This is an exercise from R. Courant's book: How to prove $\sqrt3 + \sqrt[3]{2}$ is a irrational number?

The solution is to construct a equation to prove, but is there any other method to prove this, like by contradiction?

  • $\begingroup$ @Steven. You might as well repost this question ,because everybody has changed $\sqrt{3}$ into $\sqrt{2}$ $\endgroup$ – imranfat May 14 '16 at 16:27
  • $\begingroup$ @imranfat Sorry, I fixed it. For some strange reason, all of us were too influenced by Winther's comment. $\endgroup$ – Aritra Das May 14 '16 at 16:28
  • $\begingroup$ Sorry about that guys:) It should have read 'consider $(r-\sqrt{3})^3$' and trying to show that it implies $\sqrt{3}$ is irrational. I'll remove the comment to avoid confusion. $\endgroup$ – Winther May 14 '16 at 16:40
  • $\begingroup$ Coool. It didn't change the proving methodology. $\endgroup$ – imranfat May 14 '16 at 16:48
  • $\begingroup$ Related: math.stackexchange.com/questions/1316335 $\endgroup$ – Watson Nov 23 '18 at 14:21

Developing the hint by Winther. Assume it is rational $r=\sqrt{3}+\sqrt[3]{2}$. So we have


Regrouping and dividing by $3r^2+3\neq 0$ we get

$$\sqrt{3}={r^3+9r-2\over 3r^2+3}$$

This means $\sqrt{3}$ is rational. Contradiction


If you're familiar with field extensions, you could do it like this:

  • The extension $\mathbb{Q}(\sqrt{3})/\mathbb{Q}$ has degree two (minimal polynomial $X^2-2$)
  • The extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ has degree three (minimal polynomial $X^3-2$)

This implies (by considering field towers) that the extension $\mathbb{Q}(\sqrt{3},\sqrt[3]{2})/\mathbb{Q}$ has degree 6, with basis $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2,\sqrt{3},\sqrt{3}.\sqrt[3]{2},\sqrt{3}.(\sqrt[3]{2})^2\}$. So in particular, $\sqrt{3}+\sqrt[3]{2}$ is written out as an element of this basis and is not in $\mathbb{Q}$.

This might not make any sense to you (maybe one day it will), but you asked for another method, and this is a very general one.


Note that the intersection $ K = \mathbb{Q}(\sqrt[3]{2}) \cap \mathbb{Q}(\sqrt{3}) $ is a subfield of both fields, so it has degree 1 over $ \mathbb{Q} $ (since $ 2, 3 $ are coprime) and $ K = \mathbb{Q} $. If $ \sqrt{3} + \sqrt[3]{2} \in \mathbb{Q} $, then we would have $ \sqrt{3} \in K = \mathbb{Q} $, impossible.

Alternatively, consider $ \mathbb{Q}(\sqrt{3}) $ and note that the $\mathbb{Q}$-automorphism $ \sigma : \sqrt{3} \to -\sqrt{3} $ extends to an automorphism $ \varphi $ of the splitting field of $ X^3 - 2 $ over $ \mathbb{Q}(\sqrt{3}) $. It is then easily seen that $ \sqrt{3} + \sqrt[3]{2} = q $ cannot be rational, as $ \varphi $ can only map $ \sqrt[3]{2} $ to itself (otherwise we have that $\zeta_3$ is real) which implies that $ -\sqrt{3} + \sqrt[3]{2} = q $ as well, or $ \sqrt{3} = 0 $, impossible.

  • $\begingroup$ Your first solution is very simple, nice job! About your second one: do you think any automorphism of the splitting field of $X^3-2$ must fix $\sqrt[3]{2}$? Because I think this is not true... $\endgroup$ – M. Van May 14 '16 at 16:56
  • 1
    $\begingroup$ It must fix it if $ \sqrt{3} + \sqrt[3]{2} $ is rational, because otherwise it would map it to a conjugate (a multiple by a primitive root of unity) and we would be able to isolate the root of unity in the resulting equation as a real number. $\endgroup$ – Starfall May 14 '16 at 16:57
  • $\begingroup$ So, you mean, if $\sqrt{3} + \sqrt[3]{2}$ was rational, then $-\sqrt{3}+\zeta\sqrt[3]{2} = \sqrt{3}+\sqrt[3]{2}$, from which one sees $\zeta$ is real, so $\zeta=1$, and we arrive at $\sqrt{3} = 0$? $\endgroup$ – M. Van May 14 '16 at 17:09
  • $\begingroup$ Precisely. Now I will type more stuff so SE lets me post. $\endgroup$ – Starfall May 14 '16 at 17:10

Assume that it is rational. Hence, $r=\sqrt{3} + \sqrt[3]{2}$ where $r = \frac{p}{q}$ for integral $p,\ q$ such that $q\not = 0$ and $p, \ q$ are coprime.

