# What exactly IS a square root?

It's come to my attention that I don't actually understand what a square root really is (the operation). The only way I know of to take square roots (or nth root, for that matter) it to know the answer! Obviously square root can be rewritten as $x^{1/2}$ , but how does one actually multiply something by itself half a time?

How do calculators perform the operation?

• This link might be helpful. – Trevor Wilson May 12 '15 at 6:47
• For those of us old enough to have used common logarithms in calculations, multiplication could be transformed to addition, and adding something half a time has a natural interpretation. – Henry May 12 '15 at 6:58
• The definition of "the square root of $x$" is "the positive number $y$ such that $y^2$ equals $x$". This is an implicit definition, and so it makes sense that it takes a little thought as to how to actually compute it (Trevor Wilson's link is a good one). But as Paul Picard said, don't let the notation $x^{1/2}$ (or similarly $x^\pi$) fool you into thinking that you're multiplying $x$ by itself half a time (or similarly $\pi$ times). The function $x^y$ has that interpretation when $y$ is a positive integer, but must be defined other ways when it isn't. – Greg Martin May 12 '15 at 7:11
• To answer the second part, calculators use the general binomial theorem which generates a converging infinite series that the calculator then uses to get the root to a given number of decimal places – imulsion May 12 '15 at 13:51
• An interesting exercise is to try to construct a square that has area $x$. If you succeed, the length of the side is the square root ($x^{1/2}$). – jxh May 12 '15 at 16:42

How does one actually multiply something by itself half a time ?

Zen Buddhism has a similar question: What is the sound of one hand clapping ? When my father told me, in passing, one day, that $\sqrt x=x^{1/2}$ I had pretty much the same reaction. But then I started thinking to myself: What is the fundamental property of an n-th root, $\sqrt[\Large^n]x$ ? It is basically the number which, when multiplied n times with itself, yields the desired value. At the same time, $x^{1/n}$ multiplied n times with itself, also yields the same value, since $\big(x^a\big)^b=x^{ab}$.

• To answer the Zen Buddhism with a typical mathematical attitude, "define clapping". :-) – Asaf Karagila May 12 '15 at 7:18
• The question is now reduced to why not $\sqrt x =-x^{1/2}$ – biziclop May 12 '15 at 13:18
• @CarlWitthoft: We do not $($a priori$)$ “define” $x^0$, but rather we deduce it, by logically extending the properties of exponentiation for non-zero natural exponents. – Lucian May 13 '15 at 19:05
• @AsafKaragila Or, to answer it with the pragmatism of an engineer, this is the sound of one hand clapping :) – Roman Starkov May 13 '15 at 20:02
• @MontyHarder: There's a difference between “n times with itself”, and “n more times with itself” or “n times to itself”. – Lucian May 13 '15 at 21:26

You appear to understand what a square root is just fine; $\sqrt{x}$ is any number with a square equal to $x$.

If you're asking “how do I compute that?”, well there's several possible algorithms for that. As @TrevorWilson points out, Wikipedia has an entire page on the subject.

One method is to pick a number, square it, and see if the answer is too big or too small. Once you have one number that's too big and another that's too small, you know the exact solution is somewhere between the two. You can then recursively subdivide the range into smaller and smaller chunks.

Alternatively, a computer might use Newton's method, which is good for solving all sorts of equations. $\sqrt{c}$ is the solution to $x^2 - c = 0$, which we can solve by

$$x_{n+1} = x_n - \frac{x_n^2 - c}{2x_n}$$

That is, given that $x_n$ is a (possibly very poor) estimate of $\sqrt{c}$, then $x_{n+1}$ is a better estimate. Repeat until you have sufficient accuracy.

