# Is sum of roots of 2 always irrational?

Define the sequence $r(k)= \sum_{n=2}^k 2^{\frac{1}{n}}$

Is $r(k)$ irrational for every natural $k\geq 2$?

Yes. Let we assume $k>4$ and let $p$ the greatest prime $\leq k$. Then $2^{\frac{1}{p}}$ is an algebraic number of degree $p$ over $\mathbb{Q}$ and is the only term of the sum with such a property. It follows that $r(k)$ is an algebraic number over $\mathbb{Q}$ with degree $pm\geq p$ and in particular $r(k)\not\in\mathbb{Q}$.

• Are you implicitly using Bertrand's postulate here? Commented Feb 11, 2017 at 22:36
• Would you mind elaborating why the sum has degree $pm\geq p$? Reasoning of this kind seems to apply equally well to $\sqrt{2}+\sqrt[3]{2}+(-\sqrt{2}-\sqrt[3]{2})$ - there is precisely one term of degree $3$, but the sum is not of degree divisible by $3$. Commented Feb 11, 2017 at 22:37
• @Wojowu: if $[\mathbb{Q}(\alpha):\mathbb{Q}]=p$ and $[\mathbb{Q}(\beta):\mathbb{Q}]=q$ with $\gcd(p,q)=1$, then the degree of $\alpha+\beta$ is simply $pq$. Commented Feb 11, 2017 at 22:40
• Ah, so you are actually using that there is only one term with degree divisible by $p$. I see how that helps, thanks. Commented Feb 11, 2017 at 22:42
• @JackD'Aurizio No, it's not. I just think it is worth mentioning. Commented Feb 11, 2017 at 23:08

Yes.

The number $r(k)$ is an element of the splitting field $F$ of the irreducible polynomial $f(X)=X^{N}-2$ where $N=\operatorname{lcm}(1,2,\ldots,k)$). Indeed, if $\alpha$ is the unique positive root of $f$, then we have $$r(k)=\sum_{n=2}^k\alpha^{N/n}$$ Let $p$ be a prime with $\frac k2<p\le k$ (such a prime exists by Bertrand's postulate), and let $\beta=\alpha\cdot e^{2\pi i/p}$. Then $p\mid N$ and so $f(\beta)=0$. Hence there exists an automorphism $\phi$ of $F$ that maps $\alpha\mapsto \beta$. Note that $\beta^{N/n}=\alpha^{N/n}$ for most $n$. Indeed, we have $\beta^{N/n}\ne\alpha^{N/n}$ if and only if $N/n$ is not a multiple of $p$. As $p^2\nmid N$, this is the case precisely for the case $n=p$. It follows that $$\phi(r(k))=r(k)+\beta^{N/p}-\alpha^{N/p}\ne r(k)$$ and therefore $r(k)\notin\Bbb Q$.

Yes, because the field trace of $\mathbf Q(2^{1/2}, 2^{1/3}, \ldots, 2^{1/n})/\mathbf Q$ kills this number, and the only rational number which is killed by the field trace is $0$ (and the number in the OP is clearly nonzero).

• The same argument shows that if $a_k,b_k\in\mathbb Q$ and $\sum a_k^{b_k}\in\mathbb Q$ with each term irrational, then $\sum a_k^{b_k}=0$. Nice! Commented Feb 11, 2017 at 23:23
• If you assume $a_k > 0$ for all $a_k$ and take the real roots, then yes, this holds. (To see why this is necessary, consider that the sum of the conjugates of $\zeta_3$ is $-1$, for instance...) Commented Feb 11, 2017 at 23:25
• Oops, indeed. Or more generally if $|a_k^{b_k}|\notin\mathbb Q$, with arbitrary choices of the roots. Commented Feb 11, 2017 at 23:58
• @barto That doesn't work, because the trace of $(-4)^{1/4}$ is either $2$ or $-2$, depending on your choice of complex root. Commented Feb 12, 2017 at 0:08
• It follows from the transitivity of the trace, and the fact that $X^n - 2$ is irreducible, so that the trace of $2^{1/n}$ in $\mathbf Q(2^{1/n})$ is $0$. Commented Feb 12, 2017 at 6:38