Simple question about pointing out my mistake in denesting this radical Could someone please help me spot my mistake here :
I want to denest $\sqrt{1+\sqrt{2}}$, I did the following 
$\sqrt{1+\sqrt{2}}=\sqrt{d}+\sqrt{e}$
Squaring both sides gives $1+\sqrt{2}=d+e+2\sqrt{de}$
Equating radical and no radicals we get 
$1=d+e, \Rightarrow d=1-e$
$\sqrt{2}=2\sqrt{de}\Rightarrow 2=4de \Rightarrow 1/2=de \Rightarrow 1/2=(1-e)e \Rightarrow 1/2=e-e^2 \Rightarrow e^2-e+1/2=0$
Now solving for e we obtain , $e=\tfrac{1^+_-i}{2}$,
which gives $d=\tfrac{1^+_-i}{2}$.
But 1) I read online that when denesting we're always supposed to choose a real e , which is not possible , 2) I don't think d is meant to equal to e is it ?
 A: I think, it's better the following way.
Let $\sqrt{1+\sqrt{2}}=\sqrt{c}+\sqrt{d}$, where $\{c,d\}\subset\mathbb Q$.
Thus, $$1+\sqrt2=c+d+2\sqrt{cd},$$ which gives
$$(1-c-d)^2+2+2(1-c-d)\sqrt2=4cd.$$
Now, if $1-c-d\neq0$ we obtain $\sqrt2\in\mathbb Q,$ which is a contradiction.
But if $1-c-d=0$ we obtain $2=4cd$, which gives $c\not\in\mathbb Q$ and $d\not\in\mathbb Q$, which is a contradiction again.
Id est, denesting is impossible.
A: This is curious. Well, for this to be possible we must have $$d+e=1,\,\, de=1/2.$$
From this we get that $(d-e)^2=-1,$ which proves that no such real $d,e$ as we seek exist.
I say it's curious because given any binomial $\sqrt x+\sqrt y,$ where $x,y\ge 0,$ we have, by squaring, $$x+y+2\sqrt{xy}.$$ It doesn't look like anything stops this from being reversed, yet you just provided one counterexample. I'm curious about knowing what prevents this (I must be missing something obvious). Could someone please explain?

Addendum:

Well, I looked at this a bit more closely, and found that the answer to the question in my third paragraph was already contained in the first.
So, suppose we have the binomial $$\sqrt a+\sqrt b,$$ where $a$ and $b$ are nonnegative rationals, and we want to express its square root as a binomial of the same form. Then if we suppose this to be $\sqrt x +\sqrt y,$ then by squaring the latter and equating to the former, and squaring to remove square roots we obtain the system $$(x+y)^2=a,\,\, 4xy= b.$$
Thus, it must be the case that $(x-y)^2=a-b.$ It follows that a necessary condition for us to be able to meet our goal is that $a-b$ be nonnegative. As explained in the first paragraph of this answer, the binomial $1+\sqrt 2$ proposed by OP doesn't meet this condition, hence cannot be written in the desired form. For necessary and sufficient conditions, we'll have to go in cases and I skip it since it's probably not very relevant here.