• $\sqrt{x}$ is any positive number whose square is to equal $x$. There'll be at most one solution. – lhf May 12 '15 at 16:13
• @lhf Depends on context. Most people seem to mean the positive solution, but sometimes people seem to mean any solution. (E.g., when dealing with complex numbers.) – MathematicalOrchid May 12 '15 at 16:17
• I agree with this answer. A definition doesn't need to be operational to be a definition. – Sudoku Polo May 12 '15 at 17:14
• @MathematicalOrchid Yes. In Italy sometimes I heard them called in two different ways: "arithmetical square root" when it is meant "the positive solution", whereas "algebraic square root" when "any solution" is meant (word by word translation from Italian, I don't know if that terminology makes sense in English). The first definition is by far the one I see used more frequently. – Lorenzo Donati -- Codidact.com May 13 '15 at 0:26
• @paqogomez Fixed. – MathematicalOrchid May 13 '15 at 8:06

"how does one actually multiply something by itself half a time?"

You answer that indirectly, from the definition of the square. The square of the square root of a number is that same number, or $$\left(\sqrt x\right)^2=x.$$

It turns out that this definition is consistent with a fractional exponent. For integer exponents, the following rule holds:

$$\left(x^a\right)^b=x^{ab}.$$

Then it is natural to write

$$\sqrt x=x^{1/2},$$ as this gives

$$\left(x^{1/2}\right)^2=x^{2/2}=x^1=x.$$

Assuming you didn't know how to compute a square root, you could do that by trials and errors. Let us take $\sqrt 2$ as an example.

$1^2=1<2$

$2^2=4>2$

Try the first decimal:

$1.1^2=1.21<2$

$1.2^2=1.44<2$

$1.3^2=1.69<2$

$1.4^2=1.96<2$

$1.5^2=2.25>2$

Try the second decimal:

$1.41^2=1.9881<2$

$1.42^2=2.0164>2$

$\cdots$

Some calculators still use this method. Others use the iterative Newton method. There is also the logarithm/antilogarithm approach, but this is a more advanced concept.

The method of Heron (a special case of Newton's method) was known early in history. It is based on the observation that if you know an approximation of the square root $a\approx\sqrt x$, then $\dfrac x a$ is also an approximation of the square root, and their mean $\dfrac12\left(a+\dfrac xa\right)$ is even better.

$$1\to\frac12\left(1+\frac21\right)=\frac32=1.5$$ $$\frac32\to\frac12\left(\frac32+\frac43\right)=\frac{17}{12}=1.41666\cdots$$ $$\frac{17}{12}\to\frac12\left(\frac{17}{12}+\frac{24}{17}\right)=\frac{577}{408}=1.414215\dots$$

Let $$n$$ be an integer that is not the square of any integer. Two versions of the question have already been answered, namely

1. What is the definition of $$\sqrt{n}$$, i.e. what does it mean for a given number to be a square root (or the positive square root) of $$n$$?

2. How can we approximate $$\sqrt{n}$$, i.e. how can we find rational numbers (or decimals) whose square is close to $$\sqrt{n}$$?

I will try to answer yet another version of the question that I think is lurking in the "is".

1. What kind of object is it whose square is $$n$$?

By our assumption such an object cannot be an integer. Also, it cannot be a rational number. So we need some new type of number. Here are a few approaches:

1. Dedekind cuts. If we define real numbers as Dedekind cuts, then we can simply define $$\sqrt{n}$$ as the set $$\{a \in \mathbb{Q} : a^2 < n\}$$. It takes some argument to see why it makes sense to consider such a set as a kind of number, however.

2. Cauchy sequences. An algorithm for computing square roots gives a sequence of rational numbers (which we can think of as better and better approximations for $$\sqrt{n}$$, whatever that is.) Such a sequence is a Cauchy sequence, meaning roughly that it looks like it should be converging to something, even if we don't know what that thing is yet. The sequences given by different algorithms might not be equal, but they are equivalent in the sense that they look like they must be converging to the same thing (again this can be made precise) and so we can uniquely define $$\sqrt{n}$$ as this equivalence class of Cauchy sequences.

