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What's the nice 'trick' to showing that the following expression is irrational?

$2^{1/3} + 2^{2/3}$

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$2-1=(2^{1/3}-1)(2^{2/3}+2^{1/3}+1)$ Now if $2^{1/3}-1$ is irrational then so is $2^{2/3}+2^{1/3}+1$ because a product of a rational and irrational is irratonal, which then implies that $2^{2/3}+2^{1/3}$ is irrational.If you can prove $2^{1/3}-1$ is irrational than that's it

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    $\begingroup$ To prove that $2^{1/3} - 1$ is irrational, it suffices to prove that $2^{1/3}$ is irrational. But that's easy. $\endgroup$ Nov 29, 2016 at 18:07
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Let $x=\sqrt[3]{2}+\sqrt[3]{4}$ then raising to the $3^{rd}$ power and expanding the binomial on the right:

$$x^3 = (\sqrt[3]{2})^3 + 3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{4} \cdot (\sqrt[3]{2}+\sqrt[3]{4}) + (\sqrt[3]{4})^3 = 6 x + 6$$

By the rational root theorem, the equation $x^3-6x-6=0$ can only have divisors of $6$ as rational roots, but it's easily verified that none of the divisors is in fact a root. Therefore $x$ is irrational.

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    $\begingroup$ That was exactly the idea I had in mind. You won me :P $\endgroup$
    – Xam
    Nov 29, 2016 at 18:10
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    $\begingroup$ You can also appeal to Eisenstein's criterion to get that $x^3-6x-6$ is irreducible over $\mathbb{Q}$, and hence has no rational roots. $\endgroup$
    – Ken Duna
    Nov 29, 2016 at 18:13
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    $\begingroup$ @Xammm Thank you. This is a useful technique in many cases like this. $\endgroup$
    – dxiv
    Nov 29, 2016 at 18:23
  • $\begingroup$ @KenDuna Good point. In the case of this very simple polynomial it's also easy to see that $6 | x^3$ therefore $6 | x$ which leads to the contradiction $36 | 6$ (which amounts to applying Eisenstein without calling it so). $\endgroup$
    – dxiv
    Nov 29, 2016 at 18:25
  • $\begingroup$ @dxiv you're right. $\endgroup$
    – Xam
    Nov 29, 2016 at 18:28
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Here's a slightly more 'sledgehammer' approach: since $x^3-2$ is irreducible over the rationals, the minimal polynomial of its root $z=2^{1/3}$ must be of degree three. But if $2^{1/3}+2^{2/3}$ were rational, say $\frac ab$, then that would imply that $z+z^2=\frac ab$, or $bz^2+bz-a=0$, contradicting the minimal-degree statement above.

OTOH, we get a nice prize for all this 'heavy machinery': the exact same argument shows in one fell swoop that $a2^{1/3}+b2^{2/3}$ is irrational for all (nonzero) rational $a,b$; in other words, $2^{1/3}$ and $2^{2/3}$ are linearly independent over $\mathbb{Q}$.

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