# Where can square roots come from when they are not distances?

In a recent survey "Supergeometry in Mathematics and Physics", Kapranov points out cases in which observable quantities of immediate interest are represented as bilinear combinations of more fundamental quantities which may be physically mysterious and may not even be observable by themselves. Such `mysterious square roots' underpin supersymmetry, and Kapranov explains how they all arise from actions of the first levels of the sphere spectrum $\pi_1^{\mathrm{st}}= \mathbb{Z}/2$ and $\pi_2^{\mathrm{st}}= \mathbb{Z}/2$.

This made me curious, because the only other non-contrived cases of square roots that I can think of in mathematics are distances. For example, the $1/\sqrt{n}$ in the Central Limit Theorem (note that $n$ has immediate meaning as a number of samples but $\sqrt{n}$ doesn't) comes from a standard deviation which is a distance.

Question: Is every square root in mathematics, in which an object of immediate interest is replaced by its square root which is perhaps more fundamental but less readily interpreted, either a distance or an action of $\pi_1^{\mathrm{st}}$ or of $\pi_2^{\mathrm{st}}$ or as one of Kapranov's examples? Where else can square roots come from?

These are Kapranov's basic examples (the last is a simple case of supersymmetry):

• The wave function $\psi(x)$ of a particle, which cannot be measured, although $\left\vert \psi(x)\right\vert^2= \bar{\psi}(x)\cdot \psi(x)$ represents the probability density of the particle which is real, non-negative, and measurable.
• The Laplace operator on forms on a smooth Riemannian manifold is non-negative definite and real (self-adjoint), but is defined as $\Delta= d \circ d^\ast + d^\ast \circ d$ where $d$ is the exterior derivative and $d^\ast$ is its adjoint with respect to the Riemannian metric.
• Spinors as square roots of vectors.
• For $X$ be a smooth projective curve over a finite field $\mathbb{F}_q$, the étale cohomology group $H^1(X \otimes \bar{\mathbb{F}}_q,\mathbb{Q}_l)$ is acted upon by the Frobenius element $Fr$ generating $\mathrm{Gal}(\bar{\mathbb{F}}_q/\mathbb{F}_q)$. The image of each eigenvalue of $Fr$ in each complex embedding has absolute value a square root of $q$. This example motivated the Weil conjectures.
• The differential operator $Q =\frac{\partial}{\partial\xi}+\xi \frac{\partial}{\partial t}$ in $\mathbb{C}[t] \otimes \Lambda[\xi]$ is the square root of $\frac{\partial }{\partial t}$.

UPDATE: Thank you for the nice answers, which came up with quite a few examples! Are some of these manifestations of one another, or are they distances in disguise? I will list the ones I understand, together with what I don't understand about them:

• Square roots of line bundles.
• Volume elements, which are square roots of determinants of metric tensors.
• The $\sqrt{\pi}$ term in Gaussian integrals/ the Stirling approximation.
• The size of a square array with $n$ elements, as in the longest monotone subsequence of a sequence of length n.
• The square root of a two qubit operation such as SWAP. Is this a manifestation of one of the other examples?
• As a balancing term for errors behaving like $y$ and like $1/y$ (although, as pointed out by Elkies, this might be a distance in disguise).
• Values of periodic regular continued fractions.
• Grover's quadratic speedup and Tsirelson's bound.
• In the isoperimetric inequality, bounding the length of the boundary of planar regions of given area.
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The five examples you cite from Kapranov, do not they answer your question? Then what exactly are you asking now? – Alexandre Eremenko Mar 13 at 22:16
@AlexandreEremenko The last example is supersymmetry. The others are examples- so I am asking for other examples. Thanks! I edited the question to make it ask this. – Daniel Moskovich Mar 13 at 23:12
A trivial example: involutions are square roots of the identity. – Terry Tao Mar 13 at 23:48
In a similar vein: chain complexes arise from square roots of zero. – Terry Tao Mar 14 at 0:20
Every natural can be represented as the square root of a number from this sequence in a exactly one way. – PyRulez Mar 14 at 6:17

$i=\sqrt{-1}$ has no apparent relation with any distance.