3. Algebraic field extensions. Dedekind cuts and Cauchy sequences are overkill for defining $$\sqrt{n}$$ in the sense that they also add uncountably many other numbers to $$\mathbb{Q}$$, which we might not be interested in. A more economical way to add a square root of $$n$$ to $$\mathbb{Q}$$ is to add an indeterminate $$x$$ to get the ring of polynomials in $$x$$ with rational coefficients, and then take a quotient of this ring by setting the polynomial $$x^2-n$$ equal to zero, giving us a field containing $$\mathbb{Q}$$ and also containing an element $$x$$ that acts like a square root of $$n$$.

• By the way, each of the approaches to constructing a number that has the property $x^2=a$ also gives a corresponding synthetic approach, where a certain axiom of the reals implies existence of square roots. (1) is Dedekind- or order-completeness of the $<$ relation on $\Bbb R$ (a bounded nonempty set has a supremum); (2) is metric completeness of $\Bbb R$ or $\Bbb C$ (a Cauchy sequence converges); and (3) is the fundamental theorem of algebra (although usually this is proved from (1) or (2), not postulated). – Mario Carneiro May 12 '15 at 21:08

Asking what a square root "is", to a mathematician's ear, means asking how it is defined: if I may read into what you're saying, that's not what you're trying to figure out here. But knowing how some other things are defined will help.

So, let's start with the real numbers: what you know as streams of decimal digits are generally defined in some way or another as all of the spaces between rational numbers. Rational numbers are the whole-number-fractions that you learned about in school. If you want to read up more on how to formally do this, those spaces are called Dedekind cuts.

There are two other key observations about squaring numbers which makes it easier to produce a square root. The first one is: if $0 < a < b$, then $a^2 < b^2$, this has one of those scary math-names (monotonic). The second one is that if two rationals are near each other, then their squares are near each other (continuous). (This latter property is actually even better because squaring is not just continuous but differentiable, which is a complicated word that means that, when you zoom in on the graph, it becomes a straight line everywhere -- it's like a circle, not like a fractal.)

Since it's continuous, the square root of any positive real number is always a well-defined positive real number: Given the positive gap-between-rationals which you want to take the square root of, the square root is the positive gap-between-rationals such that any rational greater than the square-root-gap squares to a rational greater than the square-gap, similarly for numbers below the gaps. So every real number has a real square root.

And that leads to the first and easiest way to find it, called bisection: suppose you know that the unknown gap $p$ corresponding to a known square $p^2$ is between two rationals $r_1$ and $r_2$, so $r_1^2 < p^2 < r_2^2$. Then take the rational halfway between those two, $r_m = (r_1 + r_2) / 2$, and see whether $r_m^2 < p^2$ (in which case it replaces $r_1$) or whether $r_m^2 > p^2$ (in which case it replaces $r_2$). This cuts your search space in half! After three or four of these you usually know another decimal digit.

Concrete example: we want the square root of 18. We choose a perfect square less than it, and one greater than it. If you're doing it by hand in decimal, it helps to know the square numbers up to 100 and then use the fact that square roots distribute over multiplication to multiply the grid by 100: so we use a grid of {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 400, 900, 1600, ...}. So we start off knowing that the value is between 16 and 25, so the square root is between 4 and 5.

So we try 4.52 = 20.25, which is too big. The square root is therefore between 4 and 4.5. We try 4.252 = 18.0625, which is still too big, so the square root is between 4 and 4.25. We try 4.1252 = 17.015625, which is finally too small, so the square root is between 4.125 and 4.25. So now we try 4.18752 = 17.5..., then 4.218752 = 17.79..., and at this point we can very firmly state that the number is approximately 4.2. The next lower bound is 4.2343752 = 17.9299..., the next lower bound is 4.24218752 = 17.996... . Then we'll start reducing the upper bound again; you can find eventually that the answer is 4.242640687119... and you'll get one decimal digit for every three or four of these operations you do.

There is a better way. If you followed that example you'll notice that we discovered that 4.252 was unusually close to 18, so if you were doing this by hand you probably would have guessed not 4.125 but 4.24 instead. And yeah, maybe you will sometimes be wrong with these guesses, but it beats doing all of that extra work if you're not a computer!

