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Assume that $a$ and $b$ are rational numbers with $a\neq b$ such that $\sqrt{a}-\sqrt{b}$ is rational. The each of $\sqrt{a}$ and $\sqrt{b}$ must be rational. Because if $\sqrt{a}-\sqrt{b}$ is rational so it is its reciprocal which is $$\frac{\sqrt{a}+\sqrt{b}}{a-b}.$$ This implies that $\sqrt{a}+\sqrt{b}$ must be rational as must $2{\sqrt a}$ and $2{\sqrt b}$, and hence each of $\sqrt{a}$ and $\sqrt{b}$ must be rational.

I am trying to prove the following assertion:

Assume that $a$ and $b$ are rational numbers such that $\sqrt[3]{a}-\sqrt[3]{b}$ is rational. The each of $\sqrt[3]{a}$ and $\sqrt[3]{b}$ must be rational.

More generally, if $a$ and $b$ are rational numbers such that $\sqrt[n]{a}-\sqrt[n]{b}$ is rational for any positive integer $n$, then each of $\sqrt[n]{a}$ and $\sqrt[n]{b}$ must be rational.

For the case $n=3$, I was able to conclude that since $$a-b =(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b^2})$$ is rational, it must be the case that $$\sqrt[3]{a^2}+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b^2}$$ must be rational. Not sure where to go from here!

student
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  • Maybe a Kummer theory argument can work. – Kenta S Jan 14 '22 at 01:19
  • Sorry! Not sure what that means!! – student Jan 14 '22 at 01:33
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    In case $n$ is prime we have that if $\sqrt[n]{a}-\sqrt[n]{b}=r$ then $a=(r+\sqrt[n]{b})^n$ is an $n-1$th degree equation for $\sqrt[n]{b}$ which is impossible as $x^n-b$ is irreducible if $n$ is prime. A modification can get the case where $n$ is composite. – Rene Schipperus Jan 14 '22 at 01:33
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    I assume we only consider $a \neq b$? Otherwise the statement is trivially false. – VTand Jan 14 '22 at 01:34
  • Sorry! I should assume that $a\neq b$. Am editing the question. – student Jan 14 '22 at 01:36
  • Rene Schipperus, can you explain the logic of your argument, and if I am looking to prove that $x^n - b$ is irreducible would Eisenstein's Irrducibility Criterion/Gauss's Lemma work? But before this I want to understand the logic of your argument. Do I reach a contradiction if I assume $\sqrt[n]{b}$ is irrational? Sorry to miss something surely obvious! – student Jan 14 '22 at 01:40
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    @ReneSchipperus FYI, using Approach0, I found a somewhat similar approach to what you suggest was used in this answer, and also a quite similar answer from the same OP, to prove for positive rationals $a$ and $b$ that if $\sqrt[n]{a}+\sqrt[n]{b}$ is rational, then so is $\sqrt[n]{a}$ and $\sqrt[n]{b}$, – John Omielan Jan 14 '22 at 01:41
  • In the composite case consider the example $\sqrt[6]{9}-\sqrt[6]{8}=\sqrt[3]{3}-\sqrt[2]{2}=r$ gives a quadratic equation for $\sqrt[3]{3}$ – Rene Schipperus Jan 14 '22 at 01:51
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    @JohnOmielan Note that there is an error in the other solution $x^n-a$ may be reducible, $n=6$, $a=9$ or $a=8$. – Rene Schipperus Jan 14 '22 at 01:55
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    Related: Linear independence of roots over $\mathbb Q$. – dxiv Jan 14 '22 at 03:23

1 Answers1

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Easiest is to say the following. Assume $x^k-a$ and $x^l-b$ are irreducible over $\mathbb{Q}$ then $\sqrt[k]{a}-\sqrt[l]{b}$ is irrational, for otherwise if $k\leq l$, $$a=(\sqrt[l]{b}+r)^k$$ is an $k$th degree equation for $\sqrt[l]{b}$, or $k-1$ degree if they are equal.

Note that if $$x^n-a=f(x)g(x)$$ is reducible with the degree of $f$ being $k$ then the roots of $f$ are of the form $\epsilon \sqrt[n]{a}$ and their product, the constant coefficient of the form $\eta(\sqrt[n]{a})^k\in \mathbb{Q}$ which implies that $(\sqrt[n]{a})^k\in \mathbb{Q}$ and so $a=q^{n/k}$. This is one case of a much more general theorem proved by other means.

Rene Schipperus
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