Student question to me involved "standard" form of radical expressions. They already know what a radical in simplest form (no perfect square factors, no fractions, no radicals in denominator). My thoughts after combing my literature was to tell him that standard form of a radical expression was something like this.

$$p+\sum_{k=1}^n{q_k\sqrt{a_k}}$$ where $p,q_i$ are rational numbers and $a_1<a_2<...<a_n$.

I feel like this would answer the question of "standard" while keeping the idea simple.

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    $\begingroup$ Also, why is the above reduction standard? I actually prefer $\frac{1}{\sqrt{2}}$ in many cases. Personally, I would avoid using the terminology "standard" as that implies a "correct" form to students, of which there really isn't one. It really depends on what you are doing. $\endgroup$ – Chris C Apr 22 '15 at 14:09
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    $\begingroup$ It is not quite clear what you are saying. Is it that for any positive integer x that $\sqrt[n]{x}$ can be written in that form? I don't think that result is true, for instance $\sqrt[3]{2}$ is an odd extension, while I believe the above result is even with no odd divisors. $\endgroup$ – Chris C Apr 22 '15 at 14:21
  • $\begingroup$ I'm strictly dealing in square roots, not $n$-th roots. Basically this. If I have $\sqrt{3}+\sqrt{11}-\sqrt{5}$, is it more "appropriate" to write $\sqrt{3}-\sqrt{5}+\sqrt{11}$. I'm trying to answer the student's question about "ordering" the final answer. $\endgroup$ – Eleven-Eleven Apr 22 '15 at 14:45
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    $\begingroup$ The choice of whether to write $\sqrt{3} + \sqrt{11} - \sqrt{5}$ or $\sqrt{3} - \sqrt{5} + \sqrt{11}$ is completely arbitrary, and I see no reason to prefer one over the other. $\endgroup$ – Daniel Hast Apr 22 '15 at 16:04
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    $\begingroup$ It seems to me that this whole business with "standard" forms is overemphasized. It's a convenient way to prohibit obviously bad answers on exams --- answers that ought to be simplified. But it burdens the student with remembering obscure and ultimately useless rules to determine what is standard. $\endgroup$ – Andreas Blass Apr 22 '15 at 16:10

The form you propose seems reasonable to me (given your constraints). The rational term at the start could be thought of as corresponding to $\sqrt{1}$ and then the numbers in the square-roots increase.

I agree with commenters that too much emphasis on "standard" forms might not be good, but when pressed to give an answer I would go with what you propose.

One thing to consider, though, is that $\sqrt{3}- \sqrt{2}$ is arguably more pleasant than $- \sqrt{2}+\sqrt{3}$, but after all we are seeking a standard form not an optimal form, so this is not that much of a problem.

Let me also add that it makes sense to talk about a standard form in this case as given (distinct) square-free $a_k$ the coefficients $q_k$ (and $p$) will be uniquely determined by the sum; in other words it is not possible to have $p+\sum_{k=1}^n{q_k\sqrt{a_k}} = p'+\sum_{k=1}^n{q_k'\sqrt{a_k}}$ unless $p=p'$ and $q_k=q_k'$ for all $k$.

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    $\begingroup$ Squarefree is unnecessarily strict. Rather, they will be $\,\Bbb Q\,$- inearly independent as long as they are multiplicatively independent, i.e. no subset of the radicals has product $\in\Bbb Q.\,$ See here for a proof. $\endgroup$ – Bill Dubuque Apr 22 '15 at 16:48
  • $\begingroup$ Thank you for this additional information; a motivation to restrict to the squarefree setting was that this is the property that is already mentioned (indirectly with "no perfect square factors") in OP. $\endgroup$ – quid Apr 22 '15 at 16:54
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    $\begingroup$ One of the reasons that i used my notation above as "standard" was to help the students collect and combine like terms. If they are "ordered" by radicand, then they can more easily identify commutatatively like terms and combine. $\endgroup$ – Eleven-Eleven Apr 22 '15 at 20:06

There is no standard form for such expressions. However, some forms may prove more convenient for certain applications, e.g. transforming the radicands to be squarefree helps to ascertain multiplicative independence, hence linear independence (cf. this theorem). But even that is not necessary since it can also be tested without squarefree normalization by instead verifying that no subset of the radicals has rational product.

The fact that there is no computationally effective normal form for such expressions is a reflection of the innate complexity of their arithmetic. The complexity of even simpler problems involving radicals is currently unknown. For example, no polynomial time algorithm is known for determining the sign of a sum of real radicals $\,\sum c_i q_i^{1/r_i},\,$ where $\,c_i, q_i\,$ are rationals and $\,r_i\,$ is a positive integer. Such sums play an important role in various geometric problems (e.g. Euclidean shortest paths and traveling salesman tours). Testing whether such a sum of radicals is zero can be decided in polynomial time, but this is of no help in determining the sign (it only shows that if sign testing is in NP then it is already in NP $\cap $ co-NP).

Generally the context will determine what type of normal or standard forms prove most convenient. For example, denesting of radicals is governed by a structure theorem that employs a certain normalized form, e.g.

$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$ normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$


$$ \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$

normalises to

$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$

See here for more on this Denesting Structure Theorem.

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For more information about nested and how to "denest" radicals, with references to Susan Landau's work on this topic consult:


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