Nested radical

Some nested radicals can be rewritten in a form that is not nested. For example,

$${\displaystyle {\sqrt {3+2{\sqrt {2}}}}=1+{\sqrt {2}}\,,}$$

${\displaystyle {\sqrt {2}}+{\sqrt {5-2{\sqrt {6}}}}={\sqrt {3}},\quad }$[1]

$${\displaystyle {\sqrt[{3}]{{\sqrt[{3}]{2}}-1}}={\frac {1-{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}{\sqrt[{3}]{9}}}\,.}$$

Another simple example,

$${\displaystyle {\sqrt[{3}]{\sqrt {2}}}={\sqrt[{6}]{2}}}$$

Rewriting a nested radical in this way is called denesting. This is not always possible, and, even when possible, it is often difficult.

In the case of two nested square roots, the following theorem completely solves the problem of denesting.[2]

If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that

$${\displaystyle {\sqrt {a+{\sqrt {c}}}}={\sqrt {x}}\pm {\sqrt {y}}}$$

if and only if ${\displaystyle a^{2}-c~}$ is the square of a rational number d.

If the nested radical is real, x and y are the two numbers

$${\displaystyle {\frac {a+d}{2}}~}$$

and ${\displaystyle ~{\frac {a-d}{2}}~,~}$ where ${\displaystyle ~d={\sqrt {a^{2}-c}}~}$ is a rational number.

In particular, if a and c are integers, then 2x and 2y are integers.

This result includes denestings of the form

$${\displaystyle {\sqrt {a+{\sqrt {c}}}}=z\pm {\sqrt {y}}~,}$$

as z may always be written ${\displaystyle z=\pm {\sqrt {z^{2}}},}$ and at least one of the terms must be positive (because the left-hand side of the equation is positive).

A more general denesting formula could have the form

$${\displaystyle {\sqrt {a+{\sqrt {c}}}}=\alpha +\beta {\sqrt {x}}+\gamma {\sqrt {y}}+\delta {\sqrt {x}}{\sqrt {y}}~.}$$

However, Galois theory implies that either the left-hand side belongs to ${\displaystyle \mathbb {Q} ({\sqrt {c}}),}$ or it must be obtained by changing the sign of either ${\displaystyle {\sqrt {x}},}$ ${\displaystyle {\sqrt {y}},}$ or both. In the first case, this means that one can take x = c and ${\displaystyle \gamma =\delta =0.}$ In the second case, ${\displaystyle \alpha }$ and another coefficient must be zero. If ${\displaystyle \beta =0,}$ one may rename xy as x for getting ${\displaystyle \delta =0.}$ Proceeding similarly if ${\displaystyle \alpha =0,}$ it results that one can suppose ${\displaystyle \alpha =\delta =0.}$ This shows that the apparently more general denesting can always be reduced to the above one.

Proof: By squaring, the equation

$${\displaystyle {\sqrt {a+{\sqrt {c}}}}={\sqrt {x}}\pm {\sqrt {y}}}$$

is equivalent with

$${\displaystyle a+{\sqrt {c}}=x+y\pm 2{\sqrt {xy}},}$$

and, in the case of a minus in the right-hand side,

(square roots are nonnegative by definition of the notation). As the inequality may always be satisfied by possibly exchanging x and y, solving the first equation in x and y is equivalent with solving

$${\displaystyle a+{\sqrt {c}}=x+y\pm 2{\sqrt {xy}}.}$$

This equality implies that ${\displaystyle {\sqrt {xy}}}$ belongs to the quadratic field ${\displaystyle \mathbb {Q} ({\sqrt {c}}).}$ In this field every element may be uniquely written ${\displaystyle \alpha +\beta {\sqrt {c}},}$ with ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ being rational numbers. This implies that ${\displaystyle \pm 2{\sqrt {xy}}}$ is not rational (otherwise the right-hand side of the equation would be rational; but the left-hand side is irrational). As x and y must be rational, the square of ${\displaystyle \pm 2{\sqrt {xy}}}$ must be rational. This implies that ${\displaystyle \alpha =0}$ in the expression of ${\displaystyle \pm 2{\sqrt {xy}}}$ as ${\displaystyle \alpha +\beta {\sqrt {c}}.}$ Thus

$${\displaystyle a+{\sqrt {c}}=x+y+\beta {\sqrt {c}}}$$

for some rational number ${\displaystyle \beta .}$ The uniqueness of the decomposition over 1 and ${\displaystyle {\sqrt {c}}}$ implies thus that the considered equation is equivalent with

$${\displaystyle a=x+y\quad {\text{and}}\quad \pm 2{\sqrt {xy}}={\sqrt {c}}.}$$

It follows by Vieta's formulas that x and y must be roots of the quadratic equation

$${\displaystyle z^{2}-az+{\frac {c}{4}}=0~;}$$

its ${\displaystyle ~\Delta =a^{2}-c=d^{2}>0~}$ (≠ 0, otherwise c would be the square of a), hence x and y must be

$${\displaystyle {\frac {a+{\sqrt {a^{2}-c}}}{2}}~}$$

and ${\displaystyle ~{\frac {a-{\sqrt {a^{2}-c}}}{2}}~.}$

Thus x and y are rational if and only if ${\displaystyle d={\sqrt {a^{2}-c}}~}$ is a rational number.

