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From Galois theory, we know that $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \dots \sqrt{p_k}) : \mathbb{Q}] = 2^k$. Suppose I plug in rational approximations to the square roots, then of course the classical theorem would no longer hold since rationals are linearly dependent over rationals. Is there a way to define some sort of "bounded linear dependence" -- maybe by enforcing magnitude conditions on linear dependence and declaring a set of rational numbers to be linearly independent if no such dependence exists (at this point I'm not even sure if this is well-defined) -- What is the right notion to capture this requirement and has this been investigated before? My gut feeling is that this should have something to do with Diophantine approximation, but since I'm unfamiliar with the area, I don't know where to start looking.

I admit my question is a little vague, but I'd be grateful for any relevant pointers!

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  • $\begingroup$ There are algorithms for searching for linear integer relations among given quantities. en.wikipedia.org/wiki/Integer_relation_algorithm will get you started. You can plug in your rational numbers, and see how high the algorithms have to go to find linear dependencies, and use that as a measure of linear dependence. $\endgroup$
    – Gerry Myerson
    Aug 8, 2019 at 0:24

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For arbitrarily high $q$, each root can be approximated by $p/q$ to within $1/q^2$. (This is Dirichlet's approximation theorem.) So with coefficients of at most $q$ we can establish a linear dependence to accuracy of $1/q$.

E.g. with an accuracy of $.0001$, \begin{align} \sqrt{2}\sim \frac{99}{70},\ \sqrt{3}\sim &\frac{97}{56},\ \sqrt{5}\sim \frac{161}{72}\\ \\ (70)\frac{99}{70} + (56)&\frac{97}{56} - (72)\frac{161}{72} - 35 = 0\\ \\ (70)\sqrt{2} + (56)&\sqrt{3} - (72)\sqrt{5} - 35 \simeq .007 \end{align}

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  • $\begingroup$ Thanks! Shouldn't this somehow depend on the degree of the extension $2^k$? What you say works for rational approximations of arbitrary algebraic numbers. I'm looking for something that is an appropriate version of multiplicative property of field extensions. $\endgroup$
    – Nikhil
    Jul 8, 2019 at 21:35
  • $\begingroup$ Yes, but the more elements in the extension, the easier it is to find small linear dependencies. $\endgroup$
    – user44143
    Jul 8, 2019 at 22:42
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    $\begingroup$ I think one can do better. There is a variant of Dirichlet's Theorem saying for any real numbers $\alpha_1,\dots,\alpha_n$ and any positive integer $M$ there exist integers $x_1,\dots,x_{n+1}$ satisfying $$|\alpha_1x_1+\cdots+\alpha_nx_n+x_{n+1}|<M^{-n}$$ with $1\le|x_i|\le M$ for $1\le i\le n$. $\endgroup$
    – Gerry Myerson
    Jul 9, 2019 at 8:17

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