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?
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I don't see any way to simplify it. – 5xum Sep 22 '16 at 10:29
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1What do you mean by simplify? In which sense is it not simple? – Ittay Weiss Sep 22 '16 at 10:32
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@IttayWeiss I just mean I don't want to see a square root inside another square root. Is there a more universal definition of simplified form? – Zirui Wang Sep 22 '16 at 11:35
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See also: Denesting radicals at Wikipedia, Strategies to denest nested radicals (and the posts linked there). – Martin Sleziak Apr 20 '18 at 11:07
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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.
egreg
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Is $p\sqrt a+q\sqrt b$ of the most general form? Why is $p\sqrt a+q\sqrt b+r\sqrt c$ not possible? – Zirui Wang Sep 22 '16 at 11:50
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@ZiruiWang I wanted to keep it simple. With advanced tools it can be shown that also your more general form isn't possible. – egreg Sep 22 '16 at 12:46
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@egreg: I know that, but what is the easiest method you have in mind? $\sqrt{1+\sqrt{2}}$ is in a quartic extension of the rationals, but it is also possible for $p\sqrt{a}+q\sqrt{b}+r\sqrt{c}$ to be, despite $a,b,c$ being square-free and distinct, just that that would mean $abc$ is a square. – user21820 Dec 07 '17 at 04:35
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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.
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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.
David Quinn
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