For explicitly choosing the various signs, one must consider only positive real square roots, and thus assuming c > 0. The equation ${\displaystyle a^{2}=c+d^{2}}$ shows that |a| > √c. Thus, if the nested radical is real, and if denesting is possible, then a > 0. Then the solution is

$${\displaystyle {\begin{aligned}{\sqrt {a+{\sqrt {c}}}}&={\sqrt {\frac {a+d}{2}}}+{\sqrt {\frac {a-d}{2}}},\\[6pt]{\sqrt {a-{\sqrt {c}}}}&={\sqrt {\frac {a+d}{2}}}-{\sqrt {\frac {a-d}{2}}}.\end{aligned}}}$$

Srinivasa Ramanujan demonstrated a number of curious identities involving nested radicals. Among them are the following:[3]

$${\displaystyle {\sqrt[{4}]{\frac {3+2{\sqrt[{4}]{5}}}{3-2{\sqrt[{4}]{5}}}}}={\frac {{\sqrt[{4}]{5}}+1}{{\sqrt[{4}]{5}}-1}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right),}$$

$${\displaystyle {\sqrt {{\sqrt[{3}]{28}}-{\sqrt[{3}]{27}}}}={\tfrac {1}{3}}\left({\sqrt[{3}]{98}}-{\sqrt[{3}]{28}}-1\right),}$$

$${\displaystyle {\sqrt[{3}]{{\sqrt[{5}]{\frac {32}{5}}}-{\sqrt[{5}]{\frac {27}{5}}}}}={\sqrt[{5}]{\frac {1}{25}}}+{\sqrt[{5}]{\frac {3}{25}}}-{\sqrt[{5}]{\frac {9}{25}}},}$$

and

${\displaystyle {\sqrt[{3}]{\ {\sqrt[{3}]{2}}\ -1}}={\sqrt[{3}]{\frac {1}{9}}}-{\sqrt[{3}]{\frac {2}{9}}}+{\sqrt[{3}]{\frac {4}{9}}}.\quad }$[4]

In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested.[5] Earlier algorithms worked in some cases but not others. Landau's algorithm involves complex roots of unity and runs in exponential time with respect to the depth of the nested radical.[6]

In trigonometry, the sines and cosines of many angles can be expressed in terms of nested radicals. For example,

$${\displaystyle \sin {\frac {\pi }{60}}=\sin 3^{\circ }={\frac {1}{16}}\left[2(1-{\sqrt {3}}){\sqrt {5+{\sqrt {5}}}}+{\sqrt {2}}({\sqrt {5}}-1)({\sqrt {3}}+1)\right]}$$

and

$${\displaystyle \sin {\frac {\pi }{24}}=\sin 7.5^{\circ }={\frac {1}{2}}{\sqrt {2-{\sqrt {2+{\sqrt {3}}}}}}={\frac {1}{2}}{\sqrt {2-{\frac {1+{\sqrt {3}}}{\sqrt {2}}}}}.}$$

The last equality results directly from the results of § Two nested square roots.

Nested radicals appear in the algebraic solution of the cubic equation. Any cubic equation can be written in simplified form without a quadratic term, as

$${\displaystyle x^{3}+px+q=0,}$$

whose general solution for one of the roots is

$${\displaystyle x={\sqrt[{3}]{-{q \over 2}+{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}+{\sqrt[{3}]{-{q \over 2}-{\sqrt {{q^{2} \over 4}+{p^{3} \over 27}}}}}.}$$

In the case in which the cubic has only one real root, the real root is given by this expression with the radicands of the cube roots being real and with the cube roots being the real cube roots. In the case of three real roots, the square root expression is an imaginary number; here any real root is expressed by defining the first cube root to be any specific complex cube root of the complex radicand, and by defining the second cube root to be the complex conjugate of the first one. The nested radicals in this solution cannot in general be simplified unless the cubic equation has at least one rational solution. Indeed, if the cubic has three irrational but real solutions, we have the casus irreducibilis, in which all three real solutions are written in terms of cube roots of complex numbers. On the other hand, consider the equation

$${\displaystyle x^{3}-7x+6=0,}$$

which has the rational solutions 1, 2, and −3. The general solution formula given above gives the solutions

$${\displaystyle x={\sqrt[{3}]{-3+{\frac {10{\sqrt {3}}i}{9}}}}+{\sqrt[{3}]{-3-{\frac {10{\sqrt {3}}i}{9}}}}.}$$

For any given choice of cube root and its conjugate, this contains nested radicals involving complex numbers, yet it is reducible (even though not obviously so) to one of the solutions 1, 2, or –3.

• Exponentiation
• Sum of radicals