Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?

Question

Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?

We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is \begin{align*} 2ab &= 1 \\ a^2 + 2b^2 &= 1 \end{align*} We write $b = \frac{1}{2a}$ and substitute this in the second equation. \begin{align*} a^2 + 2\left(\frac{1}{2a}\right)^2 &= 1 \\ a^2 + \frac{1}{2a^2} &= 1 \\ 2a^4 - 2a^2 + 1 &= 0 \end{align*} Let $z = a^2$, so $z^2 = a^4$. The equation is then \begin{equation} 2z^2 - 2z + 1 = 0 \end{equation} Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$. We then find $b$ using $a$ in our original system of equations. \begin{align*} \frac{1 \pm i}{2} + 2b^2 &= 1 \\ 1 \pm i + 4b^2 &= 2 \\ \pm i + 4b^2 &= 1 \\ 4b^2 &= 1 \pm i \\ 2b &= \sqrt{1 \pm i} \\ b &= \frac{\sqrt{1 \pm i}}{2} \\ \end{align*}

Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions. What do I need to reconsider? How can I improve my answer?

Answer 1

The truth is that $\sqrt{1+\sqrt2}$ cannot be expressed as $a+b\sqrt2$ where $a,b$ are integers or rationals. If they were reals then the problem becomes trivial.

(Another way to see the first fact above is that the minimal polynomial of $\sqrt{1+\sqrt2}$ is degree-$4$, whereas if it were expressible as $a+b\sqrt2$ with $a,b\in\mathbb Q$ the minimal polynomial would only be quadratic.)

Answer 2

Your start of the question Find $\sqrt{1+ \sqrt 2}$ is misleading.

You're supposing that $\alpha = \sqrt{1+ \sqrt 2}$ belongs to the field extension $\mathbb Q(\sqrt 2) / \mathbb Q$... And you proved that it's not the case.

In fact $\alpha$ is an element of degree $2$ over $\mathbb Q(\sqrt 2)$ because $\alpha$ is a root of the polynomial $p \in \mathbb Q(\sqrt 2) [x]$

$$p(x) = x^2 - (1 +\sqrt 2).$$

And $p$ is irreducible over $\mathbb Q(\sqrt 2)$.

Answer 3

With a couple of corrections, your answer seems consistent:

$z=\dfrac{1\pm i}2\implies a=\color{red}\pm\sqrt{\dfrac{1\pm i}2}$

$\pm i+4b^2=1\implies 4b^2=1\color{red}\mp i\implies b=\color{red}\pm\dfrac{\sqrt{1\color{red}\mp i}}{2}$

Answer 4

Note \begin{align} \sqrt{\sqrt{2}+1}&=\sqrt{(\sqrt2-1)(\sqrt2+1)} \cdot\sqrt{\sqrt{2}+1} = \sqrt{\sqrt{2}-1} (1+\sqrt2)\\ \end{align} with $a=b = \sqrt{\sqrt{2}-1}$, thus no rational simplification.