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I am asked to prove that $(3+\sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.

This is what I have so far:

$ x = (3+\sqrt{2})^{2/3} $

$ x^3 = (3+\sqrt{2})^{2} $

$ x^3 - 11 - 6\sqrt{2} = 0 $

From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= \frac{c}{d}$ $c,d, \in \mathbb{Z} $ would need to have $d= \pm 1$ and $c$ divide $-11 -6\sqrt{2}$.

So we have

$ -11 -6\sqrt{2} = z c$, $z \in \mathbb{Z}$

$-6\sqrt{2} = 11 +zc$

Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $\sqrt{3}$ is irrational.

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    $\begingroup$ $(x^3 - 11)^2 = 72$ from your calculations, so $x^6 - 22 x^3 + 49 = 0.$ If this integer sextic has no rational roots, you are done. $\endgroup$ – Will Jagy Nov 30 '18 at 23:08
  • $\begingroup$ Would it count to use the argument: if it were rational, then its cube $(3+\sqrt{2})^2 = 11 + 6\sqrt{2}$ would be rational, implying $\sqrt{2}$ would have to be rational, then apply the rational zeroes theorem to finish from there? $\endgroup$ – Daniel Schepler Nov 30 '18 at 23:09
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From $x^3-11-6\sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $\pm1$, $\pm7$, and $\pm49$. However, none of them is.

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You don't need the RZT and you already did all the hard work: from

$$x^3-11-6\sqrt2=0$$

we can deduce as follows: if $\;x\in\Bbb Q\;$ then also $\;x^3\in\Bbb Q\;$ and thus also $\;x^3-11=6\sqrt2\in\Bbb Q\;$ , from where we get the straightforward contradiction that $\;\sqrt2\in\Bbb Q\;$ ...

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    $\begingroup$ Yes, but the questions asks to use RZT. $\endgroup$ – pmac Nov 30 '18 at 23:16
  • $\begingroup$ @pmac In that case Santos' answer seems to be the most fit, since you need an integer polynomial to apply the RZT. $\endgroup$ – DonAntonio Nov 30 '18 at 23:16
  • $\begingroup$ But $\sqrt{2}$ is a root of $t^2 - 2 \in \Bbb Z[t]$, and RZT shows that polynomial has no rational solutions. $\endgroup$ – Travis Willse Dec 1 '18 at 11:01
  • $\begingroup$ @Travis Yes and yes...so what? I don't understand what you're trying to tell... $\endgroup$ – DonAntonio Dec 1 '18 at 11:21
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Note $(3 + \sqrt 2)^{\frac 23}$ doesn't have to be the only root.

You got that it is a solution to

$x^3 - (11 + 6\sqrt 2) = 0$.

So it is a solution to $(x^3 - (11 + 6\sqrt 2))(x^3 - (11 - 6\sqrt 2)) =$

$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.

Now by the rational root test the only possible rational roots are $\pm 1, \pm 7, \pm 49$ and not of them are roots.

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D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6\sqrt2=0 \iff (x^3 -11) = 6\sqrt 2 \implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.

They both work and have the same result but in terms of ease in seeing and teaching... His is better.

(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)

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  • $\begingroup$ Ah. If only I had factored out the negative sign maybe I would have seen this too. Thanks. $\endgroup$ – pmac Nov 30 '18 at 23:31
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As an alternative, we have that

$$x=(3 + \sqrt{2})^2=13+6\sqrt 2\not \in \mathbb{Q}$$

now suppose by contradiction that

$$\sqrt[3] x =y\in \mathbb{Q} \iff y^3=x$$

which is impossible.

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