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Math Help - sqrt 2, sqrt 3, sqrt 5 can't be in the arithmetic/geometric progression

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    sqrt 2, sqrt 3, sqrt 5 can't be in the same arithmetic/geometric progression

    Prove that \sqrt{2}, \sqrt{3} and \sqrt{5} can't be terms in the same
    • arithmetic
    • geometric

    progression.
    Last edited by james_bond; January 5th 2008 at 10:31 PM.
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by james_bond View Post
    Prove that \sqrt{2}, \sqrt{3} and \sqrt{5} can't be terms in the same
    • arithmetic
    • geometric

    progression.
    Calculate r and d. You'll see they're not the same between \sqrt{2}, \sqrt{3} and \sqrt{3}, \sqrt{5}
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    Quote Originally Posted by janvdl View Post
    Calculate r and d. You'll see they're not the same between \sqrt{2}, \sqrt{3} and \sqrt{3}, \sqrt{5}
    Trivial with a calculator of course. But proving \sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3} is a bit more interesting (for a student at this level) without one .....
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    Quote Originally Posted by mr fantastic View Post
    Trivial with a calculator of course. But proving \sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3} is a bit more interesting (for a student at this level) without one .....
    It is not hard. Assume that is is then 2\sqrt{3} = \sqrt{5} + \sqrt{2}. Square and there is a condatriction.

    In general given, p_1,...,p_n distinct primes and p_{n+1} is also distinct. Then \sqrt{p_{n+1}} cannot be expressed as a linear combination of \sqrt{p_1},...,\sqrt{p_n} over \mathbb{Q}.
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    But they don't have to be neighbors!
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    Exclamation

    Quote Originally Posted by ThePerfectHacker View Post
    It is not hard. Assume that is is then 2\sqrt{3} = \sqrt{5} + \sqrt{2}. Square and there is a condatriction.

    In general given, p_1,...,p_n distinct primes and p_{n+1} is also distinct. Then \sqrt{p_{n+1}} cannot be expressed as a linear combination of \sqrt{p_1},...,\sqrt{p_n} over \mathbb{Q}.
    The proof of this theorem (at least for the concrete case arising in this question) is the interesting bit (for a student at this level).
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    It is not hard. Assume that is is then 2\sqrt{3} = \sqrt{5} + \sqrt{2}. Square and there is a condatriction.
    I just went through the trouble of doing so, and then I saw this post...

    EDIT: But i'll show what i did anyway
    Last edited by janvdl; January 6th 2008 at 01:05 AM.
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    Suppose \sqrt{2},\sqrt{3},\sqrt{5} are not consecutive terms of an arithmetic or geometric progression.

    For the arithmetic progression:

    Let \sqrt{2}=a_m=a_1+(m-1)r (1)
    \sqrt{3}=a_n=a_1+(n-1)r (2)
    \sqrt{5}=a_p=a_1+(p-1)r (3)
    Substracting (2) from (1) and (3) from (2) we get
    \sqrt{3}-\sqrt{2}=(n-m)r
    \sqrt{5}-\sqrt{3}=(p-n)r
    Then \displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{n-m}{p-n}
    But, the left side member is irational and the right side member is rational.


    For the geometric progression:

    Let \sqrt{2}=b_1q^{m-1} (1)
    \sqrt{3}=b_1q^{n-1} (2)
    \sqrt{5}=b_1q^{p-1} (3)
    Then \displaystyle q=\left(\frac{2}{3}\right)^{\frac{m-n}{2}}=\left(\frac{3}{5}\right)^{\frac{n-p}{2}}
    \displaystyle \Rightarrow 2^{\frac{m-n}{2}}5^{\frac{n-p}{2}}=3^{\frac{m-p}{2}}\Rightarrow 2^{m-n}5^{n-p}=3^{m-p}
    The last equality is not true.
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    Quote Originally Posted by red_dog View Post
    Suppose \sqrt{2},\sqrt{3},\sqrt{5} are not consecutive terms of an arithmetic or geometric progression.

