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I'm looking for a simple proof or a reference to any proof that

For $j \in ℤ$, $0, when each $k^j \notin ℚ$ and each $d_j \in ℚ$, $d_j \neq 0$, then $\sum_{j=1}^{m-1} d_j k^j \notin ℚ$

My searches have turned up a few papers that appear to make use of this property, but none that prove it, refer to a source, or even give it a name.

EDIT:

I'm working with $k$ restricted to positive integer roots of rational numbers: $k=r^{1/m}=(a/b)^{1/m}$ with $r$ and $m$ chosen so that $m$ is the least positive integer for which $k^m \in ℚ$ (so when $k \in ℚ$, $m=1$).

The set of positive integer roots of rational numbers (set $$) is not closed under addition (see the $\sqrt{2}-1$ case in comments) Given $l$, $h$ members of $$, $\left(l\pm h\right) = l \left(1\pm{h\over l}\right) = l \left(1+k\right) \notin $ occurs when $\left(1+k\right) \notin $.

The sum before the edit is a rearrangement of the binomial expansion of $\left(1+k\right)^n$ where a different choice of $n$ changes the values of $d_j$, and excluding the rational $j=0$ term. If the sum is irrational, then $\left(1+k\right)^n \notin ℚ$ for any $n$, $\left(1+k\right) \notin $, and $\left(l\pm h\right) \notin $ occurs when $\pm {h\over l} = k \notin ℚ$.

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    More conditions are necessary for this to be true: $d_1 = 2$, $d_1 = -1$, $k^1 = \sqrt{2}$, $k^2 = 2 \sqrt{2}$, but $d_1 k_1 + d_2 k_2 = 0 \in \mathbb{Q}$.2012-06-08
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    Do you think that it holds for $k=\sqrt{2}-1$ and $m=3$?2012-06-08
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    There is something missing here, because you can set the final polynomial equal to zero and use this as the definition of $k$. This will give a counterexample if you choose an irreducible polynomial.2012-06-08
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    I don't understand. If $m=3$, $k=\varphi=\frac{1+\sqrt{5}}{2}$ and $d_j=(-1)^j$ then $k^j \notin ℚ$ for $0 \le j $, $d_j \in ℚ$, $d_j \neq 0$ but $\sum_{j=1}^{m-1} d_j k^j = \varphi^2 - \varphi = 1 \in ℚ$.2012-06-08
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    @copper.hat, your example violates the precondition that $k^j \notin ℚ$ for $0; You use $m=3$ with $k=\sqrt{2}$, but $k^2=(\sqrt{2})^2=2 \in ℚ$ requires $m \le 2$2012-06-08
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    I see. I though the $\ ^j$ was an index, not a power.2012-06-08
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    @WimC, I think it holds for $k=\left(\sqrt{2}-1\right)$ for any positive $m$. For $m=3$ the sum evaluates to $d_1 \left(\sqrt{2}-1\right) + d_2 \left(3-2 \sqrt{2}\right) = d_1\sqrt{2} - d_1 + 3 d_2 - 2 d_2 \sqrt{2} = \left(3 d_2-d_1\right)+\left(d_1-2d_2\right)\sqrt{2}$, which is irrational. All cases of $k=p^{1/n}\pm 1$ work similarly for any $m$.2012-06-08
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    @Phira, I don't understand your comment. Obviously $0 \in ℚ$, so that would be a counterexample, but I don't know what you mean about the definition of $k$.2012-06-08
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    So $m$ is unconstrained? Just trying to understand the question: why do you have an $m-1$ term (vs. an $m$) in the sum?2012-06-08
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    @Polyergic I set $\sum\dots = 0$ which gives me a value of $k$. If this polynomial is irreducible (apart from the factor $k$), then the powers of $k$ will not be rational. Say, you take the largest zero of $k(k^3+k-1)=0$ as $k$.2012-06-08
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    @Henry, thank you for the counterexample. I forgot that the $k$ I'm working with are not arbitrary irrationals, I'll edit the question to specify the constraints.2012-06-08
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    This was answered (counterexample) in a [very recent](http://math.stackexchange.com/questions/155726/reference-for-linear-independence-of-a-set-of-pairwise-independent-irrationals) answer to a similar question you asked.2012-06-08
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    If $k$ is [transcendental](http://en.wikipedia.org/wiki/Transcendental_number) then your statement is true2012-06-08
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    @Anonymous, $m$ is (at most) the least positive integer such that $k^m \in ℚ$. I picked $0 to isolate one period of the fractional part of $j/m$ power on $r^{j/m}$, wherein each $r^{j/m}=k^j \notin ℚ$.2012-06-09
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    @Phira, are you saying that any irreducible polynomial has nonzero real roots which when raised to any integer power yield an irrational number? If so, counterexample achieved for the original question, but if $k$ is required to be an integer root of a rational (see edit), does that imply that if this polynomial is irreducible it also has no rational solutions on such $k$?2012-06-09
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    @AndréNicolas, Are the questions actually equivalent? I haven't found a source that says so, and I'd be interested to see one. I thought I'd made a mistake in putting the first question in terms of independence, but after having to edit this one I think the bigger mistake was trying to avoid asking a very long question.2012-06-09

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The result that you want can be found in a fair number of places. The one I know is Lang's Algebra. Here is a link. It starts at page $297$, in the Galois theory chapter.

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    That link gives me a page that is "unavailable for viewing," and my library doesn't seem to have that book. I imagine I could find it in another book with a chapter on Galois theory, is there any subheading I should look for?2012-06-09
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    Strange, I just tested link again, worked fine for me. Unfortunately can't lend you my copy of the book!2012-06-09
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    Tried again and I can see it this time. It's over my head enough I can't just read it and understand, but it does look like what I was looking for. Thanks!2012-06-10
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The current question is as follows:

Let $k=(a/b)^{1/m}$, where $a,b,m$ are integers and $m$ is minimal for such a representation of $k$. (This implies that $k^j$ is not rational, for $0.)

$k$ is a root of the polynomial $bx^m-a$.

For most values of $a$ and $b$, this polynomial is irreducible, so there cannot be a polynomial of smaller degree such that $k$ is a root which implies your conclusion.

(In particular, if $a$ is square-free, and different from $0,\pm 1$, then the Eisenstein criterion guarantees that the polynomial is irreducible.)

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    This is very close, but for what values of $a$ and $b$ would the polynomial be reducible? $a$ is not necessarily square-free, for example, if $k = \left({2^4\over 3}\right)^{1\over5}$2012-06-09