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In algebra, nested radicals are radical expressions that contain another radical expression. Examples include ## Infinitely nested radicals

### Square roots

### Cube roots

## References

- $sqrt\{5-2sqrt\{5\}\; \}$

which arises in discussing the regular pentagon,

- $sqrt\{5+2sqrt\{6\}\; \},$

or more complicated ones such as

- $sqrt[3]\{2+sqrt\{3\}+sqrt[3]\{4\}\; \}.$

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:

- $sqrt\{a+b\; sqrt\{c\}\; \}\; =\; sqrt\{d\}+sqrt\{e\},$

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

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.

Under certain conditions infinitely nested square roots such as

- $x\; =\; sqrt\{2+sqrt\{2+sqrt\{2+sqrt\{2+cdots\}\}\}\}$

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

- $x\; =\; sqrt\{2+x\}.$

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:

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

The same procedure also works to get

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

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

- $\{n\}\; =\; \{x^2\}\; +\; \{x\}.\; ,$

In certain cases, infinitely nested cube roots such as

- $x\; =\; sqrt[3]\{6+sqrt[3]\{6+sqrt[3]\{6+sqrt[3]\{6+cdots\}\}\}\}$

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

- $x\; =\; sqrt[3]\{6+x\}.$

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

- $sqrt[3]\{n+sqrt[3]\{n+sqrt[3]\{n+sqrt[3]\{n+cdots\}\}\}\}$ 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

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

- Decreasing the Nesting Depth of Expressions Involving Square Roots
- Simplifying Square Roots of Square Roots

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Last updated on Tuesday September 16, 2008 at 01:07:26 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday September 16, 2008 at 01:07:26 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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