    For the arithmetic progression:

    Let \sqrt{2}=a_m=a_1+(m-1)r (1)
    \sqrt{3}=a_n=a_1+(n-1)r (2)
    \sqrt{5}=a_p=a_1+(p-1)r (3)
    Substracting (2) from (1) and (3) from (2) we get
    \sqrt{3}-\sqrt{2}=(n-m)r
    \sqrt{5}-\sqrt{3}=(p-n)r
    Then \displaystyle\frac{\sqrt{3}-\sqrt{2}}{\sqrt{5}-\sqrt{3}}=\frac{n-m}{p-n}
    But, the left side member is irational and the right side member is rational.


    For the geometric progression:

    Let \sqrt{2}=b_1q^{m-1} (1)
    \sqrt{3}=b_1q^{n-1} (2)
    \sqrt{5}=b_1q^{p-1} (3)
    Then \displaystyle q=\left(\frac{2}{3}\right)^{\frac{m-n}{2}}=\left(\frac{3}{5}\right)^{\frac{n-p}{2}}
    \displaystyle \Rightarrow 2^{\frac{m-n}{2}}5^{\frac{n-p}{2}}=3^{\frac{m-p}{2}}\Rightarrow 2^{m-n}5^{n-p}=3^{m-p}
    The last equality is not true.
    My solution was a little more humble...

    Arithmetic:

    Assume they are equal.

    \sqrt{3} - \sqrt{2} = \sqrt{5} - \sqrt{3}

    Set \sqrt{3} = x

    x - \sqrt{2} = \sqrt{5} - x

    2x = \sqrt{5} + \sqrt{2}

    Square both sides

    4x^2 = 25 + 2( \sqrt{5} \cdot \sqrt{2} ) + 4

    4(\sqrt{3})^2 = 29 + 2 \sqrt{10}

    12 - 29 = 2 \sqrt{10}

    -17 \neq 2 \sqrt{10}

    Therefore \sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}

    ----

    Geometric

    Once again, to make it easier, set \sqrt{3} = x

    Assume equality

    \frac{x}{ \sqrt{2} } = \frac{ \sqrt{5} }{x}

    x^2 = \sqrt{10}

    x^4 = 10

    (\sqrt{3})^4 \neq 10

    Therefore \frac{x}{ \sqrt{2} } \neq \frac{ \sqrt{5} }{x}
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    Quote Originally Posted by janvdl View Post
    My solution was a little more humble...

    Arithmetic:

    Assume they are equal.

    \sqrt{3} - \sqrt{2} = \sqrt{5} - \sqrt{3}

    Set \sqrt{3} = x

    x - \sqrt{2} = \sqrt{5} - x

    2x = \sqrt{5} + \sqrt{2}

    Square both sides

    4x^2 = 25 + 2( \sqrt{5} \cdot \sqrt{2} ) + 4

    4(\sqrt{3})^2 = 29 + 2 \sqrt{10}

    12 - 29 = 2 \sqrt{10}

    -17 \neq 2 \sqrt{10}

    Therefore \sqrt{3} - \sqrt{2} \neq \sqrt{5} - \sqrt{3}

    ----

    Geometric

    Once again, to make it easier, set \sqrt{3} = x

    Assume equality

    \frac{x}{ \sqrt{2} } = \frac{ \sqrt{5} }{x}

    x^2 = \sqrt{10}

    x^4 = 10

    (\sqrt{3})^4 \neq 10

    Therefore \frac{x}{ \sqrt{2} } \neq \frac{ \sqrt{5} }{x}
    You are again assuming they are consecutive terms of the progressions. See red_dog's solution, he does not assume it
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by Isomorphism View Post
    You are again assuming they are consecutive terms of the progressions. See red_dog's solution, he does not assume it
    TPH said it was possible to do it this way. And if it's not wrong, what's your problem with me to do it this way?
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    Quote Originally Posted by janvdl View Post
    TPH said it was possible to do it this way. And if it's not wrong, what's your problem with me to do it this way?
    It is just that you were doing a different problem than Red_dog. You were saying that sqrt(2),sqrt(3),sqrt(5) cannot be right next to each other, and what you did is correct. But Red_dog did a stronger problem he showed that you cannot have these three numbers in any arithmetic progession. Meaning you cannot have sqrt(2) as a 3rd term, sqrt(3) as a 10th term, and sqrt(5) as a 29th term.
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