Also $\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}.$

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Thanks! These were also the first answers I considered, although I convinced myself that they were not counterexamples. The imaginary unit i is a wave function in a 1-point space, I think. As for the Gaussian integral, I'd convinced myself that it is a distance because of the classical proof (polar coordinates), but probably you are right that it isn't really. So nice example! – Daniel Moskovich Mar 13 at 23:08
Closely related to the second example: $\Gamma(1/2) = \sqrt{\pi}$. – Terry Tao Mar 13 at 23:49
@Terry Tao: and closely related to this is the volume and area of the $n$-sphere which has a suspicious $\sqrt{\pi}$. – Alexandre Eremenko Mar 14 at 12:54
Isn't $i = \sqrt{0^2 - 1^2}$ the spacetime distance for an interval of time? – Owen Mar 15 at 23:18
@AlexandreEremenko The formula for the volume of a sphere has a $\sqrt\pi$ only if you write it that way. Otherwise, its a rational multiple of an integer power of $\pi$. In other words, there is a relation between $\Gamma(n+\frac12)$ and $\sqrt\pi$ - but that's already apparent from the answer and Terry Tao's comment. – Sebastian Goette Mar 17 at 12:26

The Dirichlet-to-Neumann map on a half-space is the square root of the (positive definite) Laplacian on the boundary of the half-space (see the previous MathOverflow question Characterisation of the square root of the Laplacian as a Dirichlet to Neumann mapping ). (This is related to the Hodge decomposition example in the OP.)

Dirac operators are also square roots of (negative definite) Laplacians. (This is related to the spinor example in the OP, of course.)

(One could argue, though, that both these examples are measuring some sort of "distance" in frequency space.)

The Pfaffian of a skew-symmetric matrix is the square root of its determinant. (Presumably this falls under the category of "supersymmetry" somehow.)

In geometric quantisation, it is convenient to work with half-densities - square roots of densities. In particular, this allows one to work with the Hilbert space $L^2(M)$ on a manifold $M$ without needing a reference measure. (This is of course related to the wave function example in the OP.)

In Bayesian probability, the Jeffreys prior is the square root of the (determinant of the) Fisher information.

If one accepts a half-dimensional subspace as a "square root" of the space it is contained in, then Lagrangian subspaces can be viewed as square roots of symplectic vector spaces, and real vector spaces can be viewed as square roots of their complexifications.

Square roots of the canonical line bundle on an algebraic curve are known as theta characteristics.

UPDATE: I found a table at this nlab page which lists three of the above examples and adds two more: metalinear structures (which seem connected both to theta characteristics and to spinors, the latter in that they both involve double covers of classical Lie groups, which I guess is the Lie group analogue of a square root) and the partition function from self-dual higher gauge theory (which I don't understand too well, but also seems linked to theta characteristics and spinors).

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These are nice examples! Small questions- can we argue that the Jeffreys prior is a distance, for example, because by Cramer-Rao, Fisher information bounds one over the standard deviation squared? Also, maybe theta characteristics fall under the category of supersymmetry? – Daniel Moskovich Mar 14 at 5:23
I like especially the example of the Pfaffian (I wanted to contribute with it, unitl I read this). Don Knuth says that there are no determinants, there are only Pfaffians. – Denis Serre Mar 14 at 8:30
@Denis_Serre do you have a reference to Knuth's remark? How do you define it for odd dimensional matrices? – John Jiang Mar 14 at 21:04
The Fisher information is certainly a "quadratic" quantity (as are most of the other things one is taking square roots of in these examples), but I don't see how to naturally interpret the Jeffrey's prior as a distance, particularly in the multivariate case where there is also this determinant to deal with. – Terry Tao Mar 14 at 21:54
Square root of the Laplacian also arises in Radon's transform. – Alexandre Eremenko Mar 15 at 2:12

The longest monotone subsequence of a sequence of length $n$ has length at least $\lceil \sqrt n \, \rceil$, and this is sharp (special case of the Erdős–Szekeres theorem). The longest increasing subsequence of a random permutation of $\{1,2,\ldots,n\}$ grows as $2\sqrt n$, the difference approaching a scaled Tracy-Widom distribution.

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In the proof $\sqrt n$ really arises as the size of a square array with $n$ entries, so it feels like a legitimate example. – Noam D. Elkies Mar 14 at 3:35

The square root of SWAP (which is the only two-qubit operation needed to realize a universal quantum computation) has no "distance" interpretation I can think of.

