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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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    @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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    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