# How to prove $\sqrt3 + \sqrt[3]{2}$ is a irrational number?

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?

• @Steven. You might as well repost this question ,because everybody has changed $\sqrt{3}$ into $\sqrt{2}$ – imranfat May 14 '16 at 16:27
• @imranfat Sorry, I fixed it. For some strange reason, all of us were too influenced by Winther's comment. – Aritra Das May 14 '16 at 16:28
• 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. – Winther May 14 '16 at 16:40
• Coool. It didn't change the proving methodology. – imranfat May 14 '16 at 16:48
• – 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

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

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.

• 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... – M. Van May 14 '16 at 16:56
• 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. – Starfall May 14 '16 at 16:57
• 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 $\sqrt{3} = 0$? – M. Van May 14 '16 at 17:09
• Precisely. Now I will type more stuff so SE lets me post. – 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.

• 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 :) – Steven Liu May 15 '16 at 12:52
• @StevenLiu why not ?? It is perfectly logical. – Aritra Das May 15 '16 at 12:58
• 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 $2 - (r^3 + 9r) = 0$? – Steven Liu May 16 '16 at 16:28
• @StevenLiu Yes of course since if 2 numbers are equal, their rational parts must be equal and their irrational parts must be equal. – 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!

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