# Methods of computing square roots

Methods of computing square roots are numerical analysis algorithms for finding the principal, or non-negative, square root (usually denoted S, 2S, or S1/2) of a real number. Arithmetically, it means given S, a procedure for finding a number which when multiplied by itself, yields S; algebraically, it means a procedure for finding the non-negative root of the equation x2 - S = 0; geometrically, it means given the area of a square, a procedure for constructing a side of the square.

Every real number has two square roots.[Note 1]. The principal square root of most numbers is an irrational number or repeating decimal, so that any computed result is a finite approximation to the square root. For some numbers, the square root is finitely representable in the base (i.e. decimal or binary); for these numbers, including integers known as perfect squares, the square root is exact. However, even if the square root is exact, the procedure used to compute it may return an approximation.

The most common analytical methods are iterative and consist of two steps: finding a suitable starting value, followed by iterative refinement until some termination criteria is met. The starting value can be any number, but fewer interations will be required the closer it is to the final result. The most familiar such method, most suited for programmatic calculation, is Newton's method, which is based on a property of the derivative in the calculus. A few methods like paper-and-pencil synthetic division and series expansion, do not require a starting value. In some applications, an integer square root is required, which is the square root rounded or truncated to the nearest integer (a modified procedure may be employed in this case).

The method employed depends on what the result is to be used for (i.e. how accurate it has to be), how much effort one is willing to put into the procedure, and what tools are at hand. The methods may be roughly classified as those suitable for mental calculation, those usually requiring at least paper and pencil, and those which are implemented as programs to be executed on a digital electronic computer or other computing device. Algorithms may take into account convergence (how many iterations are required to achieve a specified precision), computational complexity of individual operations (i.e. division) or iterations, and error propagation (the accuracy of the final result).

Procedures for finding square roots (particularly the square root of 2) have been known since at least the period of ancient Babylon in the 17th century BCE. Modern analytic methods began to be developed after introduction of the Arabic numeral system to western Europe in the early Renaissance. Today, nearly all computing devices have a fast and accurate square root function, either as a programming language construct, a compiler intrinsic or library function, or as a hardware operator, based on one of the described procedures.

## Initial estimate

Many iterative square root algorithms require an initial seed value. If the initial seed value is far away from the actual square root, the algorithm will require more iterations. It is therefore useful to have a rough estimate, which may have limited accuracy but is easy to calculate. In general, the better the initial estimate, the faster the convergence.

Methods of estimating include scalar, linear, hyperbolic and logarithmic. A decimal base is usually used for mental or paper-and-pencil estimating. A binary base is more suitable for computer estimates. In estimating, the exponent and mantissa are usually treated separately, as the number would be expressed in scientific notation.

Simple estimates require no calculation - one can use the initial number, or the initial number divided by 2, since we know that the square root is smaller than the number itself (at least for numbers > 1). However, these are very inaccurate and likely to cost more effort to reach a usable result.

### Decimal estimates

Typically the number $S$  is expressed in scientific notation as $a\times 10^{2n}$  where $1\leq a<100$  and n is an integer, and the range of possible square roots is ${\sqrt {a}}\times 10^{n}$  where $1\leq {\sqrt {a}}<10$ .

Scalar methods divide the range into intervals, and the estimate in each interval is represented by a single scalar number. If the range is considered as a single interval, the arithmetic mean (5) or geometric mean (${\sqrt {10}}\approx 3.16$ ) times $10^{n}$  are plausible estimates. The absolute and relative error for these will differ. In general, a single scalar will be very inaccurate. Better estimates divide the range into two or more intervals, but scalar estimates have inherently low accuracy.

For two intervals, divided geometrically, the square root ${\sqrt {S}}={\sqrt {a}}\times 10^{n}$  can be estimated as[Note 2]

${\sqrt {S}}\approx {\begin{cases}2\cdot 10^{n}&{\text{if }}a<10,\\6\cdot 10^{n}&{\text{if }}a\geq 10.\end{cases}}$

This estimate has maximum absolute error of $4\cdot 10^{n}$  at a = 100, and maximum relative error of 100% at a = 1.

For example, for $S=125348$  factored as $12.5348\times 10^{4}$ , the estimate is ${\sqrt {S}}\approx 6\cdot 10^{2}=600$ . ${\sqrt {125348}}=354.0$ , an absolute error of 246 and relative error of almost 70%.

A better estimate, and the standard method used, is a linear approximation to the function $y=x^{2}$  over a small arc. If, as above, powers of the base are factored out of the number $S$  and the interval reduced to [1,100], a secant line spanning the arc, or a tangent line somewhere along the arc may be used as the approximation, but a linear regression line intersecting the arc will be more accurate.

### Binary estimates

When working in the binary numeral system (as computers do internally), by expressing $S$  as $a\times 2^{2n}$  where $0.1_{2}\leq a<10_{2}$ , the square root ${\sqrt {S}}={\sqrt {a}}\times 2^{n}$  can be estimated as ${\sqrt {S}}\approx 2^{n}$ , since the geometric mean of the lowest and highest possible values is ${\sqrt {{\sqrt {0.1_{2}}}\cdot {\sqrt {10_{2}}}}}={\sqrt[{4}]{1}}=1$ .

