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I am solving a math problem for fun and it amounts to proving that a specified (finite) set of numbers, each defined by an infinite series involving polynomials and the factorials, is a rationally independent set. Ie. if the numbers are $a_1 \dots a_n$, then proving that if $r_1, \dots r_n \in \mathbb{Q}$ and $\sum_{i=1}^n r_i a_i = 0$, then $r_i = 0 \forall i$.

I'm not asking for a solution to this problem (nor going to tell you what this problem is!) but I would like to learn more about some methods (whether specialized or general) for proving that a given set of numbers is rationally-independent. For this problem, I think examining continued fractions of the $a_i$ might be fruitful and worth pursuing, so some discussion of continued fraction methods would definitely be appreciated - but all in all I would like to learn about methods of proving some numbers are rationally independent. Eg. A naive proof that $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \sqrt{11}, \sqrt{13}$ are rationally independent by squaring would still result in three radicals on each side of the equation...

Faraz Masroor
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  • "Ie. if the numbers are $a_1 \dots a_n$, then proving that if $r_1, \dots r_n \in \mathbb{Q}$ and $\sum_{i=1}^n r_i a_i = 0$, then $r_i = 0 \forall i$ " : can you better explain this affirmation ? – G Cab Jun 08 '22 at 16:42
  • Umm, not really. But here's an example: if r1, r2, r3 are rational numbers and r1sqrt(2) + r2sqrt(3) + r3*sqrt(5) = 0, then r1=r2=r3=0. (r1,r2,r3 being rational here is key - of course r1 = -sqrt(3), r2 = sqrt(2), r3 = 0 works but those arent rational numbers) – Faraz Masroor Jun 09 '22 at 02:05

3 Answers3

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When the numbers in question are algebraic, the theory of algebraic field extensions and Galois theory (etc) can come in handy. I'm going to assume familiarity with abstract algebra in my response, but if there are any details you'd like me to expound on let me know and I can.

To give a concrete example, one can show that the set $\{\sqrt{n}:\text{$n$ is a squarefree integer}\}$ is linearly independent over $\mathbb Q$ (which greatly generalizes your six square root example). To see this, we need to establish some tools:

If $K\supseteq F$ is a Galois field extension of finite degree, then let $\text{Aut}_F(K)$ denote the group of field automorphisms of $K$ fixing $F$. Given such an automorphism group (call it $G$ for brevity), we can define the corresponding trace via $T(\alpha)=\tfrac{1}{|G|}\sum_{g\in G} g(\alpha)$, i.e. the average of the orbit of $\alpha$ with respect to $G$. One can easily verify that the trace is a linear operator and fixes elements of the base field $F$.

Claim: If $\alpha$ is a square root of a rational number that is not itself rational and $K\supseteq \mathbb Q$ is a Galois field extension of finite degree with $\alpha\in K$, then the corresponding trace satisfies $T(\alpha)=0$.

Sketch of proof: The minimal polynomial of $\alpha$ is $x^2-\alpha^2$, which in particular has $\pm \alpha$ as roots. Since the extension is Galois, elements of the automorphism group send $\alpha$ to each of $\pm \alpha$ equally often, for an overall sum of $0$.

With this in hand, we're ready to prove that the set $S=\{\sqrt{n}:\text{$n$ is a squarefree integer}\}$ is linearly independent over $\mathbb Q$. To do this, suppose we have a finite linear combination $0=\sum_{i=1}^n c_i \sqrt{b_i}$ where each $c_i\in \mathbb Q$ and each $\sqrt{b_i}\in S$ (and hence each $b_i\in \mathbb Q$). Multiplying everything by $\sqrt{b_1}$ gives $0=c_1b_1+\sum_{i=2}^n c_i\sqrt{b_ib_1}$. One can easily check each of these latter terms remains irrational, so taking the corresponding trace gives $0=c_1b_1$, so since $b_1\neq 0$, we know $c_1=0$. Applying the same argument yields that all of the $c_i$ are $0$, as desired.

When the numbers in question are transcendental... good luck (e.g. it's unknown whether or not $e+\pi$ is rational).

Connor Gordon
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  • Unfortunately, none of these numbers are algebraic! They are simply infinite sums - Wolfram alpha says these may be values of hypergeometric functions but I am not fluent with those at all. – Faraz Masroor May 31 '22 at 02:24
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A dynamical approach would be to consider the linear flow on the $n$-torus $\mathbb{T}^n$ along the vector field $a=(a_1,a_2,...,a_n)$. The components of $a$ are rationally independent iff the orbit of $0$ (or any other point) is dense in $\mathbb{T}^n$. (see Dynamics on the torus )

Of course in practice this perspective may not be as useful to verify the rational independence of a given tuple, especially if $n$ is large. Though it seems to me one could still computationally visualize projections of orbits to subtori of $\mathbb{T}^n$ to make a guess.

Alp Uzman
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(too long for a comment)

So, given the vector $$ \overline {\bf a} = \left( {\sqrt 2 ,\sqrt 3 ,\sqrt 5 , \cdots } \right) $$ (the overbar indicates the transpose)
there is not a (fully) rational vector $\bf r$ orthogonal to $\bf a$, i.e. such that $$ \overline {\bf a} \cdot {\bf r} = 0 $$

Now, instead, putting that $\bf a , \bf r$ are vectors with all their components rational,
you want to find:

  • a (set) of ${\bf r}_k$ which are/are not orthogonal to $\bf a$ ?, or
  • an $\bf a$ which is not orthogonal to a given set $\{ {\bf r}_k \}$ ?
G Cab
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