Unnesting $\sqrt{1+\sqrt2}$

Question

I feel like this is an extremely simple question but somehow I cannot solve it.

Is it possible to unnest $$\sqrt{1+\sqrt2}\quad?$$ This was motivated by the Nested Radical Constant.

If we let $x^2=1+\sqrt2\implies x^4-2x^2-1=0$ and after solving with substitution $u=x^2$, the argument is circular.

What do you mean by "exact value"? Since the number is not periodic, it is hard write it down completely. You can, however, find a decimal approximation. Wolframalpha, for example, returns 1.553773974030037307344158953063146948164583499410307836332... — YukiJ, Apr 20 '18 at 07:31
I would say that $\sqrt{1+\sqrt2}$ is the exact value. Maybe they want you to simplify it somewhat, but I don't see how that should be done. It isn't the square root or fourth root of any rational number. Maybe it's the sum of two rational square roots, and that's what they're after? — Arthur, Apr 20 '18 at 07:31
@YukiJ Simply put, I would like it to be expressed as a non-nested radical. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Apr 20 '18 at 07:32
@Arthur Just something of the form $\text{rational number} + \text{finite sum of radicals}$ — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Apr 20 '18 at 07:35
Not nested radical :$\sqrt{1+\sqrt{2}}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{ 4+\cfrac{1}{6+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cdots}}}}}}}}$ — Alex D, Apr 20 '18 at 07:37
@TheSimpliFire Of course not, it's not a rational number after all. However, it does get periodic. — Arthur, Apr 20 '18 at 07:47
The title with "exact value" is misleading. You should explicitly speak of unnesting. — , Apr 20 '18 at 07:59
See also: Simplify $\sqrt{1+\sqrt 2}$. You could also have a look at: Denesting radicals at Wikipedia, Strategies to denest nested radicals (and the posts linked there). — Martin Sleziak, Apr 20 '18 at 11:06
There is a complex denesting: $\sqrt{1+\sqrt2}=\sqrt{1-i }/\sqrt2+1/\sqrt{1-i}$ — Michael Hoppe, Apr 20 '18 at 15:26
@TheSimpliFire: How do you get $x^2=1+\sqrt2\implies x^4-2x^2-1=0$? Isn't it $x^4 = 3 + 2\sqrt(2)$? — Jim, Apr 04 '22 at 21:53
@Jim $x^2=1+\sqrt2\implies x^2-1=\sqrt2\implies(x^2-1)^2=2\implies x^4-2x^2+1=2$ — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Apr 05 '22 at 08:36

score 4 · Accepted Answer · 2018-04-20T08:08:14.667

Assume that you want to unnest under the form

$$a+b\sqrt c$$ where $a,b,c$ are rationals.

Then

$$\left(a+b\sqrt c\right)^2=a^2+b^2c+2ab\sqrt c=1+\sqrt2.$$

The only way to achieve this is by having $c$ equal to twice a perfect square, let $c=2d^2$.

Now by identification

$$a^2+2b^2d^2=1,\\2abd=1$$ or, with $e:=bd$, $$a^2+2e^2=1,\\4a^2e^2=1.$$

By eliminating $e$,

$$4a^4-4a^2+2=0,$$ which has no real solution.

This proves that no unnesting of the given form is possible. (But we can't exclude the possibility of other forms...)

If we try solutions that are sums of a rational and a finite number of square roots of rationals, the square of such a number is also a sum of rationals and square roots of rationals.

But as the square roots of non-perfect-squares are linearly independent, it won't be possible to cancel out the square roots that have factors different from $2$, and these forms are also excluded.

Unnesting $\sqrt{1+\sqrt2}$

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