Well a long time ago (in, like, Babylonian days) we discovered exactly what you should guess. The idea is that if you have a guess $g$ that is smaller than the square root of N by a small amount $s$, then you don't know $s$ but you do know that $$N = (g + s)^2 = g^2 + 2 g s + s^2$$. Ignoring the $s^2$ as being (small) * (small) = (too small) gives an approximation for $s$:$$s \approx s_g = (N - g^2) / (2 g)$$ which suggests that you should make your next guess $$g + s_g = \frac{g^2 + N}{2 g} = \frac{g + N/g}{2}$$. This guess even has the nice property that when you start with an underestimate then you get a slight overestimate, and vice versa.

So for that, to get the square root of 18, we start by guessing 4. Our next guess is (4 + 18/4)/2 = 17/4, the same "good guess" we got before: we now know that the number is between 4.0 and 4.25. We do it again to find (17/4 + 18*4/17)/2, which works out to 577/136 or 4.24264..., so we already have our next two digits (because its being between 4.24... and 4.25 means that it's 4.24...). If you do the next step you get 665,857/156,944, which lets you know that it's 4.24264..., and if you do the next step you get 886,731,088,897 / 209,004,522,016, which is so close to the square root that my computer actually can't tell the difference (at double precision), so with one step more you can prove that it's that value (to double precision).

Lazy people should use be careful to avoid having to perform a division in each step of Newton's method. They should consider the equation:

$$\frac{1}{x^2}- \frac{1}{y} = 0$$

This then yields the recursion:

$$x_{n+1} = \frac{x_n}{2}\left(3-\frac{x_n^2}{y}\right)$$

Here, you only have division by fixed numbers in each step.

• Nice. You might add that one could precompute $\frac{1}{y}$ once so that one never needs to perform any division thereafter. Although technically speaking, division has exactly the same computational complexity as multiplication so it's only useful if you're doing it by hand and via decimals. – user21820 May 13 '15 at 3:23
• In case anyone else was thrown by the $y$ in this answer, it is a constant: you're doing Newton-Raphson on $f(x) = x^{-2} - y^{-1}$ to find $x - (x^{-2} - y^{-1}) / (-2 x^{-3})$. – CR Drost May 13 '15 at 4:55

It's come to my attention that I don't actually understand what a square root really is (the operation).

I think your intuition is telling you, correctly, that "square rooting" really isn't an operation—except for under some very special circumstances.

As you probably know, when $n$ is a whole number, an $n$th root of $x$ is defined as a number whose $n$th power is $x$. For example,

• $2$ and $-2$ are both square roots of $4$.
• In the world of complex numbers $1$, $i$, $-1$ and $-i$ are all fourth roots of $1$.

As you can see, a number typically has more than one $n$th root. In fact, in the world of complex numbers, every number has exactly $n$ $n$th roots. (This can be proven, with a bit of work, from the fundamental theorem of algebra.)

Because of this, "find an $n$th root of $x$" isn't an operation in the way that "find the negative of $x$" or "find the cube of $x$" is: it's not an instruction that leads you, inexorably, to a single predetermined number.

If you restrict yourself to working with positive real numbers, however, a miracle happens: every number has exactly one $n$th root! To put it more formally,

For every positive real number $x$, there's exactly one positive real number whose $n$th power is $x$. This number is often called $x^{1/n}$.

So, if you restrict yourself to working with positive real numbers, square rooting becomes an operation, just like negating and cubing.

The only way I know of to take square roots (or nth root, for that matter) it to know the answer!

Very astute! In a broad sense, this is correct: the only way to find an $n$th root of $x$ is to check lots of numbers until you find one whose $n$th power is $x$. Since many useful classes of numbers are infinite, this seems like a hopeless task. How does your calculator manage to do it on command?

