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Math Help - Sum of nth powers

  1. #1
    MHF Contributor Drexel28's Avatar
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    Sum of nth powers

    Hello friends.

    I was just wondering if there is an easy proof for the following:

    For ever n>1, \sum_{j=1}^{n}\sqrt[j]{j}\notin\mathbb{Q}
    I may have found a horrid proof using the rational root theorem, but I have not checked the details.

    Any input is appreciated!
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  2. #2
    Senior Member roninpro's Avatar
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    You can probably try to show that the set \{1, \sqrt{2}, \sqrt[3]{3},\ldots, \sqrt[n]{n}\} is linearly independent over the field \mathbb{Q}. The result will follow immediately.

    It seems to be fairly straightforward if you know some field theory. (But I haven't actually tried it, so I could be wrong.)
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by roninpro View Post
    You can probably try to show that the set \{1, \sqrt{2}, \sqrt[3]{3},\ldots, \sqrt[n]{n}\} is linearly independent over the field \mathbb{Q}. The result will follow immediately.

    It seems to be fairly straightforward if you know some field theory. (But I haven't actually tried it, so I could be wrong.)
    I am sorry, it was my bad for not saying. I was aware there was an algebraic proof, but you see my cousin asked me and he's in calc III, I was hoping there may be an "elementary" number theoretic proof. Thanks anyways, I appreciate it!
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  4. #4
    Super Member Bacterius's Avatar
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    Wouldn't it be possible to somehow get rid of the roots by taking the inverses of the exponents while keeping the sum equivalent ? Dealing with powers seems simpler somehow.

    This is just an idea, I guess it would be possible by messing around with logs and using their irrational properties.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bacterius View Post
    Wouldn't it be possible to somehow get rid of the roots by taking the inverses of the exponents while keeping the sum equivalent ? Dealing with powers seems simpler somehow.

    This is just an idea, I guess it would be possible by messing around with logs and using their irrational properties.
    I'm sorry, I'm not quite sure what you mean.
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  6. #6
    Super Member Bacterius's Avatar
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    I was dreading it

    I meant, expressing your statement using a sum of powers instead of a sum of roots. Some changes would be required, but for instance :

    \sqrt[y]{x} = b^\frac{\log_b{(x)}}{y} - or even better - \sqrt[j]{j} = b^\frac{\log_b{(j)}}{j}

    Seems easier than dealing with \sqrt[j]{j} somehow.

    But this is not gonna be "elementary" anyway
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