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Is it possible to express $\sqrt[6]{2}$ as rational combination of $1, \sqrt{2}, \sqrt[4]{2}$?

So can we find $a,b,c \in \mathbb{Q}$ such that $\sqrt[6]{2}=a+ b\sqrt{2}+c \sqrt[4]{2}$?

My guess is no and one should try to prove by contradiction. And the contradiction should show that either at least one number from ${a, b, c}$ is irrational or at least one number from ${\sqrt{2}, \sqrt[4]{2}}$ is rational.

However, my attempts failed because they lead to some unmanageable (for me at least) long computations.

Can you please prove some smart way to handle this?

Thanks a lot!

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  • $\begingroup$ This is question about field extensions of the rational numbers. Are you familiar with this subject? $\endgroup$ – Hans Engler Oct 29 '17 at 16:35
  • $\begingroup$ Yes, but I need elementary proof for that or at least scheme of the proof. $\endgroup$ – Hedgehog Oct 29 '17 at 16:36
  • $\begingroup$ If you could, then you could double te cube with an unmarked sraightedge and compasses. Uhhh... no. $\endgroup$ – Oscar Lanzi Oct 29 '17 at 16:40
  • $\begingroup$ @OscarLanzi I hope this problem is simpler since we are much more limited compared with sraightedge and compasses constructions. $\endgroup$ – Hedgehog Oct 29 '17 at 16:46
  • $\begingroup$ Raise everything to the sixth power; then the right hand side is a ${\mathbb Q}$-linear combination of the powers of $\sqrt[4]{2}$. $\endgroup$ – franz lemmermeyer Oct 29 '17 at 16:58
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By Eisenstein's Criterion, you know that $x^6-2$ is irreducible over $\mathbb{Q}$. Observe that $\sqrt[6]{2}$ is a root of this polynomial.

Suppose (for contradiction) that $\sqrt[6]{2}=a+b\sqrt{2}+c\sqrt[4]{2}$ for some rational $a$, $b$, and $c$. Let $y=a+b\sqrt{2}+c\sqrt[4]{2}$. If you look at $y^0,\cdots,y^4$ as vectors in the span of $1$, $\sqrt[4]{2}$, $\sqrt{2}$, and $\sqrt[4]{8}$, then this is a collection of $5$ vectors in a vector space of dimension at most $4$, so there is some nontrivial combination of these which equals zero. This tells you that there is a nonzero polynomial $p$ of degree at most $4$ with rational coefficients with $y$ as a root (since $y^0,\dots,y^4$ is a linear dependent set).

Therefore, $\sqrt[6]{2}$ is a root of both $x^6-2$ and $p$, so it is a root of their gcd (which cannot be $1$ because the gcd needs a root). This, however, is impossible because such a gcd would need to be of degree at most $4$ and at least $1$ and divide $x^6-2$, but $x^6-2$ is irreducible. This is a contradiction.

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  • $\begingroup$ Thanks a lot! It is more delicate than it looks, so I have to check everything carefully. $\endgroup$ – Hedgehog Oct 29 '17 at 17:59
  • $\begingroup$ Yes, I understand the proof now, it is terrific! $\endgroup$ – Hedgehog Oct 29 '17 at 18:12
  • $\begingroup$ The proof by @robjohn is quite slick and it's worth taking the time to understand it as well. $\endgroup$ – Michael Burr Oct 30 '17 at 11:54
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Hint: By Eisenstein's Criterion, $x^{12}-2$ is irreducible over $\mathbb{Q}$. Therefore, there are no $a,b,c,d\in\mathbb{Z}$ so that $ax^6+bx^3+cx^2+d=0$ where $x=2^{1/12}$.

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