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I understand proving that $\sqrt{7}-\sqrt {2}$ is irrational, but how does the answer change if its cube root of $7$ instead of square root?

the way I solve $\sqrt{7}-\sqrt {2}$ is by assuming its rational, and multiplying both sides by $\sqrt{7}+\sqrt {2}$, and deriving a contradiction

Thank you!

lone student
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user904299
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    Sketch: say it is rational. So $\sqrt 2 = r +\sqrt[3] 7$. Square and derive a contradiction. Or write $\sqrt[3] 7 = s+\sqrt 2$, cube and derive a contradiction (that way is faster). – lulu Apr 16 '21 at 19:27
  • @user904299 Could you expand a bit on the method you know to prove that $\sqrt{7} - \sqrt{2}$ is irrational which would probably allow us to tailor an answer better? – Daniel Schepler Apr 16 '21 at 19:31
  • @lulu is s going to be in the form m/n? (where both are integers, and we go from there) – user904299 Apr 16 '21 at 19:31
  • If you read what the original poster wrote, you can understand that actually he wanted to write $\sqrt[3]7-\sqrt2$, indeed he wrote “how does the answer change if its square root of $7$ instead of cube root?” – Angelo Apr 16 '21 at 19:32
  • Yes, $r,s$ are both rational ($r=-s$, not that it matters). – lulu Apr 16 '21 at 19:33
  • @Angelo The OP is indicating that they know how to handle the case of $\sqrt 7 - \sqrt 2$ (presumably by squaring) and they are asking if the same technique works in this higher degree case. – lulu Apr 16 '21 at 19:34
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    Very similar problem: Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational. – Martin R Apr 16 '21 at 19:37
  • I'm not sure that complex numbers have any relation with your problem @user904299 – PNT Apr 17 '21 at 15:41

3 Answers3

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Let’s go for an elementary proof (no rational root theorem and no field extension)

Assume $r=\sqrt[3]{7}-\sqrt{2}$ is rational. This means $\sqrt[3]{7}=r+\sqrt{2}$. Cubing one gets

$$7=r^3+3r^2\sqrt{2}+6r+2\sqrt{2}$$

This leads to

$$\sqrt{2}={7-r^3-6r\over 3r^2+2}$$

$r$ being rational this means $\sqrt{2}$ is rational. A contradiction

marwalix
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$\mathbb{Q}(\sqrt[3]{7})$ is a degree 3 extension of $\mathbb{Q}$.

Suppose that $\sqrt[3]{7} - \sqrt{2}$ is rational (call it $a$). Then, $\sqrt[3]{7} = \sqrt{2} + a$. Therefore, $\mathbb{Q}(\sqrt[3]{7}) = \mathbb{Q}(\sqrt{2})$.

However, $\mathbb{Q}(\sqrt{2})$ is a degree 2 extension of $\mathbb{Q}$. This is a contradiction, as the same extension cannot be both degree 2 and degree 3.

David Lui
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Let $\alpha=\sqrt[3] 7 -\sqrt 2$.

Then $\alpha$ is a root of $x^6 - 6 x^4 - 14 x^3 + 12 x^2 - 84 x + 41=0$. Since this is a monic polynomial with integer coefficients (the actual polynomial is not important), the rational root theorem tells you that $\alpha$ is either irrational or an integer.

Now

$\quad 1.9 <\sqrt[3] 7 < 2.0 $

$\quad 1.4 < \sqrt 2 < 1.5 $

gives

$\quad 0.4 < \alpha < 0.6 $

which proves that $\alpha$ is not an integer and so must be irrational.

We can avoid these fine estimates: if $\alpha$ is an integer then it must divide $41$ (now the polynomial is important), but clearly $0 < \alpha < \sqrt[3] 7 < 2 $ and $\alpha \ne 1$.

lhf
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