For $S=125348=1\;1110\;1001\;1010\;0100_{2}=1.1110\;1001\;1010\;0100_{2}\times 2^{16}\,$  the binary approximation gives ${\sqrt {S}}\approx 2^{8}=1\;0000\;0000_{2}=256\,.$

## Babylonian method

Graph charting the use of the Babylonian method for approximating a square root of 100 (±10) using starting values x0 = 50, x0 = 1, and x0 = −5. Note that a positive starting value yields the positive root, and a negative starting value the negative root.

Perhaps the first algorithm used for approximating ${\sqrt {S}}$  is known as the Babylonian method, despite there being no direct evidence, beyond informed conjecture, that the eponymous Babylonian mathematicians employed exactly this method. The method is also known as Heron's method, after the first-century Greek mathematician Hero of Alexandria who gave the first explicit description of the method in his AD 60 work Metrica. The basic idea is that if x is an overestimate to the square root of a non-negative real number S then S/x will be an underestimate, or vice versa, and so the average of these two numbers may reasonably be expected to provide a better approximation (though the formal proof of that assertion depends on the inequality of arithmetic and geometric means that shows this average is always an overestimate of the square root, as noted in the article on square roots, thus assuring convergence).

More precisely, if x is our initial guess of ${\sqrt {S}}$  and e is the error in our estimate such that S = (x+ e)2, then we can expand the binomial and solve for

$e={\frac {S-x^{2}}{2x+e}}\approx {\frac {S-x^{2}}{2x}},$  since $e\ll x$ .

Therefore, we can compensate for the error and update our old estimate as

$x+e\approx x+{\frac {S-x^{2}}{2x}}={\frac {S+x^{2}}{2x}}={\frac {{\frac {S}{x}}+x}{2}}\equiv x_{\text{revised}}$

Since the computed error was not exact, this becomes our next best guess. The process of updating is iterated until desired accuracy is obtained. This is a quadratically convergent algorithm, which means that the number of correct digits of the approximation roughly doubles with each iteration. It proceeds as follows:

1. Begin with an arbitrary positive starting value x0 (the closer to the actual square root of S, the better).
2. Let xn + 1 be the average of xn and S/xn (using the arithmetic mean to approximate the geometric mean).
3. Repeat step 2 until the desired accuracy is achieved.

It can also be represented as:

$x_{0}\approx {\sqrt {S}},$
$x_{n+1}={\frac {1}{2}}\left(x_{n}+{\frac {S}{x_{n}}}\right),$
${\sqrt {S}}=\lim _{n\to \infty }x_{n}.$

This algorithm works equally well in the p-adic numbers, but cannot be used to identify real square roots with p-adic square roots; one can, for example, construct a sequence of rational numbers by this method that converges to +3 in the reals, but to −3 in the 2-adics.

### Example

To calculate S, where S = 125348, to six significant figures, use the rough estimation method above to get

{\begin{aligned}{\begin{array}{rlll}x_{0}&=6\cdot 10^{2}&&=600.000\\[0.3em]x_{1}&={\frac {1}{2}}\left(x_{0}+{\frac {S}{x_{0}}}\right)&={\frac {1}{2}}\left(600.000+{\frac {125348}{600.000}}\right)&=404.457\\[0.3em]x_{2}&={\frac {1}{2}}\left(x_{1}+{\frac {S}{x_{1}}}\right)&={\frac {1}{2}}\left(404.457+{\frac {125348}{404.457}}\right)&=357.187\\[0.3em]x_{3}&={\frac {1}{2}}\left(x_{2}+{\frac {S}{x_{2}}}\right)&={\frac {1}{2}}\left(357.187+{\frac {125348}{357.187}}\right)&=354.059\\[0.3em]x_{4}&={\frac {1}{2}}\left(x_{3}+{\frac {S}{x_{3}}}\right)&={\frac {1}{2}}\left(354.059+{\frac {125348}{354.059}}\right)&=354.045\\[0.3em]x_{5}&={\frac {1}{2}}\left(x_{4}+{\frac {S}{x_{4}}}\right)&={\frac {1}{2}}\left(354.045+{\frac {125348}{354.045}}\right)&=354.045\end{array}}\end{aligned}}

Therefore, 125348 ≈ 354.045.

### Convergence

Suppose that x0 > 0 and S > 0. Then for any natural number n, xn > 0. Let the relative error in xn be defined by

$\varepsilon _{n}={\frac {x_{n}}{\sqrt {S}}}-1>-1$

and thus

$x_{n}={\sqrt {S}}\cdot (1+\varepsilon _{n}).$

Then it can be shown that

$\varepsilon _{n+1}={\frac {\varepsilon _{n}^{2}}{2(1+\varepsilon _{n})}}\geq 0.$

And thus that

$\varepsilon _{n+2}\leq \min \left\{{\frac {\varepsilon _{n+1}^{2}}{2}},{\frac {\varepsilon _{n+1}}{2}}\right\}$

and consequently that convergence is assured, and quadratic.

