Denesting square roots

You may know to simplify $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ by clearing the radicals out of the denominator, but how do you simplify $\sqrt{1+\sqrt{3}}$ or $\sqrt{2+\sqrt{3}}$ ?

Did you figure out that $\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}$ ? If not, you can double check by squaring both sides. But how would you do this without guess-and-check? The process is known as denesting radicals, since $\sqrt{2+\sqrt{3}}$ has nested square roots, while $\frac{\sqrt{6}+\sqrt{2}}{2}$ does not.

Problem. Can $\sqrt{1+\sqrt{3}}$ be denested? What about $\sqrt{4+3\sqrt{2}}$ ?

We will answer both of these questions, but first let’s solve a special case:

Case 1: The norm is a square

The expression $a^2-b^2c=(a+b\sqrt{c})(a-b\sqrt{c})$ is called the norm of $a+b\sqrt{c}$ . It turns out that $\sqrt{a+b\sqrt{c}}$ can be denested whenever the norm of $a+b\sqrt{c}$ is a perfect square. Let $A=\sqrt{a+b\sqrt{c}}$ and $B=\sqrt{a-b\sqrt{c}}$ . Then we will always find that $AB=r$ is rational and $(A+B)^2=s$ is rational. (In fact, $s=2a+2r$ .) Therefore, we need to solve the system of equations

$\displaystyle AB=r$ ,
$\displaystyle A+B=\sqrt{s}$ .

This system of equations can be solved using elimination of variables or Vieta’s formulas, but either way we find that $A$ and $B$ are the roots of the quadratic $x^2-\sqrt{s}x+r=0$ , which are

$\displaystyle \sqrt{a\pm b\sqrt{c}}=\frac{\sqrt{s}\pm\sqrt{s-4r}}{2}$ .

So $\sqrt{a+b\sqrt{c}}$ denests!

Example. Suppose we want to denest $A=\sqrt{2+\sqrt{3}}$ . Let $B=\sqrt{2-\sqrt{3}}$ , and notice $AB=1$ and $(A+B)^2=6$ , so $A+B=\sqrt{6}$ . Then $A,B$ are the roots of $x^2-\sqrt{6}x+1=0$ , so by the quadratic formula,

$\displaystyle \sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}$ .

Example. We had asked about $\sqrt{1+\sqrt{3}}$ and $\sqrt{4+3\sqrt{2}}$ , but neither of these can be denested using this strategy, since both of them have norm $-2$ , which is not a square. So we need a little more work.

Case 2: The norm is not a square

In fact, we can prove that neither of these can be denested in terms of square roots:

Theorem. Suppose $a,b,c$ are rational numbers, and $c$ is not a perfect square. Then $\sqrt{a+b\sqrt{c}}$ can be expressed as a rational linear combination of square roots if and only if the norm $a^2-b^2c$ is a perfect square.

We first need to understand what the theorem statement means. A rational linear combination of square roots is an expression of the form

$\displaystyle k_1\sqrt{n_1}+k_2\sqrt{n_2}+\cdots$ ,

where $k_1,k_2,\ldots$ are rational numbers, $n_1,n_2,\ldots$ are squarefree integers (integers which are not divisible by any perfect square except $1$ ), and the sum has finitely many terms. We can also assume that $n_1,n_2,\ldots$ are all distinct, because if two were equal, we may as well group them together.

Theorem. If a number can be expressed as a rational linear combination of square roots, then it can be expressed uniquely in this form.

The proof is an interesting exercise. We will omit it, but try it yourself to brush up on your proof skills!

We tend to say that a rational linear combination of square roots is in simplest form. For example, $\frac{1}{\sqrt{2}}$ is not in simplest form because it can be expressed as a rational linear combination of square roots, $\frac{1}{2}\sqrt{2}$ , which corresponds to what we learned in school.

Now we have already proven that if $\sqrt{a^2-b^2c}$ is rational, then $\sqrt{a+b\sqrt{c}}$ can be expressed as a rational linear combination of square roots. Let’s prove the converse: that $\sqrt{a^2-b^2c}$ is rational if $\sqrt{a+b\sqrt{c}}$ can be expressed as a rational linear combination of square roots.

