# Prove $x = \sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ is irrational

Prove $x = \sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ is irrational.

I can prove that $x$ is irrational by showing that it's a root of a polynomial with integer coefficients and use rational root theorem to deduce it must be either irrational or an integer and then show it's not an integer, therefore, it must be irrational.

I was wondering what are the other methods to prove $x$ is irrational. I'd be very interested in seeing alternative proofs.

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It's not an answer, but to be honest it's the only proof I could come up with. – Patrick Da Silva Dec 6 '11 at 23:47
There's a direct proof assuming the freshman's dream. :-) – msh210 Dec 7 '11 at 0:16
Your method is fine. The hard part seems to be proving that $x$ is not an integer. How do you do that without evaluating $x$? BTW, $x$ is almost an integer: $x \approx 2.00013$. – lhf Dec 7 '11 at 0:44
What polynomial do you think it is not a root of? – Henry Dec 7 '11 at 1:09
@lhf: It is easy to prove $x$ is not an integer: since $0\lt \sqrt{2} \lt \sqrt{3} \lt 2^{99}$ you can find $0 \lt \sqrt[100]{\sqrt{3} - \sqrt{2}} \lt 1 \lt \sqrt[100]{\sqrt{3} + \sqrt{2}} \lt 2$ so $1 \lt x \lt 3$ leaving perhaps the possibility $x=2$. But $x=y + 1/y$ where $y=\sqrt[100]{\sqrt{3} + \sqrt{2}}$, and the only solution of $2=y + 1/y$ is $y=1$ while $\sqrt[100]{\sqrt{3} + \sqrt{2}} \not= 1$. – Henry Dec 7 '11 at 1:15

Let $y = \sqrt[100]{\sqrt{3} + \sqrt{2}}$. Then $x = y + {1 \over y}$. Suppose $x$ were some rational number $q$. Then $y^2 - qy + 1 = 0$. This means ${\mathbb Q}(y)$ is a field extension of ${\mathbb Q}$ of degree two, and every rational function of $y$ is in this field extension. This includes $y^{100} = \sqrt{3} + \sqrt{2}$, and $y^{-100} = \sqrt{3} - \sqrt{2}$. So their sum and difference is also in ${\mathbb Q}(y)$. Hence ${\mathbb Q}(\sqrt{2},\sqrt{3}) \subset {\mathbb Q}(y)$. But ${\mathbb Q}(\sqrt{2},\sqrt{3})$ is an extension of ${\mathbb Q}$ of degree 4, a contradiction.

You can make the above more elementary by showing successive powers of $y$ are always of the form $q_1y + q_2$ with $q_1$ and $q_2$ rational and eventually showing some $q_3\sqrt{2} + q_4\sqrt{3}$ must be rational, a contradiction.

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If $y + y^{-1} = x$, then $y^n + y^{-n}$ is a polynomial $F_n(x)$ with integer coefficients in $x$. In particular, if $x$ is rational, then so is: $$F_{100}(x) = y^{100} + y^{-100} = (\sqrt{3}+\sqrt{2}) + (\sqrt{3}-\sqrt{2}) =2 \sqrt{3}.$$

The polynomials $F_{n}(x)$ are (up to scaling and renormalization) the Chebyshev polynomials. One has $F_0(x) = 2$, $F_1(x) = x$, $F_2(x) = x^2 - 2$, and $F_{n}(x) = x F_{n-1}(x) - F_{n-2}(x)$.

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HINT $\rm\quad \alpha+\bar\alpha,\ \alpha\:\bar\alpha\:\in \mathbb Q\ \Rightarrow\ \alpha^{\:n}+\bar\alpha^{\:n}\in \mathbb Q\$ by induction, since

$$\rm\ \alpha^{\:n+1}\!+\:\bar\alpha^{\:n+1}\ =\ (\alpha+\bar\alpha)\ (\alpha^n\!+\:\bar\alpha^n)-\alpha\:\bar\alpha\ (\alpha^{\:n-1}\!+\:\bar\alpha^{\:n-1})$$

In your case $\rm\ \alpha\:\bar\alpha = 1\in\mathbb Q\$ so $\rm\ \alpha + \bar\alpha\ \in\mathbb Q\ \Rightarrow\ \alpha^{\:100} + \bar\alpha^{\:100} =\ 2\:\sqrt{3}\in\mathbb Q\ \Rightarrow\Leftarrow$