#### Worst case for convergence

If using the rough estimate above with the Babylonian method, then the least accurate cases in ascending order are as follows:

{\begin{aligned}S&=1;&x_{0}&=2;&x_{1}&=1.250;&\varepsilon _{1}&=0.250.\\S&=10;&x_{0}&=2;&x_{1}&=3.500;&\varepsilon _{1}&<0.107.\\S&=10;&x_{0}&=6;&x_{1}&=3.833;&\varepsilon _{1}&<0.213.\\S&=100;&x_{0}&=6;&x_{1}&=11.333;&\varepsilon _{1}&<0.134.\end{aligned}}

Thus in any case,

$\varepsilon _{1}\leq 2^{-2}.\,$
$\varepsilon _{2}<2^{-5}<10^{-1}.\,$
$\varepsilon _{3}<2^{-11}<10^{-3}.\,$
$\varepsilon _{4}<2^{-23}<10^{-6}.\,$
$\varepsilon _{5}<2^{-47}<10^{-14}.\,$
$\varepsilon _{6}<2^{-95}<10^{-28}.\,$
$\varepsilon _{7}<2^{-191}<10^{-57}.\,$
$\varepsilon _{8}<2^{-383}<10^{-115}.\,$

Rounding errors will slow the convergence. It is recommended to keep at least one extra digit beyond the desired accuracy of the xn being calculated to minimize round off error.

#### Scaling Heron’s algorithm

Efficiency of Heron algorithm can be enhanced by prescaling.

If $S\in [0.5,2.0]$  then, using a starting guess $x_{0}=1.0$ , Heron algorithm proceeds with quadratic convergence and 5 iterations of the algorithm are sufficient to compute ${\sqrt {S}}$  with 16 digit accuracy.

Prescaling consists in finding an integer number k such that $S=2^{2k}{\tilde {S}}$ , where ${\tilde {S}}\in [0.5,2.0]$ , then solve the intermediate problem $z={\sqrt {\tilde {S}}}$  by performing a fixed number of iterations of the Heron method starting from $z_{0}=1.0$ , finally obtain the solution $x=2^{k}z$ .

This technique is particularly effective when working with floating point numbers in binary format (e.g. IEEE-754), as the determination of k and ${\tilde {S}}$  is trivial. In fact, these numbers are represented in the format $1.xxx\dots 2^{p}$  where mantissa and exponent are immediately available. Then if p is even, then ${\tilde {S}}=1.xxx\dots$  and $k=p/2$ , otherwise ${\tilde {S}}=1.xxx\dots /2.0$  and $k=(p+1)/2$ . Note that there is no floating point mathematical operation involved here, as multiplication and division by powers of 2 can be done with trivial integer operations on the exponent. Similarly, no floating point operation is required when computing the final solution $x=2^{k}z$ .

## Bakhshali method

This method for finding an approximation to a square root was described in an ancient Indian mathematical manuscript called the Bakhshali manuscript. It is equivalent to two iterations of the Babylonian method beginning with x0. Thus, the algorithm is quartically convergent, which means that the number of correct digits of the approximation roughly quadruples with each iteration. The original presentation, using modern notation, is as follows: To calculate ${\sqrt {S}}$ , let x02 be the initial approximation to S. Then, successively iterate as:

$a_{n}={\frac {S-x_{n}^{2}}{2x_{n}}},$
$b_{n}=x_{n}+a_{n},$
$x_{n+1}=b_{n}-{\frac {a_{n}^{2}}{2b_{n}}}.$

Written explicitly, it becomes

$x_{n+1}=(x_{n}+a_{n})-{\frac {a_{n}^{2}}{2(x_{n}+a_{n})}}.$

Let x0 = N be an integer which is the nearest perfect square to S. Also, let the difference d = S - N2, then the first iteration can be written as:

${\sqrt {S}}\approx N+{\frac {d}{2N}}-{\frac {d^{2}}{8N^{3}+4Nd}}={\frac {8N^{4}+8N^{2}d+d^{2}}{8N^{3}+4Nd}}={\frac {N^{4}+6N^{2}S+S^{2}}{4N^{3}+4NS}}={\frac {N^{2}(N^{2}+6S)+S^{2}}{4N(N^{2}+S)}}.$

This gives a rational approximation to the square root.