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Fantastic example! Thanks! – Daniel Moskovich Mar 13 at 23:15
To be fair, "which is the only two-qubit operation needed to realize a universal quantum computation" is true of almost every two-qubit operation. i.e., there's nothing special about the square root of SWAP in this regard at all. – Nathaniel Johnston Mar 14 at 16:02
true, it's the physics that makes the $\sqrt{SWAP}$ special, as the two-qubit operation that is most easily implemented via the Heisenberg Hamiltonian. – Carlo Beenakker Mar 14 at 16:34

The number of partitions of $n$ is asymptotic to $$\frac1{4n\sqrt{3}} \exp \left( \pi \sqrt{\frac{2n}{3}} \right)$$ as $n \to \infty$ [Hardy-Ramanujan 1918; refinements replace the $n$ in $\sqrt{2n/3}$ by $n-\frac1{24}$].

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The square root arises by balancing multiplies of $ny$ and $1/y$ in a stationary-phase estimate, so it's not a distance (unless you insist that it's the inverse of the distance of the contour of integration from the real axis...) – Noam D. Elkies Mar 14 at 4:27
distance from the hyperbola $xy=n$ to the origin? :-) – Steven Stadnicki Mar 14 at 17:47
Well, if anything it's the distance from that hyperbola to $x+y=0$ (scaled by $\sqrt2$). – Noam D. Elkies Mar 16 at 2:27

Fixed points of fractional linear transformations $x \mapsto \frac{ax + b}{cx + d}$, where $a, b, c, d$ are integers, are usually quadratic surds. Equivalently, values of periodic regular continued fractions. I'm not convinced these are really best thought of as distances, even though geometric means are in the same ballpark.

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Thanks! Could you point me to a reference where this example is explained in more detail? – Daniel Moskovich Mar 16 at 13:27
@DanielMoskovich The first sentence is a triviality by the quadratic formula, but generally (as you may well recall) the periodicity of the continued fraction of $\sqrt{n}$ is intimately bound up with the rich literature on Pell's equation en.wikipedia.org/wiki/Pell's_equation#Continued_fractions Perhaps not a deep observation but nevertheless pleasant is that this periodicity and density of quadratic surds is one ingredient in the observation that $x \mapsto \text{frac}(1/x)$ (frac = fractional part) on irrationals in $(0, 1)$ is a prototypical example of a chaotic dynamical system. – Todd Trimble Mar 16 at 13:43

A third category would be counts of things. I think probability and computer science should have many examples. Noam Elkies already mentioned longest increasing subsequence here.

• Birthday paradox: If we throw $k$ balls uniformly into $n$ bins, and choose $k$ such that we expect one collision, then $k = \Theta(\sqrt{n})$. More precisely one chooses $n = {k \choose 2}$.

• More generally, to test whether a given distribution is uniform on $n$ coordinates or $\epsilon$-far from uniform, it is necessary and sufficient to draw order $\frac{\sqrt{n}}{\epsilon^2}$ samples.

• Grover's algorithm for quantum search inverts a function on domain of size $n$ with $O(\sqrt{n})$ queries (and this is the optimal order). I think there are probably many examples of algorithms along these lines once one starts looking. Pollard's rho algorithm perhaps.

• Along the lines of the above, $\sqrt{n}$ arises often when balancing the size of two terms. Noam Elkies again mentions similar reasoning here. I'm struggling to think up good examples on the spot, but here are two. If you're going to search through up to $n$ entries by restarting $k$ times and each time inspecting $n/k$ items, then you can minimize $\max\{k,n/k\}$ by choosing $k = \sqrt{n}$ and this is something you sometimes want to do. In online learning theory, optimal regret bounds look like $\sqrt{T}$ when there are $T$ rounds. Interestingly the lower bound comes from the central limit theorem, but the upper arises from choosing $\eta > 0$ to minimize $\eta T + \frac{1}{\eta}$. I don't think regret is a "distance", whereas a key special case of regret is a count of number of mistakes.

