# Prove $\sqrt{1 + \sqrt[3]{2}}$ is irrational using the theorem about rational roots of a polynomial

I'm having trouble with this specific problem at the moment. The theorem states that if $n/m$ is a rational root of a polynomial with integer coefficients, the leading coefficient is divisible by m and the free coefficient is divisible by n.

Using this theorem, I'm supposed to prove that $\sqrt{1 + \sqrt[3]{2}}$ is irrational. I don't have any idea where to start on this one.

Any help or hints are appreciated.

You want to use the rational root theorem.

Hint: Let $x= \sqrt{1 + \sqrt[3]{2}}$, then, $x^2 = 1+ \sqrt[3]{2}$, so $(x^2-1) = \sqrt[3]{2}$. Hence, $(x^2-1)^3 = 2$.

• So basically, I assume that the number is rational and then I construct a polynomial that has it as it's root the way you did it, just with the 2 subtracted. Then I check all possible n/m pairs of the resulting polynomial and conclude that our number isn't among them? Is this at least close to right? – Luka Horvat Jan 11 '13 at 0:10
• There's no assuming that the number is rational. I'm just forming an equation where $x$ is a root. Since it's hard to guess what equation might work, I showed how to make such an equation. The rest is correct, as in Clayton's solution. – Calvin Lin Jan 11 '13 at 0:32

Using Calvin Lin's hint above, we can expand the polynomial to $$(x^2-1)^3-2=x^6-3x^4+3x^2-3=0.$$ The Rational Root Theorem implies that the only possible rational roots are $\{\pm3,\pm1\}$. Checking these values shows that no roots are rational. By construction of the polynomial, we know in particular that $\sqrt{1+\sqrt[3]{2}}$ is irrational.

Here is a twist on the use of the Rational Root Theorem.

$$\sqrt{1 + \sqrt[3]{2}}$$ is a root of $$(x^2-1)^3-2$$. The Rational Root Theorem implies that rational roots of a monic polynomial with integer coefficients must be integers.

But $$1 < \sqrt[3]{2} < 2$$ implies $$1 < \sqrt 2 < \sqrt{1 + \sqrt[3]{2}} < \sqrt 3 < 2$$ and so $$\sqrt{1 + \sqrt[3]{2}}$$ cannot be an integer.

Note that $x = \sqrt{1 + 2^{\frac{1}{3}}}$ satisfies $(x^2 - 1)^3 - 2 = 0$. One may use the Eisenstein criterion to conclude that this polynomial is irreducible over the rationals. But if $x$ were rational this polynomial would have a linear factor with rational coefficients, so $x$ must be irrational.

If $$x\in\mathbb{Q}$$ then $$x^2-1\in\mathbb{Q}$$, so it suffices to prove $$\sqrt[3]{2}\notin\mathbb{Q}$$. You can do this with the rational root theorem by considering the polynomial $$x^3-2$$.

• BTW the polynomial $x^6-3x^4+3x^2-3$ meets Eisenstein's Criterion. – DanielWainfleet Jan 12 '17 at 1:14