I solved the problem by way of contradiction. Suppose $x = \sqrt{2} + \sqrt[3]{2}$ is rational. Then we have $$2 = (x - \sqrt{2})^3 = x^3 - 3\sqrt{2}x^2 + 6x - 2\sqrt{2} = (x^3 + 6x) - \sqrt{2}(3x^2 - 2)$$ I've already shown that $\sqrt{2}$ is irrational, and it's easy to show a rational number plus an irrational number is irrational and that the product of an irrational and a rational is irrational. Given those facts, we have $(x^3 + 6x) \in \mathbb{Q}$ and $\sqrt{2}(3x^2 - 2) \not\in \mathbb{Q}$, so $$2 = a\in\mathbb{Q} + b\not\in\mathbb{Q} \implies 2 \not\in\mathbb{Q}$$ a contradiction.

However, the book does it differently: it says

Show that $x$ satisfies an equation of the type $$x^6 + a_1x^5 + \ldots + a_6 = 0$$ where $a_1,\ldots,a_6$ are integers; prove that $x$ is then either irrational or an integer.

  1. Is my way correct as well?
  2. I don't understand how the book does it. How does one obtain that equation, and once obtained, how does it prove $x$ is irrational?
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    $\begingroup$ For 2. start with $x-\sqrt 2 =\root 3 \of 2$, then cube both sides. Proceed in the obvious manner to obtain your polynomial. Use the Rational Roots theorem at the end. $\endgroup$ – David Mitra Jun 7 '15 at 23:06
  • $\begingroup$ I did cube both sides, but it only gives you a 3rd order polynomial. The book says to use a 6th order. $\endgroup$ – Elliot Gorokhovsky Jun 7 '15 at 23:07
  • $\begingroup$ You should be able to rewrite it with only a product of a square root with a polynomial on one one side.and a polynomial on the other Square both sides. $\endgroup$ – David Mitra Jun 7 '15 at 23:08
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    $\begingroup$ Your reasoning looks perfectly valid. You show clearly that if $x$ is rational then $\sqrt{2}$ must also be rational. $\endgroup$ – Winther Jun 7 '15 at 23:10
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    $\begingroup$ The proof has a gap: the argument fails if $\,3x^2-2 =0.\ $ Do you see how to fix that? $\endgroup$ – Bill Dubuque Jun 7 '15 at 23:16

You've done half the job for their way: from $$2 = (x - \sqrt{2})^3 = x^3 - 3\sqrt{2}x^2 + 6x - 2\sqrt{2} = (x^3 + 6x) - \sqrt{2}(3x^2 - 2)$$ you deduce:$$(x^3 + 6x-2)^2=2(3x^2+2)^2,$$ whence $$x^6-6x^4-4x^3+12x^2-24x-4=0.$$ By the Rational roots theorem, we know that a rational root has to be an integer, and a divisor of $4$, i. e. it can be only $\pm 1, \pm 2, \pm 4$. Just test them to check none is a root.

  1. Yes.
  2. It is a more general way. We look for an equation written up by integers that $x$ satisfies. The first steps are the same: to arrive to the degree 6 polynomial, put the member with $\sqrt2$ to one side and all the others to the other side in your final equation, and square to get integer coeffiecients.
    If a solution was rational, it would be integer because of the lead coefficient and would divide $a_6$.

Square $\sqrt2\,(3x^2-2)=x^3+6x-2$.

  • $\begingroup$ Your first claim is incorrect - see my comment to the question. $\endgroup$ – Bill Dubuque Jun 7 '15 at 23:21
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    $\begingroup$ Hey, I didn't know that you can hide things here! Thanks for the hint. $\endgroup$ – Martin Brandenburg Jun 7 '15 at 23:25
  • $\begingroup$ @BillDubuque But once it's shown that $3x^2 - 2 \neq 0$, the proof is correct? $\endgroup$ – Elliot Gorokhovsky Jun 7 '15 at 23:27
  • $\begingroup$ @RenéG Yes, since then we infer $\,\sqrt2 = (x^3\!+6x-2)/(3x^2\!-2)\in\Bbb Q,\,$ contradiction. $\endgroup$ – Bill Dubuque Jun 7 '15 at 23:29
  • $\begingroup$ @BillDubuque yes indeed, you are right, $3x^2-2$ is missing. $\endgroup$ – Berci Jun 7 '15 at 23:47

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