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Don't these come from Stirling's approximation, whose square root is the square root from the Gaussian integral in disguise? (Alexandre Eremenko's answer). – Daniel Moskovich Mar 14 at 5:54
@DanielMoskovich, none of the proofs have anything to do with Stirling's, except one lower bound I mentioned. – usul Mar 14 at 14:54
The birthday paradox is one line: There are ${k \choose 2}$ pairs, each with probability $\frac{1}{n}$ of collision, hence the expected number is ${k \choose 2}/n = 1$ for $k \approx \sqrt{n}$. The upper bound for distribution testing has the same thing going on; although slightly more complex, there is nothing asymptotic or Gaussian. The lower bound comes from analyzing a particular counterexample and is again non-asymptotic. Proofs of Grover's and Pollard's rho are purely combinatorial/straightforward. And I mentioned how one obtains the upper-bound for the proof for no-regret learning. – usul Mar 14 at 15:05
For how it follows from Stirling's approximation, see solipsys.co.uk/new/TheBirthdayParadox.html Although the version you give is simpler. – Daniel Moskovich Mar 14 at 18:54
@DanielMoskovich, thanks for the pointer. In the big picture, I think the important point is that the number being measured is a number of samples drawn, which has the order of $\sqrt{\text{support}}$, so I think it's a count which is very different from a distance. – usul Mar 14 at 19:16

Suppose a "geometry" consists on $n$ points and a collection of "blocks" all of size $k.$ Various conditions force $k \ge \sqrt{n}$ more or less. In the first two cases, suppose every pair of points is in at least one common block.

• If $k \le \sqrt{n}$ then some two blocks are disjoint. So one condition is no two blocks are disjoint. It is possible to have projective planes with $n=q^2+q+1$ and $k=q+1 \approx \sqrt{n}+\frac12$ for $q$ a prime power.

• Blocks may be disjoint. If $k \lt \sqrt{n}$ then there is a point $p$ and block $B$ so that two blocks $B'$ and $B''$ on $p$ are disjoint from $B.$ So one condition is a point $p$ not on a block $B$ may belong to at most one disjoint block. Here affine planes with $n=q^2$ and $k=q=\sqrt{n}$ satisfying Playfair's axiom are possible.

Another example (not exactly the first in disguise, but close): Suppose $A \subset \{0,1,2,\cdots ,n-1\}$ is such that $|A| \le \sqrt{n}$, Then $A$ is disjoint from at least one of the $n$ sets $A+i$ (considered $\mod n$). It is possible to have $n=q^2+q+1$ and $|A|=q+1$ so that the sets $A+i$ are the lines of a projective plane (hence each pair intersecting in a singleton.)

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The Gauss sum attached to a primitive Dirichlet character mod $q$ has absolute value $\sqrt q$.

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(But maybe you're going to count any absolute value in $\bf C$ as a length by definition?) – Noam D. Elkies Mar 14 at 3:36
This seems like an elementary version of the 'eigenvalues of Frobenius' bullet point, although I'm not sure the exact relation – Kevin Casto Mar 14 at 6:58
Gauss sums are indeed special cases of eigenvalues of Frobenius. So are Kloosterman sums (though this connection is not as elementary). – Noam D. Elkies Mar 14 at 13:56

The most natural example I can think of is the Fibonacci numbers, and the golden ratio, which I am sure most of you knows more about than me.

To recap: The golden ratio is the positive solution to $x^2=x+1$, $x=\frac12(\sqrt{5}+1)$, and gives the asymptotic growth rate of the Fibonacci numbers. This ratio appear in many places in nature and stuff with five-fold symmetry.

I am not sure if one can explain this square root using either distance, quantum physics or probability.

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well, perhaps not quantum physics or probability, but the golden ratio can of course be interpreted as distance between two non-adjacent vertices of the regular pentagon with unit edge length... – Moritz Firsching Mar 14 at 20:06
@MoritzFirsching: That is true! However, is that the underlying explanation for shell spirals and sunflowers? I thought it was more related to some type of optimization theorem... – Per Alexandersson Mar 14 at 21:50
@PerAlexandersson one would think that the regular pentagon, once you draw all the diagonals, exhibits the (sort of) same kind of self-similarity as the spirals you mention.. – Vladimir Dotsenko Mar 15 at 15:01
As a matrix guy, I would say that the square root here arises as an eigenvalue of the transformation $\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}$, so a more general answer in this theme would be as eigenvalues. This may just be my skew view on things, though. – Federico Poloni Mar 17 at 8:22
@FedericoPoloni As eigenvalues of 2x2 matrices, in other words, as roots of polynomials of degree 2. Or do you have in mind a more general collection of matrices producing square roots as eigenvalues? – Sebastian Goette Mar 17 at 12:32

The solution to the optimization problem

$$\underset{x\in(0,1)}{\text{minimize}} \quad \frac{a}{x}+\frac{b}{1-x}$$

is $x^*=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}$, with a minimum value of the objective function of $\left(\sqrt{a}+\sqrt{b}\right)^2$. This does not involve any distances but rather follows from the derivative of $\frac1x$ being $\frac{-1}{x^2}$.

