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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. |
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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, $(x^2-1)^3 = 2$. |
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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. |
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