### Example

Using the same example as given in Babylonian method, let $S=125348.$  Then, the first iterations gives

$x_{0}=600$
$a_{1}={\frac {125348-600^{2}}{2\times 600}}=-195.543$
$b_{1}=600+(-195.543)=404.456$
$x_{1}=404.456-{\frac {(-195.543)^{2}}{2\times 404.456}}=357.186$

Likewise the second iteration gives

$a_{2}={\frac {125348-357.186^{2}}{2\times 357.186}}=-3.126$
$b_{2}=357.186+(-3.126)=354.06$
$x_{2}=354.06-{\frac {(-3.1269)^{2}}{2\times 354.06}}=354.046$

## Digit-by-digit calculation

This is a method to find each digit of the square root in a sequence. It is slower than the Babylonian method, but it has several advantages:

• It can be easier for manual calculations.
• Every digit of the root found is known to be correct, i.e., it does not have to be changed later.
• If the square root has an expansion that terminates, the algorithm terminates after the last digit is found. Thus, it can be used to check whether a given integer is a square number.
• The algorithm works for any base, and naturally, the way it proceeds depends on the base chosen.

Napier's bones include an aid for the execution of this algorithm. The shifting nth root algorithm is a generalization of this method.

### Basic principle

First, consider the case of finding the square root of a number Z, that is the square of a two-digit number XY, where X is the tens digit and Y is the units digit. Specifically:

Z = (10X + Y)2 = 100X2 + 20XY + Y2

Now using the Digit-by-Digit algorithm, we first determine the value of X. X is the largest digit such that X2 is less or equal to Z from which we removed the two rightmost digits.

In the next iteration, we pair the digits, multiply X by 2, and place it in the tenth's place while we try to figure out what the value of Y is.

Since this is a simple case where the answer is a perfect square root XY, the algorithm stops here.

The same idea can be extended to any arbitrary square root computation next. Suppose we are able to find the square root of N by expressing it as a sum of n positive numbers such that

$N=(a_{1}+a_{2}+a_{3}+\dotsb +a_{n})^{2}.$

By repeatedly applying the basic identity

$(x+y)^{2}=x^{2}+2xy+y^{2},$

the right-hand-side term can be expanded as

{\begin{aligned}&(a_{1}+a_{2}+a_{3}+\dotsb +a_{n})^{2}\\=&\,a_{1}^{2}+2a_{1}a_{2}+a_{2}^{2}+2(a_{1}+a_{2})a_{3}+a_{3}^{2}+\dotsb +a_{n-1}^{2}+2\left(\sum _{i=1}^{n-1}a_{i}\right)a_{n}+a_{n}^{2}\\=&\,a_{1}^{2}+[2a_{1}+a_{2}]a_{2}+[2(a_{1}+a_{2})+a_{3}]a_{3}+\dotsb +\left[2\left(\sum _{i=1}^{n-1}a_{i}\right)+a_{n}\right]a_{n}.\end{aligned}}

This expression allows us to find the square root by sequentially guessing the values of $a_{i}$ s. Suppose that the numbers $a_{1},\ldots ,a_{m-1}$  have already been guessed, then the m-th term of the right-hand-side of above summation is given by $Y_{m}=[2P_{m-1}+a_{m}]a_{m},$  where $P_{m-1}=\sum _{i=1}^{m-1}a_{i}$  is the approximate square root found so far. Now each new guess $a_{m}$  should satisfy the recursion

$X_{m}=X_{m-1}-Y_{m},$

such that $X_{m}\geq 0$  for all $1\leq m\leq n,$  with initialization $X_{0}=N.$  When $X_{n}=0,$  the exact square root has been found; if not, then the sum of $a_{i}$ s gives a suitable approximation of the square root, with $X_{n}$  being the approximation error.

For example, in the decimal number system we have

$N=(a_{1}\cdot 10^{n-1}+a_{2}\cdot 10^{n-2}+\cdots +a_{n-1}\cdot 10+a_{n})^{2},$

where $10^{n-i}$  are place holders and the coefficients $a_{i}\in \{0,1,2,\ldots ,9\}$ . At any m-th stage of the square root calculation, the approximate root found so far, $P_{m-1}$  and the summation term $Y_{m}$  are given by

$P_{m-1}=\sum _{i=1}^{m-1}a_{i}\cdot 10^{n-i}=10^{n-m+1}\sum _{i=1}^{m-1}a_{i}\cdot 10^{m-i-1},$
$Y_{m}=[2P_{m-1}+a_{m}\cdot 10^{n-m}]a_{m}\cdot 10^{n-m}=[20\sum _{i=1}^{m-1}a_{i}\cdot 10^{m-i-1}+a_{m}]a_{m}\cdot 10^{2(n-m)}.$

Here since the place value of $Y_{m}$  is an even power of 10, we only need to work with the pair of most significant digits of the remaining term $X_{m-1}$  at any m-th stage. The section below codifies this procedure.

It is obvious that a similar method can be used to compute the square root in number systems other than the decimal number system. For instance, finding the digit-by-digit square root in the binary number system is quite efficient since the value of $a_{i}$  is searched from a smaller set of binary digits {0,1}. This makes the computation faster since at each stage the value of $Y_{m}$  is either $Y_{m}=0$  for $a_{m}=0$  or $Y_{m}=2P_{m-1}+1$  for $a_{m}=1$ . The fact that we have only two possible options for $a_{m}$  also makes the process of deciding the value of $a_{m}$  at m-th stage of calculation easier. This is because we only need to check if $Y_{m}\leq X_{m-1}$  for $a_{m}=1.$  If this condition is satisfied, then we take $a_{m}=1$ ; if not then $a_{m}=0.$  Also, the fact that multiplication by 2 is done by left bit-shifts helps in the computation.

