Simplify $\sqrt{1+\sqrt 2}$ When I was computing $\sqrt{1+i}$, $\sqrt{1+\sqrt 2}$ came up and I could not simplify it. So the question is how to simplify $\sqrt{1+\sqrt 2}$, or is it in the most simplified form?
 A: If your aim is to rewrite $\sqrt{1+\sqrt{2}}$ as a linear combination (with rational coefficients) of square roots of integers, it's not possible. 
In some cases you can do it, for instance
$$
\sqrt{8+4\sqrt{3}}=\sqrt{6}+\sqrt{2}
$$
which you can check by squaring both sides. In the case of $\sqrt{1+\sqrt{2}}$ you can't find such an expression.
Suppose $\sqrt{1+\sqrt{2}}=p\sqrt{a}+q\sqrt{b}$, with $a,b,p,q$ rational. Then, after squaring,
$$
1+\sqrt{2}=ap^2+bq^2+2pq\sqrt{ab}
$$
You can easily prove that this implies
$$
\begin{cases}
ap^2+bq^2=1 \\[4px]
2pq\sqrt{ab}=\sqrt{2}
\end{cases}
$$
and the second equation implies $2abp^2q^2=1$. Set $x=ap^2$ and $y=bq^2$, so $x+y=1$ and $2xy=1$. In particular, $x$ and $y$ are the roots of
$$
t^2-t+\frac{1}{2}=0
$$
which has negative discriminant, so no rational solutions.
A: Radical unnesting would be done with
$$\sqrt{1+\sqrt2}=a\sqrt2+b$$
or
$$1+\sqrt2=2a^2+b^2+2\sqrt2ab.$$
By identification,
$$2a^2+b^2=1,2ab=1.$$
Then
$$2a^4+a^2b^2=a^2,\\
2a^4-a^2+\frac14=0,$$
which doesn't have real solutions.
A: If it were possible to simplify it, one might expect it to be written as $$a+b\sqrt{2}$$ where $a$ and $b$ are rational. This is like when you simplify $$\sqrt{4+2\sqrt{3}}=1+\sqrt{3}$$ for example.
However is you set $$(a+b\sqrt{2})^2=1+\sqrt{2}$$
you will get, by comparing rational and irrational parts, a pair of equations $$a^2+2b^2=1$$ and $$2ab=1$$
These equations have no real solution for $a$ and $b$, so you already have the simplest form. 