The key is the following miraculous property of exponentiation:

If the $n$th power of a complex number $y$ is close to $x$, there's a complex number close to $y$ whose $n$th power is exactly $x$.

If you say carefully what "close" means, this sentence becomes the statement that exponentiation is continuous.

This property means that, to find an $n$th root of $x$, you can start by looking for a number $y_0$ whose $n$th power is close to $x$—a tedious but not hopeless task. Once you succeed, you can look near $y_0$ for a number $y_1$ whose $n$th power is even closer to $x$. Continuing this process yields a sequence of numbers $y_0, y_1, y_2 \ldots$ which approach an actual $n$th root of $x$.

By refining your guesses in clever ways, as described in other answers, you can make this process very fast. Your calculator probably uses one of these clever guessing schemes to find square roots.

If you fix a method for refining your guesses, finding an $n$th root of $x$ becomes sort of a de facto operation, in the sense that the $n$th root you come up with depends only on your initial guess $y_0$. The relationship between the initial guess and the root found, however, can be very complicated!

If you use Newton's method for refining your guesses, plotting the relationship between the initial guess and the root found produces a picture called a Newton fractal.

If you use a positive real number as an initial guess for Newton's method, all the refinements will also be positive reals, leading you to the positive real square root.

Though we've been working mostly in the world of complex numbers, similar discussions can be had in lots of other places. Here are some examples.

• In the world of integers modulo nine, $4$ and $5$ are both square roots of $7$. In this world, numbers that look close together can have wildly different powers, so the guess-and-refine method of finding $n$th roots doesn't work!
• In the world of linear transformations of the plane, rotation by $\pi/6$, rotation by $\pi/3$, and reflection across the origin are all cube roots of reflection across the origin. Every positive-definite transformation has a unique positive-definite $n$th root, just as every positive real number has a unique positive real square root.
• In the ultra-fancy world of pseudo-differential operators, the operator $\left(\frac{d}{dx}\right)^2 - 1$ has at least one square root, whose power series begins $$\tfrac{d}{dx} - \tfrac{1}{2}\left(\tfrac{d}{dx}\right)^{-1} - \tfrac{1}{8} \left(\tfrac{d}{dx}\right)^{-3} - \tfrac{1}{16} \left(\tfrac{d}{dx}\right)^{-5} - \tfrac{5}{128} \left(\tfrac{d}{dx}\right)^{-5} - \ldots$$

No matter how much we generalize our concept of numbers, the idea of an $n$th root, and the question of when "find the $n$th root of $x$" is really an operation, always seem to stay close by!

You should study the definitions of logarithms and exponential functions to understand not only the square root but expressions like $a^x$ when $x$ is rational. Spivak's Calculus book has an amazing chapter about that. After you learn some basic definitions and useful theorems finally you will realize that if $a>0$ then for all $x$, you can write $a^x=(e^{\ln a})^x=e^{x\ln a}$. For instance, the square root of 4 can be defined as $e^{1/2\ln4}=2$.

I don't believe any computer actually does it this way, but you can compute a continued fraction to get an exact representation of a square root. For example, since $10^2$ $<$ $111$ $<$ $11^2$:

\begin{eqnarray} \sqrt{111} &=& 10 + (\sqrt{111} - 10) \\ &=& 10 + \frac{11}{\sqrt{111} + 10} = 10 + \frac{1}{\frac{\sqrt{111} + 10}{11}} \\ \frac{\sqrt{111} + 10}{11} &=& 1 + \frac{\sqrt{111} - 1}{11} \\ &=& 1 + \frac{10}{\sqrt{111} + 1} = 1 + \frac{1}{\frac{\sqrt{111} + 1}{10}} \\ \frac{\sqrt{111} + 1}{10} &=& 1 + \frac{\sqrt{111} - 9}{10} \\ &=& 1 + \frac{3}{\sqrt{111} + 9} = 1 + \frac{1}{\frac{\sqrt{111} + 9}{3}} \\ \frac{\sqrt{111} + 9}{3} &=& 6 + \frac{1}{\frac{\sqrt{111} + 9}{10}} \\ \frac{\sqrt{111} + 9}{10} &=& 1 + \frac{1}{\frac{\sqrt{111} + 1}{11}} \\ \frac{\sqrt{111} + 1}{11} &=& 1 + \frac{1}{\sqrt{111} + 10} \\ \sqrt{111} + 10 &=& 20 + \frac{1}{\frac{\sqrt{111} + 10}{11}} \end{eqnarray}