### Decimal (base 10)

Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into pairs, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the square. One digit of the root will appear above each pair of digits of the square.

Beginning with the left-most pair of digits, do the following procedure for each pair:

1. Starting on the left, bring down the most significant (leftmost) pair of digits not yet used (if all the digits have been used, write "00") and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by 100 and add the two digits. This will be the current value c.
2. Find p, y and x, as follows:
• Let p be the part of the root found so far, ignoring any decimal point. (For the first step, p = 0.)
• Determine the greatest digit x such that $x(20p+x)\leq c$ . We will use a new variable y = x(20p + x).
• Note: 20p + x is simply twice p, with the digit x appended to the right).
• Note: x can be found by guessing what c/(20·p) is and doing a trial calculation of y, then adjusting x upward or downward as necessary.
• Place the digit $x$  as the next digit of the root, i.e., above the two digits of the square you just brought down. Thus the next p will be the old p times 10 plus x.
3. Subtract y from c to form a new remainder.
4. If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration.

#### Examples

Find the square root of 152.2756.

          1  2. 3  4
/
\/  01 52.27 56

01                   1*1 <= 1 < 2*2                 x = 1
01                     y = x*x = 1*1 = 1
00 52                22*2 <= 52 < 23*3              x = 2
00 44                  y = (20+x)*x = 22*2 = 44
08 27             243*3 <= 827 < 244*4           x = 3
07 29               y = (240+x)*x = 243*3 = 729
98 56          2464*4 <= 9856 < 2465*5        x = 4
98 56            y = (2460+x)*x = 2464*4 = 9856
00 00          Algorithm terminates: Answer is 12.34


### Binary numeral system (base 2)

Inherent to digit-by-digit algorithms is a search and test step: find a digit, $\,e$ , when added to the right of a current solution $\,r$ , such that $\,(r+e)\cdot (r+e)\leq x$ , where $\,x$  is the value for which a root is desired. Expanding: $\,r\cdot r+2re+e\cdot e\leq x$ . The current value of $\,r\cdot r$ —or, usually, the remainder—can be incrementally updated efficiently when working in binary, as the value of $\,e$  will have a single bit set (a power of 2), and the operations needed to compute $\,2\cdot r\cdot e$  and $\,e\cdot e$  can be replaced with faster bit shift operations.

#### Example

Here we obtain the square root of 81, which when converted into binary gives 1010001. The numbers in the left column gives the option between that number or zero to be used for subtraction at that stage of computation. The final answer is 1001, which in decimal is 9.

             1 0 0 1
---------
√ 1010001

1      1
1
---------
101     01
0
--------
1001     100
0
--------
10001    10001
10001
-------
0


This gives rise to simple computer implementations:

short isqrt(short num) {
short res = 0;
short bit = 1 << 14; // The second-to-top bit is set: 1 << 30 for 32 bits

// "bit" starts at the highest power of four <= the argument.
while (bit > num)
bit >>= 2;

while (bit != 0) {
if (num >= res + bit) {
num -= res + bit;
res = (res >> 1) + bit;
}
else
res >>= 1;
bit >>= 2;
}
return res;
}


Using the notation above, the variable "bit" corresponds to $e_{m}^{2}$  which is $(2^{m})^{2}=4^{m}$ , the variable "res" is equal to $2re_{m}$ , and the variable "num" is equal to the current $X_{m}$  which is the difference of the number we want the square root of and the square of our current approximation with all bits set up to $2^{m+1}$ . Thus in the first loop, we want to find the highest power of 4 in "bit" to find the highest power of 2 in $e$ . In the second loop, if num is greater than res + bit, then $X_{m}$  is greater than $2re_{m}+e_{m}^{2}$  and we can subtract it. The next line, we want to add $e_{m}$  to $r$  which means we want to add $2e_{m}^{2}$  to $2re_{m}$  so we want res = res + bit<<1. Then update $e_{m}$  to $e_{m-1}$  inside res which involves dividing by 2 or another shift to the right. Combining these 2 into one line leads to res = res>>1 + bit. If $X_{m}$  isn't greater than $2re_{m}+e_{m}^{2}$  then we just update $e_{m}$  to $e_{m-1}$  inside res and divide it by 2. Then we update $e_{m}$  to $e_{m-1}$  in bit by dividing it by 4. The final iteration of the 2nd loop has bit equal to 1 and will cause update of $e$  to run one extra time removing the factor of 2 from res making it our integer approximation of the root.

Faster algorithms, in binary and decimal or any other base, can be realized by using lookup tables—in effect trading more storage space for reduced run time.

