Definitions

$sqrt\left\{5-2sqrt\left\{5\right\} \right\}$

which arises in discussing the regular pentagon,

$sqrt\left\{5+2sqrt\left\{6\right\} \right\},$

or more complicated ones such as

$sqrt\left[3\right]\left\{2+sqrt\left\{3\right\}+sqrt\left[3\right]\left\{4\right\} \right\}.$

Denesting these radicals is generally considered a difficult problem. A special class of nested radical can be denested by assuming it denests into a sum of two surds:

$sqrt\left\{a+b sqrt\left\{c\right\} \right\} = sqrt\left\{d\right\}+sqrt\left\{e\right\},$

$a+b sqrt\left\{c\right\} = d + e + 2 sqrt\left\{de\right\};$

this can be solved by the quadratic formula and by setting rational and irrational parts on both sides of the equation equal to each other.

In some cases, higher-power radicals may be needed to denest certain classes of nested radicals.

### Square roots

Under certain conditions infinitely nested square roots such as

$x = sqrt\left\{2+sqrt\left\{2+sqrt\left\{2+sqrt\left\{2+cdots\right\}\right\}\right\}\right\}$

represent rational numbers. This rational number can be found by realizing that x also appears under the radical sign, which gives the equation

$x = sqrt\left\{2+x\right\}.$

If we solve this equation, we find that x = 2 (the second solution x = −1 doesn't apply, under the convention that the positive square root is meant). This approach can also be used to show that generally, if n > 0, then:

$sqrt\left\{n+sqrt\left\{n+sqrt\left\{n+sqrt\left\{n+cdots\right\}\right\}\right\}\right\} = frac\left\{1 + sqrt \left\{1+4n\right\}\right\}\left\{2\right\}.$

The same procedure also works to get

$sqrt\left\{n-sqrt\left\{n-sqrt\left\{n-sqrt\left\{n-cdots\right\}\right\}\right\}\right\} = frac\left\{-1 + sqrt \left\{1+4n\right\}\right\}\left\{2\right\}.$

This method will give a rational $x$ value for all values of $n$ such that

$\left\{n\right\} = \left\{x^2\right\} + \left\{x\right\}. ,$

### Cube roots

In certain cases, infinitely nested cube roots such as

$x = sqrt\left[3\right]\left\{6+sqrt\left[3\right]\left\{6+sqrt\left[3\right]\left\{6+sqrt\left[3\right]\left\{6+cdots\right\}\right\}\right\}\right\}$

can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation

$x = sqrt\left[3\right]\left\{6+x\right\}.$

If we solve this equation, we find that x = 2. More generally, we find that

$sqrt\left[3\right]\left\{n+sqrt\left[3\right]\left\{n+sqrt\left[3\right]\left\{n+sqrt\left[3\right]\left\{n+cdots\right\}\right\}\right\}\right\}$ is the real root of the equation $x^3-x-n=0 ,!$ for all n where n > 0.

The same procedure also works to get

$sqrt\left[3\right]\left\{n-sqrt\left[3\right]\left\{n-sqrt\left[3\right]\left\{n-sqrt\left[3\right]\left\{n-cdots\right\}\right\}\right\}\right\}$ as the real root of the equation $x^3+x-n=0 ,!$ for all n and x where n > 0 and |x| ≥ 1.

## References

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