This yields the continued fraction $[10, \dot1, 1, 6, 1, 1, \dot{20}]$, and it looks like:

$$\sqrt{111} = 10 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{6 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{20 + \cfrac{1}{...}}}}}}}$$

Type $y = x^2$ into Wolfram Alpha. You'll see a continuous curve. Now consider the line $y = a$. The line and curve intersect at two points $(\sqrt{a},a)$ and $(-\sqrt{a},a)$. One algorithm for finding one of those points is decimal search. Let's say you want to find $\sqrt{5}$. What's the unit digit? Is it $0$, $1$, $2$, $3$ or something else? $0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 9$ so the unit digit is $2$. Now what's in the tenths position? $2.0^2 = 4, 2.1^2 = 4.41, 2.2^2 = 4.84, 2.3^2 = 5.29$ so to $2$ significant figures it's $2.2$. You can continue this search in the hundredths position. A much more efficient method which I might find time to write about is the Newton-Raphson method, which approximates the $y=x^2$ curve at some point as a line.

If you limit yourself to positive real numbers, one good way of thinking about powers in in terms of growth. At time zero you have $1\ \text{cm}^2$ of bacteria, and that bacteria grows at a constant rate so that you have $x\ \text{cm}^2$ of bacteria in one hour, $x^2$ in two hours, and so on. The area of bacteria you have after 30 minutes (half an hour), is $\sqrt{x}$.

Another place square roots occur is as distances. The distance of a point $(x_1,x_2,\ldots, x_n)$ from the origin in $\mathbb{R}^n$ is $\sqrt{x_1^2+x_2^2+\cdots + x_n^2}$ as a result of the Pythagoras Theorem.

A third place is that if a square has area $x\ \text{cm}^2$, then the length of one of its sides is $x$ cm.

Most square roots in mathematics come from one of the above sources. A list of exceptions is given in answers to this question.

It sounds to me that the OP understands the mathematical implications of a square root, and is asking where are they found in nature. These are some of the answers:

1. Given a square of area N, the square root of N is the length of a side.
2. Given two bodies orbiting in a synchronized fashion (such as the moons of Jupiter), the square root of their periods is their relative distances from the center of mass.
3. Given two supernovae, the square root of their relative brightnesses is their relative distances.
4. Given two ducks, the square root of their relative quacking loudness is their relative distances.

The point being, that the relationship of N / N^2 is commonly found in nature, everywhere from geometry of crystal formation, to orbital mechanics, to magnetic and electrical phenomenon, to the decay of energy fields. Even art and music often find such relationships common.

Square root $x$ of $a$ satisfies:

$$x^2 =a$$

For integers, you can simply iterate through all the possible choices to find it, or to find out that it doesn't exist.

You may also write it rationally:

$$(\frac{x_1}{x_2})^2=\frac{a_1}{a_2}$$

This number does not generally exist for rational numbers either, and as such does not always have a numeric decimal representation (for example, choose $a=2$). You can always get closer by using one of the algorithms, for example the Newton like algorithms which have the general form:

$$x_{n+1}=(x_n^2-a)f(x_n)$$

But as said, there may not be an exact solution. Yet rationally we can at least get as close as we would like.

And as such, for real numbers there is a solution: $\sqrt x$ (since what ever you can get arbitrarily close can be viewed as a definition of a real number, to be exact the Cauchy sequence definition). However, irrational numbers do not have a decimal representation (at least, not a finite one), so you are stuck with the symbol, approximation, or no solution at all, depending on the task at hand.