## Exponential identity

Pocket calculators typically implement good routines to compute the exponential function and the natural logarithm, and then compute the square root of S using the identity found using the properties of logarithms ($\ln x^{n}=n\ln x$ ) and exponentials ($e^{\ln x}=x$ ):

${\sqrt {S}}=e^{{\frac {1}{2}}\ln S}.$

The denominator in the fraction corresponds to the nth root. In the case above the denominator is 2, hence the equation specifies that the square root is to be found. The same identity is used when computing square roots with logarithm tables or slide rules.

## A two-variable iterative method

This method is applicable for finding the square root of $0  and converges best for $S\approx 1$ . This, however, is no real limitation for a computer based calculation, as in base 2 floating point and fixed point representations, it is trivial to multiply $S\,\!$  by an integer power of 4, and therefore ${\sqrt {S}}$  by the corresponding power of 2, by changing the exponent or by shifting, respectively. Therefore, $S\,\!$  can be moved to the range ${\frac {1}{2}}\leq S<2$ . Moreover, the following method does not employ general divisions, but only additions, subtractions, multiplications, and divisions by powers of two, which are again trivial to implement. A disadvantage of the method is that numerical errors accumulate, in contrast to single variable iterative methods such as the Babylonian one.

The initialization step of this method is

$a_{0}=S\,\!$
$c_{0}=S-1\,\!$

$a_{n+1}=a_{n}-a_{n}c_{n}/2\,\!$
$c_{n+1}=c_{n}^{2}(c_{n}-3)/4\,\!$

Then, $a_{n}\rightarrow {\sqrt {S}}$  (while $c_{n}\rightarrow 0$ ).

Note that the convergence of $c_{n}\,\!$ , and therefore also of $a_{n}\,\!$ , is quadratic.

The proof of the method is rather easy. First, rewrite the iterative definition of $c_{n}\,\!$  as

$1+c_{n+1}=(1+c_{n})(1-c_{n}/2)^{2}\,\!$ .

Then it is straightforward to prove by induction that

$S(1+c_{n})=a_{n}^{2}$

and therefore the convergence of $a_{n}\,\!$  to the desired result ${\sqrt {S}}$  is ensured by the convergence of $c_{n}\,\!$  to 0, which in turn follows from $-1 .

This method was developed around 1950 by M. V. Wilkes, D. J. Wheeler and S. Gill for use on EDSAC, one of the first electronic computers. The method was later generalized, allowing the computation of non-square roots.

## Iterative methods for reciprocal square roots

The following are iterative methods for finding the reciprocal square root of S which is $1/{\sqrt {S}}$ . Once it has been found, find ${\sqrt {S}}$  by simple multiplication: ${\sqrt {S}}=S\cdot (1/{\sqrt {S}})$ . These iterations involve only multiplication, and not division. They are therefore faster than the Babylonian method. However, they are not stable. If the initial value is not close to the reciprocal square root, the iterations will diverge away from it rather than converge to it. It can therefore be advantageous to perform an iteration of the Babylonian method on a rough estimate before starting to apply these methods.

• Applying Newton's method to the equation $(1/x^{2})-S=0$  produces a method that converges quadratically using three multiplications per step:
$x_{n+1}={\frac {x_{n}}{2}}\cdot (3-S\cdot x_{n}^{2}).$
$y_{n}=S\cdot x_{n}^{2}$ , and
$x_{n+1}={\frac {x_{n}}{8}}\cdot (15-y_{n}\cdot (10-3\cdot y_{n}))$ .

### Goldschmidt’s algorithm

Some computers use Goldschmidt's algorithm to simultaneously calculate ${\sqrt {S}}$  and $1/{\sqrt {S}}$ . Goldschmidt's algorithm finds ${\sqrt {S}}$  faster than Newton-Raphson iteration on a computer with a fused multiply–add instruction and either a pipelined floating point unit or two independent floating-point units.

The first way of writing Goldschmidt's algorithm begins

$b_{0}=S$
$Y_{0}\approx 1/{\sqrt {S}}$  (typically using a table lookup)
$y_{0}=Y_{0}$
$x_{0}=Sy_{0}$

and iterates

$b_{n+1}=b_{n}Y_{n}^{2}$
$Y_{n+1}=(3-b_{n+1})/2$
$x_{n+1}=x_{n}Y_{n+1}$
$y_{n+1}=y_{n}Y_{n+1}$

until $b_{i}$  is sufficiently close to 1, or a fixed number of iterations. The iterations converge to

$\lim _{n\to \infty }x_{n}={\sqrt {S}}$ , and
$\lim _{n\to \infty }y_{n}=1/{\sqrt {S}}$ .

Note that it is possible to omit either $x_{n}$  and $y_{n}$  from the computation, and if both are desired then $x_{n}=Sy_{n}$  may be used at the end rather than computing it through in each iteration.

