Mathematics Educators Stack Exchange is a question and answer site for those involved in the field of teaching mathematics. It only takes a minute to sign up.
Sign up to join this community
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.
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 $k$.
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 $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$
and
$$ \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$
normalises to
$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$
See here for more on this Denesting Structure Theorem.
For more information about nested and how to "denest" radicals, with references to Susan Landau's work on this topic consult:
