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How do I prove that the following is not rational?

$$x=\sqrt 2 + \sqrt[3]2$$

To prove a simpler case like $\sqrt{2}=a/b$, I can raise both sides to the power of 2 and get $a^2=2b^2$, therefore both $a$ and $b$ must be even numbers which can't be true.

Asaf Karagila
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amin
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    This is unreadable. If you are asking about $\sqrt 2 +\sqrt[3] 2$ then that is irrational so you can't prove that it is rational. – lulu Jul 29 '22 at 12:45
  • Sorry, updated the question. – amin Jul 29 '22 at 12:48
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    For algebraic numbers , the most straightford way to decide their rationality is to determine their minimal polynomial. If and only if its degree is $1$ , the given number is rational. – Peter Jul 29 '22 at 12:52
  • For the given $x$, the proof is not so easy as in the mentioned simpler example. – Peter Jul 29 '22 at 12:54
  • "therefore both a and b must be even numbers which can't be true." Why not? (I am asking the OP this) – Adam Rubinson Jul 29 '22 at 13:08
  • Are you allowed to use the precalculus/school-algebra rational root theorem? If so, then a little algebra leads you to a 6th degree polynomial whose only possible rational roots are $\pm 1,$ $\pm 2,$ $\pm 4,$ each of which is is easily seen to not be a root of that 6th degree polynomial. – Dave L. Renfro Jul 29 '22 at 13:12
  • @AdamRubinson we could assume that a/b don't have any common factor, otherwise we would simplify it to a'/b' with common factors removed. – amin Jul 29 '22 at 16:02
  • That’s right. You should have written those details in the question as it highlights where the contradiction arose in the proof if irrationality of $\sqrt{2}.$ – Adam Rubinson Jul 29 '22 at 17:56

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Let $$x=\sqrt 2 + \sqrt[3]2$$ then $$x-\sqrt 2 = \sqrt[3]2$$ $$\left(x-\sqrt 2 \right)^3=2$$ $$x^3-3x^2\left( \sqrt 2 \right)+3x(2)-2\left( \sqrt 2 \right)=2$$ $$\sqrt 2= \frac {x^3+6x-2}{3x^2+2}$$

$x$ cannot be rational because $\frac {x^3+6x-2}{3x^2+2}$ will then be rational and yet $\sqrt 2$ is irrational.

Li Kwok Keung
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