By assumption, we have

$\displaystyle a+b\sqrt{c}=(k_1\sqrt{n_1}+k_2\sqrt{n_2}+\cdots)^2$ .

When expanded, the right hand side contains terms of the form $2k_ik_j\sqrt{n_in_j}$ . By the last theorem, it must be that each of those terms is either an integer ( $n_in_j$ is a square, so $n_i=n_j$ ) or a $\sqrt{c}$ term ( $cn_in_j$ is a square, so $n_i=cn_j$ or $n_j=cn_i$ ). In other words, we must have

$\displaystyle \sqrt{a+b\sqrt{c}}=k\sqrt{n}+m\sqrt{nc}$ .

Squaring both sides, $a=n(k^2+m^2c)$ and $b\sqrt{c}=2kmn\sqrt{c}$ . Dividing, $\frac{a}{b}=\frac{k^2+m^2c}{2km}$ , or $\frac{2a}{b}=\frac{k}{m}+c\frac{m}{k}$ . Multiplying by $\frac{k}{m}$ , we find that $\frac{k}{m}$ satisfies the polynomial $x^2-\frac{2a}{b}x+c=0$ . But since $\frac{k}{m}$ is rational, that means that the discriminant $\frac{4a^2}{b^2}-4c=\frac{4a^2-4b^2c}{b^2}$ must be a square, so the norm $a^2-b^2c$ must be a perfect square. The proof is complete!

Denesting using fourth roots

According to our theorem, $\sqrt{4+3\sqrt{2}}$ cannot be expressed in terms of non-nested square roots. However, let’s try to use the same strategy anyway. Define $A=\sqrt{3\sqrt{2}+4}$ and $B=\sqrt{3\sqrt{2}-4}$ . (Notice we use $\sqrt{3\sqrt{2}-4}$ instead of $\sqrt{4-3\sqrt{2}}$ because $3\sqrt{2}>4$ .) Then

$\displaystyle AB=\sqrt{2}$
$\displaystyle (A+B)^2=8\sqrt{2}=(2\sqrt[4]{8})^2$

Therefore, $A$ and $B$ are roots of the quadratic $x^2-2\sqrt[4]{8}x+\sqrt{2}=0$ , so

$\sqrt{4+3\sqrt{2}}=\sqrt[4]{8}+\sqrt[4]{2}$ .

You can always verify it by squaring both sides! The final word on this subject is due to Borodin, Fagin, Hopcroft, and Tompa:

Theorem. Suppose $a,b,c$ are rational numbers, and $c$ is not a perfect square. Let $N=a^2-b^2c$ , the norm of $a+b\sqrt{c}$ .

If $N=n^2$ where $n$ is rational, then $\sqrt{a\pm b\sqrt{c}}$ denests (using our algorithm) as $\sqrt{\frac{a+n}{2}}\pm\sqrt{\frac{a-n}{2}}$ .
If $N=-cn^2$ where $n$ is rational, then $\sqrt{\pm a+b\sqrt{c}}$ denests (using our algorithm) as $\sqrt{\frac{b+n}{2}}\sqrt[4]{c}\pm\sqrt{\frac{b-n}{2}}\sqrt[4]{c}$ .
Otherwise, $\sqrt{a+b\sqrt{c}}$ does not denest as a linear combination of $n^\text{th}$ roots for any $n$ .

References

Borodin, Fagin, Hopcroft, and Tompa. Decreasing the Nesting Depth of Expressions Involving Square Roots. 1985
Landau. How to Tangle with a Nested Radical. 1993
Zippel. Simplification of Expressions Involving Radicals. 1985

Denesting square roots

Case 1: The norm is a square

Case 2: The norm is not a square

Denesting using fourth roots

Other problems

References

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Case 1: The norm is a square

Case 2: The norm is not a square

Denesting using fourth roots

Other problems

References

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