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Thanks! Is there a reference where I could read more? – Daniel Moskovich Mar 16 at 13:42
Hmm, I wouldn't think there's anything interesting about this. This was just a small calculation I had to do when solving a problem in my research. It's not complicated to prove though, just set the derivative to zero and you'll end up with a simple quadratic equation. – jadhachem Mar 17 at 9:29

Square roots are common in the analysis of algorithms. A standard example is the Hopcroft–Karp bipartite matching algorithm (or its generalization to non-bipartite graphs by Micali and Vazirani) which takes time $O(m\sqrt n)$; here $n$ is the number of vertices and $m$ the number of edges in a bipartite graph.

Underlying some of these algorithm time bounds are corresponding combinatorial bounds in graph theory which also involve square roots (or half-integer powers). For instance, in a graph with $m$ vertices, the maximum number of triangles is $O(m^{3/2})$, which also forms a natural time bound for triangle-finding algorithms. Relatedly, the degeneracy (or arboricity) of an $m$-edge graph is always $O(\sqrt m)$, a bound that is sometimes tight.

Another important square root in graph theory comes from the planar separator theorem and its generalizations (the fact that you can evenly partition a planar graph into two subgraphs by removing $O(\sqrt n)$ vertices). This also has ramifications in the analysis of algorithms, e.g. exact algorthms for NP-hard problems on planar graphs tend to be exponential in $\sqrt n$ rather than exponential in $n$. However to some extent you could argue that this one is the same as the square root relation between area and perimeter of figures in the plane, so not unrelated to distances.

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Sometimes you can multiply two different numbers and get a square. For instance if $f$ is a modular form and $\chi$ is a quadratic character attached to a quadratic field of absolute discriminant $d$, then the product of central values of $L$-functions $L(1/2, f, d) = L(1/2,f)L(1/2, f \otimes \chi)$ (a positive number) is essentially $b(d)^2$, where $b(d)$ is a Fourier coefficient of an associated form $g$ of half-integral weight. So $b(d)$ is essentially the square root of $L(1/2, f, d)$. Determining the sign of this square root is an interesting and nontrivial problem.

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Polar decomposition of square matrix $A$ involves square-root of $A^*A$

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This example is a bit different though, in that the positive-semidefinite Hermitian matrix component has immediate meaning also (stretch directions), where as $A^\ast A$ does not. – Daniel Moskovich Mar 15 at 11:22

There was a popular wall Street interview question about the least number of trials needed to test the maximum number of stairs a beer bottle can withstand falling through if you have two bottles. The final answer involves a square root. On a more serious note, see my conjecture about bounds on schur polynomials on the unit circle here.

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The 'frivolous' part of this answer (the beer-bottle question) can be thought of as inverting the Cantor Pairing Function; in a sense it's a rough estimate of distance between a point in $\mathbb{N}^2$ (under a certain indexing) and the origin. – Steven Stadnicki Mar 14 at 17:45

Besides in the calculation of distances, square roots appear naturally as the lower bound on the length of the boundary of planar regions of given area.

