Simplify $\sqrt{1+\sqrt 2}$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.