A second form, using fused multiply-add operations, begins

$y_{0}\approx 1/{\sqrt {S}}$  (typically using a table lookup)
$x_{0}=Sy_{0}$
$h_{0}=y_{0}/2$

and iterates

$r_{n}=0.5-x_{n}h_{n}$
$x_{n+1}=x_{n}+x_{n}r_{n}$
$h_{n+1}=h_{n}+h_{n}r_{n}$

until $r_{i}$  is sufficiently close to 0, or a fixed number of iterations. This converges to

$\lim _{n\to \infty }x_{n}={\sqrt {S}}$ , and
$\lim _{n\to \infty }2h_{n}=1/{\sqrt {S}}$ .

## Taylor series

If N is an approximation to ${\sqrt {S}}$ , a better approximation can be found by using the Taylor series of the square root function:

${\sqrt {N^{2}+d}}=N\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)n!^{2}4^{n}}}{\frac {d^{n}}{N^{2n}}}=N(1+{\frac {d}{2N^{2}}}-{\frac {d^{2}}{8N^{4}}}+{\frac {d^{3}}{16N^{6}}}-{\frac {5d^{4}}{128N^{8}}}+\cdots )$

As an iterative method, the order of convergence is equal to the number of terms used. With two terms, it is identical to the Babylonian method. With three terms, each iteration takes almost as many operations as the Bakhshali approximation, but converges more slowly.[citation needed] Therefore, this is not a particularly efficient way of calculation. To maximize the rate of convergence, choose N so that ${\frac {|d|}{N^{2}}}\,$  is as small as possible.

## Continued fraction expansion

Quadratic irrationals (numbers of the form ${\frac {a+{\sqrt {b}}}{c}}$ , where a, b and c are integers), and in particular, square roots of integers, have periodic continued fractions. Sometimes what is desired is finding not the numerical value of a square root, but rather its continued fraction expansion, and hence its rational approximation. Let S be the positive number for which we are required to find the square root. Then assuming a to be a number that serves as an initial guess and r to be the remainder term, we can write $S=a^{2}+r.$  Since we have $S-a^{2}=({\sqrt {S}}+a)({\sqrt {S}}-a)=r$ , we can express the square root of S as

${\sqrt {S}}=a+{\frac {r}{a+{\sqrt {S}}}}.$

By applying this expression for ${\sqrt {S}}$  to the denominator term of the fraction, we have

${\sqrt {S}}=a+{\frac {r}{a+(a+{\frac {r}{a+{\sqrt {S}}}})}}=a+{\frac {r}{2a+{\frac {r}{a+{\sqrt {S}}}}}}.$

Proceeding this way, we get a generalized continued fraction for the square root as

${\sqrt {S}}=a+{\frac {r|}{|2a}}+{\frac {r|}{|2a}}+{\frac {r|}{|2a}}+\cdots$

For any S a possible choice for a and r is a = 1 and r = S - 1, yielding

${\sqrt {S}}=1+{\frac {S-1|}{|2}}+{\frac {S-1|}{|2}}+{\frac {S-1|}{|2}}+\cdots$

For example, for the square root of 2, we can take a = 1 and r = 1, giving us

${\sqrt {2}}=1+{\frac {1|}{|2}}+{\frac {1|}{|2}}+{\frac {1|}{|2}}+\cdots$

Taking the first three denominators give the rational approximation of 2 as [1;2,2,2] = 17/12 = 1.41667, correct up to first three decimal places. Taking the first five denominators gives the rational approximation to 2 as [1;2,2,2,2,2] = 99/70 = 1.4142857, correct up to first five decimal places. Taking more denominators give better approximations.

As another example, for the square root of 3, we can select a = 2 and r = -1, giving us

${\sqrt {3}}=2-{\frac {1|}{|4}}-{\frac {1|}{|4}}-{\frac {1|}{|4}}-\cdots$

The first three denominators gives 3 as 1.73214, correct up to the first four decimal places. Note that it is not necessary to choose an integer valued a. For instance, we can take a = 2 and r = 1, such that

${\sqrt {3}}={\sqrt {2}}+{\frac {1|}{|2{\sqrt {2}}}}+{\frac {1|}{|2{\sqrt {2}}}}+{\frac {1|}{|2{\sqrt {2}}}}+\cdots$

We can do the same for the whole numbers as well. For instance,

$2={\sqrt {4}}=1+{\frac {3|}{|2}}+{\frac {3|}{|2}}+{\frac {3|}{|2}}+\cdots$

### Algorithm

The following iterative algorithm can be used to obtain the continued fraction expansion in canonical form (S is any natural number that is not a perfect square):

$m_{0}=0\,\!$
$d_{0}=1\,\!$
$a_{0}=\left\lfloor {\sqrt {S}}\right\rfloor \,\!$
$m_{n+1}=d_{n}a_{n}-m_{n}\,\!$
$d_{n+1}={\frac {S-m_{n+1}^{2}}{d_{n}}}\,\!$
$a_{n+1}=\left\lfloor {\frac {{\sqrt {S}}+m_{n+1}}{d_{n+1}}}\right\rfloor =\left\lfloor {\frac {a_{0}+m_{n+1}}{d_{n+1}}}\right\rfloor \!.$

Notice that mn, dn, and an are always integers. The algorithm terminates when this triplet is the same as one encountered before. The algorithm can also terminate on ai when ai = 2 a0, which is easier to implement.