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Do you have an example? – Daniel Moskovich Mar 16 at 13:53
It is a consequence of the fact that the largest area, that has a boundary of fixed length, is a circle (solution to the classical problem of the Dido), so the minimal length of the boundary of unit area is $2\pi \frac{1}{\sqrt{\pi}}$ – Manfred Weis Mar 16 at 14:45
But wait... isn't the square root there coming from the arc-length formula, so that it actually is a distance? (albiet not directly) – Daniel Moskovich Mar 16 at 15:35
@DanielMoskovich length isn't the same as distance. Being able to calculate length via integrals over infinitesimal distances in case of piecewise smooth boundaries is merely a nice property, but what about fractal boundaries? – Manfred Weis Mar 16 at 17:11
@DanielMoskovich not necessarily; if the boundary of a region is fractal, then it can't be calculated via the arc length formula, but the lower bound is still valid and contains a square root. I agree that that may be considered splitting hairs, but length isn't the same as distance, even if it can in the case of piecewise smooth curves be expressed as an integral over infinitesimal distances. – Manfred Weis Mar 16 at 17:59

In problems such as sensitivity conjecture and degree of approximation or representation of boolean functions ($n$ is number of variables) and such as find minimum number of bits in communication complexity ($n$ is rank of communication matrix) I have seen some stubborn upper bounds of order $\sqrt n$ while lower bounds turn out to be $\log^2 n$. I am not sure which of these is close to truth.

There is no intuitive way to metrize these quantities.

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Where does the square root come from in this example? – Daniel Moskovich Mar 16 at 13:43

Most things using the Laplace domain. For example, if I have a 2nd-order filter then I will often want to convert between the (s^2 + a*s + b) representation and the (s + m)*(s + n) representation. Going from the former to the latter needs a square root. This is not reliant on distance, but on the nature of the system's dynamic behaviour.

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$X^2=X$ for infinite cardinal $X$ by the Axiom of Choice. (It's about square not square root but it's far from being about distance.)

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This seems to speak about the title not the body of the question- $X^2$ is far from being mysterious. – Daniel Moskovich Mar 16 at 9:53

$P=I^2 \times R$ as in Power = The current squared times the resistance.

In this case the square root is the actual current flowing through the device. This is particularly useful in situations with low transconductance power mos-fets and the heat they would have to dissipate in high performance industrial situations, with 1000-10,000 amp welders and platers.

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Inverse square laws... here also, the square root is also a quantity of immediate issue, being the current. So this seems more a case of the square being relevant than of the square root. – Daniel Moskovich Mar 16 at 9:57

I think an abstract version of some of these examples is worth highlighting.

In a general algebraic system, one often has a basic or perhaps derived operation op() of arity two and given an element b one looks for an element a such that op(a,a)=b. Sometimes a is considered as a square root of b.

Some examples are semigroups of functions, with the operation being composition: solve for f in f(f(x))=exp(x), for example.

In group and semigroup presentations, if a^2=b is part of a presentation, the natural question arises whether a^2=a as well.

Objects which are their own square roots are idempotents, and in some systems (rings especially) these are used in representing/decomposing the structure.

Many more examples of this are found in the other posts.

If one is using Cayley graphs or a time scale for a dynamical system, one can frame these as questions of distance. However, I see the abstract version as motivated by distance but not necessarily interpreting or modeling distance. It may model a different form which simpifies/complicates the present model, which may shorten/increase the distance from now to the point of understanding/enlightenment.

Gerhard "Is Humor Orthgonal To Enlightenment?" Paseman, 2016.03.16.

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This sounds almost too obvious, but when you approximate a function locally by a parabola and then solve for its intersection with a line (which occurs in a lot of contexts, e.g. ODE solving), there would seem to really be no interpretation of the square-root except as the root of a 2nd-degree polynomial...

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I feel quite silly in posting this triviality next to all this abstract advanced math, but I haven't seen it yet in the other answers.

Probably the easiest example of a square root that does not immediately relate to a distance is the good old $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

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Indeed, it's a triviality- we can reduce to the b=0 and a=1 case, in which case it's more or less the definition of the squre root ($x^2=c$ then $x=\pm\sqrt{c}$), but also there's nothing non-immediate about the square root here- it has as much immediate meaning as its square! – Daniel Moskovich Mar 17 at 15:09

periodic continued fractions are irrational solutions of quadratic equations with rational coefficients, i.e. can be expressed as sums of rationals and square roots of rationals, but not without square roots.

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This is Todd Trimble's answer, no? mathoverflow.net/a/233557/2051 – Daniel Moskovich Mar 19 at 18:30

Other than itself, all of a number's prime factors will always be less than or equal to that number's square root.

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Yes I did corrected. – Q the Platypus Mar 16 at 5:02
That is not correct. $3 * 7 = 21$ and $7>21^{\frac{1}{2}}$ – Fred Kline Mar 19 at 8:12