Hence, $$(r-\sqrt3)^3=2$$ $$\implies r^3-3\sqrt3-3\sqrt3r(r-\sqrt3) = 2 \\ \implies r^3-3\sqrt{3} -3\sqrt3r^2+9r=2 \\ \implies (r^3+9r) +\sqrt3(-3-3r^2)=2 $$ Since $r$ is rational, $r^3+9r$ has to be rational and $\sqrt3(-3-3r^2)$ has to be irrational. Since $2$ is purely rational, $$\sqrt3(-3-3r^2)=0 \\ \implies 3r^2=-3 $$

Hence, we arrive at a contradiction. Thus, $r$ can not be rational. Hence, it is irrational.

  • $\begingroup$ I got your idea, but it seems that we cannot go from $(r^3+9r) + \sqrt{3}(-3-3r^2) = 2$ to $\sqrt{3}(-3-3r^2) = 0$, thanks anyway :) $\endgroup$ – Steven Liu May 15 '16 at 12:52
  • $\begingroup$ @StevenLiu why not ?? It is perfectly logical. $\endgroup$ – Aritra Das May 15 '16 at 12:58
  • $\begingroup$ We can go from $(r^3 + 9r) + \sqrt{3}(-3-3r^2) = 2$ to $\sqrt{3}(-3-3^2) = 2 - (r^3 + 9r)$, but can we say $2 - (r^3 + 9r) = 0$? $\endgroup$ – Steven Liu May 16 '16 at 16:28
  • $\begingroup$ @StevenLiu Yes of course since if 2 numbers are equal, their rational parts must be equal and their irrational parts must be equal. $\endgroup$ – Aritra Das May 16 '16 at 16:49

I do not know whether you know any Galois theory, but if you do: consider the splitting field over $\mathbb{Q}$ of $(X^2-3)(X^3-2)$, call this field $\Omega$. $\Omega/\mathbb{Q}$ is Galois and if $\zeta$ is a primitive root of unity of order 3, then there is a non-identity element $ \sigma \in \text{Gal}(\Omega/\mathbb{Q})$ that does not fix $\sqrt{3}+\sqrt[3]{2}$, namely take $\sigma$ the isomorphism $\Omega=\mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2 \sqrt[3]{2})(\sqrt{3}) \cong \mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2\sqrt[3]{2})(-\sqrt{3})= \Omega$ that fixes $\mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}, \zeta^2 \sqrt[3]{2})$ and maps $\sqrt{3}$ to $-\sqrt{3}$. By the fundamental theorem of Galois theory, $\alpha \in \Omega$ is in $\mathbb{Q}$ if and only if it is fixed by all elements of $\text{Gal}(\Omega/\mathbb{Q})$, but $\sigma(\sqrt{3}+\sqrt[3]{2}) = -\sqrt{3} + \sqrt[3]{2}$. It remains to show of course, that $\sqrt{3} \notin \mathbb{Q}(\sqrt[3]{2}, \zeta \sqrt[3]{2}, \zeta^2 \sqrt[3]{2})$. This follows from the fact that if it did, $S_3$ would have a normal subgroup of order 2, because $\text{Gal}(\mathbb{Q}(\sqrt[3]{2}, \zeta \sqrt[3]{2}, \zeta^2 \sqrt[3]{2})/\mathbb{Q}) \cong S_3$ and $\mathbb{Q}(\sqrt{3})/\mathbb{Q}$ is Galois, which is not the case.

If you do not know any Galois theory, I would like you to know that Galois theory makes a lot of these kind of questions easier to answer, and I would encourage you to learn it!

  • $\begingroup$ That automorphism does not fix $ \mathbb{Q}(\sqrt{3}, \zeta \sqrt[3]{2}) $. $\endgroup$ – Starfall May 14 '16 at 16:39
  • $\begingroup$ You are right. I better take the one that maps $\sqrt{3}$ to $-\sqrt{3}$, I will edit it now :) $\endgroup$ – M. Van May 14 '16 at 16:46
  • $\begingroup$ I think the way I did it is cleaner - you might want to take a look :) $\endgroup$ – Starfall May 14 '16 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.