tyen who ????????

    (\sqrt 7-\sqrt 3) \ or \ (\sqrt 6-\sqrt 2)

    Show me with the stage of the solution!

    Thank you !
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  2. #2
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    I bet your calculator knows. Are you allowed to use it?

    Spoiler:
      (\sqrt 6-\sqrt 2)> (\sqrt 7-\sqrt 3)
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  3. #3
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    Let a=\sqrt7+\sqrt3 and b=\sqrt6+\sqrt2. The squares of these two numbers are a^2=7+\sqrt{21}+3=10+\sqrt{21} and b^2=6+\sqrt{12}+2=8+\sqrt{12}.
    So it's clear that a^2>b^2 and therefore a>b.

    Now we have a(\sqrt7-\sqrt3)=7-3=4 and b(\sqrt6-\sqrt2)=6-2=4.
    Write a(\sqrt7-\sqrt3)=b(\sqrt6-\sqrt2). Since a>b we must have \sqrt7-\sqrt3<\sqrt6-\sqrt2.
    Last edited by melese; September 27th 2010 at 04:22 PM. Reason: Latex typo.
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  4. #4
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    Quote Originally Posted by yehoram View Post
    Hello to all !

    I have a little problem !!

    Who is bigger than who ????????

    (\sqrt 7-\sqrt 3) \ or \ (\sqrt 6-\sqrt 2)

    Show me with the stage of the solution!

    Thank you !
    There is no need for any squaring......

    \sqrt{7}>\sqrt{6} and \sqrt{3}>\sqrt{2}\ \Rightarrow\ \left[\sqrt{7}+\sqrt{3}\;\right]>\left[\sqrt{6}+\sqrt{2}\;\right]

    Then use the property of surd conjugates

    \left[\sqrt{a}+\sqrt{b}\;\right]\left[\sqrt{a}-\sqrt{b}\;\right]=a-b

    Since 7-3=6-2, then (surd)(surd conjugate) is equal for both.

    Hence, since \left[\sqrt{7}+\sqrt{3}\;\right]>\left[\sqrt{6}+\sqrt{2}\;\right]

    \Rightarrow\ \left[\sqrt{7}-\sqrt{3}\;\right]<\left[\sqrt{6}-\sqrt{2}\;\right]



    A more general way to look at this.....

    For any positive difference "k" and any positive increment "w"

    \left[\sqrt{x+k}+\sqrt{x}\;\right]<\left[\sqrt{x+k+w}+\sqrt{x+w}\;\right]

    \left(\sqrt{x+k}+\sqrt{x}\;\right)\left(\sqrt{x+k}-\sqrt{x}\;\right)=k

    \left(\sqrt{x+k+w}+\sqrt{x+w}\;\right)\left(\sqrt{  x+k+w}-\sqrt{x+w}\;\right)=k

    \Rightarrow\ \left[\sqrt{x+k+w}-\sqrt{x+w}\;\right]<\left[\sqrt{x+k}-\sqrt{x}\;\right]


    Yet another general way to consider it is as follows.....

    \sqrt{6}-\sqrt{2}

    \sqrt{7}-\sqrt{3}

    Generally as the numbers increase with a common difference, the difference between their square roots keeps decreasing,

    since \sqrt{x}-\sqrt{x-4}, defined for x>4, becomes insignificantly small as x increases without bound,

    while x-(x-4) is always 4.

    This means \left[\sqrt{6}-\sqrt{2}\;\right]>\left[\sqrt{7}-\sqrt{3}\;\right]>\left[\sqrt{8}-\sqrt{4}\;\right]>.....>\left[\sqrt{n}-\sqrt{n-4}\;\right] as "n" keeps increasing.

    You can use calculus to show that f(x)=\sqrt{x}-\sqrt{x-4}
    is a decreasing function as x increases.

    \displaystyle\frac{d}{dx}\left[\sqrt{x}-\sqrt{x-4}\;\right]=\frac{d}{dx}\left(x^{\frac{1}{2}}-(x-4)^{\frac{1}{2}\right)

    =\displaystyle\frac{1}{2}x^{-\frac{1}{2}}-\frac{1}{2}(x-4)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-4}}

    \displaystyle\sqrt{x-4}<\sqrt{x}\;\; \Rightarrow\frac{1}{\sqrt{x-4}}>\frac{1}{\sqrt{x}}

    Therefore the derivative is negative, so the function is a decreasing function.

    Therefore \left[\sqrt{7}-\sqrt{3}\;\right]<\left[\sqrt{6}-\sqrt{2}\;\right]

    You could also examine the function f(x)=\sqrt{x+4}-\sqrt{x}

    which is defined for x positive.
    Attached Thumbnails Attached Thumbnails Who is bigger !!-difference-square-roots.jpg  
    Last edited by Archie Meade; September 28th 2010 at 09:22 AM.
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