# For what real $t$ is $\{n^t : n \geq 1\}$ linearly independent over $\mathbb{Q}$?

It's straightforward that $t$ must be irrational. I have googled many variations of this question and browsed through some books on transcendental number theory. There is much that is said about when the base is the same, but not for when the power is the same like here.

For example, does $-\sqrt 2$ satisfy this? I would be happy even with some buzz words about what machinery covers this question, or whether it is part of some grand and long-term conjecture in transcendental number theory now far beyond our powers.

• Has to be linearly. $4^t = (2^t)^2$. – Robert Israel Apr 27 '15 at 16:56
• Yes, you’re right, it can’t be algebraic. Though I gather $\{n^t:n\ge1\}$ is linearly independent iff $\{p^t:p\text{ prime}\}$ is algebraically independent. – Emil Jeřábek Apr 27 '15 at 17:00
• There are countably many possible relations, which each rule out a set of measure $0$, so the set of reals for which the collection is linearly independent has full measure. – Douglas Zare Apr 27 '15 at 19:38
• @DouglasZare In fact each relation rules out a countable set, because any linear combinations of $n^t$ is a nonconstant analytic function, so has finitely many zeros in an interval. – Will Sawin Apr 28 '15 at 0:36
• Better yet, each relation is a generalized polynomial in $e^t$, hence it rules out a finite set. – Emil Jeřábek Apr 28 '15 at 10:56

Robert Israel's guess that this is true for $t$ an algebraic irrational does actually follow from Schanuel's conjecture.

Apply Schanuel's conjecture to $\log p$ and $t \log p$ for $n$ different primes $p$. These are linearly independent unless $t$ is the ratio of the logarithms of two integers, which it isn't because $t$ is an algebraic irrational.

So we get the transcendence degree of the following field is $2n$:

$$\mathbb Q( \log p_1,\dots \log p_n, t \log p_1, \dots, t \log p_n, p_1, \dots, p_n, p_1^t, \dots, p_n^t)$$

Now $p_1,\dots, p_n$ are in $\mathbb Q$ and $t \log p_1, \dots , t \log p_n$ are algebraic over, $\log p_1, \dots, \log p_n$, so the transcendence degree of this field is also $2n$:

$$\mathbb Q( \log p_1,\dots \log p_n, p_1^t, \dots, p_n^t)$$

Hence all these numbers are algebraically independent. If primes raised to the power $t$ are algebraically independent, it implies that natural numbers to the power $t$ are linearly independent over $\mathbb Q$ by factoring natural numbers into primes.

This is a special case of the general principle that everything follows from Schanuel's conjecture.

• +1 for the general principle. – Wojowu May 20 '15 at 18:20

I would guess that this is true if $t$ is an algebraic irrational. It would be a generalization of the Gelfond-Schneider theorem; I don't know if this particular generalization, or something that implies it, has been published as a theorem or a conjecture. I don't think it follows from Schanuel's conjecture, but I could be wrong there.

Note that if $c_1 ,\ldots, c_n$ are rational numbers with $c_1 c_n < 0$ and $m_1 < m_2 < \ldots < m_n$ are positive integers, there is some real $t$ such that $\sum_{j=1}^n c_j m_j^t = 0$. It seems likely that such $t$ will be transcendental if it's not rational, but I don't know if there's a proof even in the case $n=3$ (it is true by Gelfond-Schneider if $n=2$).