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  1. #1
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    prove

    Prove that root2 is always between a/b and (a+2b)/(a+b) for any positive whole number a and b


    Which means we have to prove that a/b is lesser than root2 and root2 is lesser than (a+2b)/(a+b)...how

    I htink you need to use contradictory method and euclid's divison lemma..
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  2. #2
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    Hi -

    Root2 is irrational. Therefore a/b is not equal to root2 for any values of a and b.

    Assume first that a/b is greater than root2. Then, squaring:

    a^2/b^2 > 2

    So a^2 > 2b^2

    Then square (a + b)/(a + 2b).

    The result can be written as 1 + 3b^2/(a^2 + 2ab + b^2)

    and this is less than

    1 + 3b^2/(3b^2 + 2ab) since we have replaced a^2 by 2b^2 in the denominator (and a^2 > 2b^2, from above).

    This in turn is less than 2, since 2ab is positive.

    So (a + b)/(a + 2b) < root2.

    QED

    Next, assume a/b is less than root2. And so on ... as before with the inequalities reversed.

    OK?
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  3. #3
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    Hello, ice_syncer!

    An interesting problem . . .


    Prove that \sqrt{2} is always between \frac{a}{b} and \frac{a+2b}{a+b}
    for any positive whole numbers a and b .

    Let \frac{a}{b} be any positive rational number less than \sqrt{2}

    We have: . \frac{a}{b} \:<\:\sqrt{2} .[1]

    . . Add 1 to both sides: . \frac{a}{b} + 1 \:<\:\sqrt{2} + 1

    . . Take reciprocals: . \frac{1}{\frac{a}{b} + 1} \:>\:\frac{1}{\sqrt{2}+1}

    . . Add 1 to both sides: . 1 + \frac{1}{\frac{a}{b}+1} \:>\:1 + \frac{1}{\sqrt{2}+1} .[2]


    The left side is: . 1 + \frac{1}{\frac{a}{b}+1} \;=\;1 + \frac{1}{\frac{a+b}{b}} \;=\;1 + \frac{b}{a+b} \;=\;\frac{a + 2b}{a+b}

    The right side is: . 1 +\frac{1}{\sqrt{2}+1}\!\cdot\!{\color{red}\frac{\s  qrt{2}-1}{\sqrt{2}-1}} \;=\;1 + \frac{\sqrt{2} - 1}{1} \;=\;\sqrt{2}

    . . Hence, [2] becomes: . \frac{a+2b}{a+b} \:> \:\sqrt{2} .[3]


    Combining [1] and [3], we have: . \frac{a}{b} \:<\:\sqrt{2} \:<\:\frac{a+2b}{a+b}

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  4. #4
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    Elegant solution, Soroban!

    Grandad
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, ice_syncer!

    An interesting problem . . .



    Let \frac{a}{b} be any positive rational number less than \sqrt{2}

    We have: . \frac{a}{b} \:<\:\sqrt{2} .[1]

    . . Add 1 to both sides: . \frac{a}{b} + 1 \:<\:\sqrt{2} + 1

    . . Take reciprocals: . \frac{1}{\frac{a}{b} + 1} \:>\:\frac{1}{\sqrt{2}+1}

    . . Add 1 to both sides: . 1 + \frac{1}{\frac{a}{b}+1} \:>\:1 + \frac{1}{\sqrt{2}+1} .[2]


    The left side is: . 1 + \frac{1}{\frac{a}{b}+1} \;=\;1 + \frac{1}{\frac{a+b}{b}} \;=\;1 + \frac{b}{a+b} \;=\;\frac{a + 2b}{a+b}

    The right side is: . 1 +\frac{1}{\sqrt{2}+1}\!\cdot\!{\color{red}\frac{\s  qrt{2}-1}{\sqrt{2}-1}} \;=\;1 + \frac{\sqrt{2} - 1}{1} \;=\;\sqrt{2}

    . . Hence, [2] becomes: . \frac{a+2b}{a+b} \:> \:\sqrt{2} .[3]


    Combining [1] and [3], we have: . \frac{a}{b} \:<\:\sqrt{2} \:<\:\frac{a+2b}{a+b}

    Thanks mate! well done.
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