I am trying to prove that $\,\sqrt[3]{2} \cdot (\sqrt[3]{2} + q)\,$ is irrational for all rational choice of $q$. However, I am completely stuck. I tried cubing it and trying the trick that you use for $\,\sqrt{2}\,$ but I can't get it.

Any suggestions?


  • $\begingroup$ Fixed the formatting for you this time around, please see the MathJax primer here. $\endgroup$ – dxiv Jan 25 '18 at 3:36
  • $\begingroup$ First prove it $\sqrt [3]2$ first. An irrational plus a rational is irrational (why) so the rest is easy. $\endgroup$ – fleablood Jan 25 '18 at 4:01

Hint:   let $\,x=\sqrt[3]{2}\,$ and suppose that $\,x(x+q)=p\,$ for some rational $\,p\,$. Then:

$$x^2 + qx - p = 0 \tag{1}$$

Multiplying by $\,x\,$ and using that $\,x^3=2\,$:

$$2 + qx^2 - px = 0 \tag{2}$$

Eliminating $\,x^2\,$ between $\,(1)\,$ and $\,(2)\,$ gives $\,(q^2+p)x - pq-2 = 0\,$, but $\,x\,$ is irrational, so $\dots$

  • $\begingroup$ Our solutions are essentially the same. $\endgroup$ – marty cohen Jan 25 '18 at 3:45
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    $\begingroup$ @martycohen Indeed. The above generalizes a bit more easily to higher degrees e.g. $\sqrt[4]{2}+a \sqrt[4]{4} + b \sqrt[4]{8}\,$. $\endgroup$ – dxiv Jan 25 '18 at 3:52
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    $\begingroup$ Also, any $n$ such that $n^{1/3}$ is irrational. Probably even wider classes. $\endgroup$ – marty cohen Jan 25 '18 at 3:54

Let $t = 2^{1/3}$.

If $t * [t + q] $ is rational, say $r$, then $r =t^2+tq $ so $r+q^2/4 =t^2+tq+q^2/4 =(t+q/2)^2 $ so $t =\sqrt{r+q^2/4}-q/2 $.

Cubing, letting $r+q^2/4 = s$,

$\begin{array}\\ 2 &=((s^{1/2}-q/2)^3\\ &=s^{3/2}-3s(q/2)+3s^{1/2}(q/2)^2-(q/2)^3\\ &=s^{1/2}(s+3(q/2)^2)-3s(q/2)-(q/2)^3\\ \end{array} $

Therefore $s^{1/2}$ is rational, so $2^{1/3}=t = \sqrt{s}-q/2$ is also rational. But it ain't.


Let $\sqrt[3]2\left(\sqrt[3]2+q\right)=r,$ where $r\in\mathbb Q$.

Thus, $$\sqrt[3]4+q\sqrt[3]2-r=0$$ or since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain: $$4+2q^3-r^3+6qr=0.$$

Now, let $q=\frac{m}{n}$ and $r=\frac{k}{n},$ where $n\in\mathbb N$ and $\{m,k\}\subset\mathbb Z$.

Thus, we obtain: $$4n^3+2m^3-k^3+6mnk=0,$$ which get a contradiction by infinite descent.

Indeed, we see that $k$ is divided by $2$, which gives $k=2k_1$, where $k_1\in\mathbb Z$.

Thus, $$2n^3+m^3-4k_1^3+6mnk_1=0,$$ which gives $m$ is divided by $2$ and we can assume $m=2m_1$, where $m_1\in\mathbb Z$.

Thus, $$n^3+4m_1^3-2k_1^3+6m_1nk_1=0.$$ Now, we see that $n=2n_1$, where $n_1\in\mathbb N$ and we obtain: $$4n_1^3+2m_1^3-k_1^3+6m_1n_1k_1=0,$$ which is the same equation with the solution $(m_1,n_1,k_1),$ where $n_1<n$.

Id est, we can make this thing again and again and we'll get infinite sequence of natural numbers: $$n>n_1>n_2>...,$$ which is impossible.


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