The expansion will repeat from then on. The sequence [a0; a1, a2, a3, ...] is the continued fraction expansion:

${\sqrt {S}}=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+\,\ddots }}}}}}$

### Example, square root of 114 as a continued fraction

Begin with m0 = 0; d0 = 1; and a0 = 10 (102 = 100 and 112 = 121 > 114 so 10 chosen).

{\begin{aligned}{\sqrt {114}}&={\frac {{\sqrt {114}}+0}{1}}=10+{\frac {{\sqrt {114}}-10}{1}}=10+{\frac {({\sqrt {114}}-10)({\sqrt {114}}+10)}{{\sqrt {114}}+10}}\\&=10+{\frac {114-100}{{\sqrt {114}}+10}}=10+{\frac {1}{\frac {{\sqrt {114}}+10}{14}}}.\end{aligned}}
$m_{1}=d_{0}\cdot a_{0}-m_{0}=1\cdot 10-0=10\,.$
$d_{1}={\frac {S-m_{1}^{2}}{d_{0}}}={\frac {114-10^{2}}{1}}=14\,.$
$a_{1}=\left\lfloor {\frac {a_{0}+m_{1}}{d_{1}}}\right\rfloor =\left\lfloor {\frac {10+10}{14}}\right\rfloor =\left\lfloor {\frac {20}{14}}\right\rfloor =1\,.$

So, m1 = 10; d1 = 14; and a1 = 1.

${\frac {{\sqrt {114}}+10}{14}}=1+{\frac {{\sqrt {114}}-4}{14}}=1+{\frac {114-16}{14({\sqrt {114}}+4)}}=1+{\frac {1}{\frac {{\sqrt {114}}+4}{7}}}.$

Next, m2 = 4; d2 = 7; and a2 = 2.

${\frac {{\sqrt {114}}+4}{7}}=2+{\frac {{\sqrt {114}}-10}{7}}=2+{\frac {14}{7({\sqrt {114}}+10)}}=2+{\frac {1}{\frac {{\sqrt {114}}+10}{2}}}.$
${\frac {{\sqrt {114}}+10}{2}}=10+{\frac {{\sqrt {114}}-10}{2}}=10+{\frac {14}{2({\sqrt {114}}+10)}}=10+{\frac {1}{\frac {{\sqrt {114}}+10}{7}}}.$
${\frac {{\sqrt {114}}+10}{7}}=2+{\frac {{\sqrt {114}}-4}{7}}=2+{\frac {98}{7({\sqrt {114}}+4)}}=2+{\frac {1}{\frac {{\sqrt {114}}+4}{14}}}.$
${\frac {{\sqrt {114}}+4}{14}}=1+{\frac {{\sqrt {114}}-10}{14}}=1+{\frac {14}{14({\sqrt {114}}+10)}}=1+{\frac {1}{\frac {{\sqrt {114}}+10}{1}}}.$
${\frac {{\sqrt {114}}+10}{1}}=20+{\frac {{\sqrt {114}}-10}{1}}=20+{\frac {14}{{\sqrt {114}}+10}}=20+{\frac {1}{\frac {{\sqrt {114}}+10}{14}}}.$

Now, loop back to the second equation above.

Consequently, the simple continued fraction for the square root of 114 is

${\sqrt {114}}=[10;1,2,10,2,1,20,1,2,10,2,1,20,1,2,10,2,1,20,\dots ].\,$  (sequence A010179 in the OEIS)

Its decimal value is approximately 10.67707 82520 31311 21....

### Generalized continued fraction

A more rapid method is to evaluate its generalized continued fraction. From the formula derived there:

${\sqrt {z}}={\sqrt {x^{2}+y}}=x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+\ddots }}}}}}=x+{\cfrac {2x\cdot y}{2(2z-y)-y-{\cfrac {y^{2}}{2(2z-y)-{\cfrac {y^{2}}{2(2z-y)-\ddots }}}}}}$

and the fact that 114 is 2/3 of the way between 102=100 and 112=121 results in

${\sqrt {114}}={\cfrac {\sqrt {1026}}{3}}={\cfrac {\sqrt {32^{2}+2}}{3}}={\cfrac {32}{3}}+{\cfrac {2/3}{64+{\cfrac {2}{64+{\cfrac {2}{64+{\cfrac {2}{64+\ddots }}}}}}}}={\cfrac {32}{3}}+{\cfrac {2}{192+{\cfrac {18}{192+{\cfrac {18}{192+\ddots }}}}}},$

which is simply the aforementioned [10;1,2, 10,2,1, 20,1,2, 10,2,1, 20,1,2, ...] evaluated at every third term. Combining pairs of fractions produces

${\sqrt {114}}={\cfrac {\sqrt {32^{2}+2}}{3}}={\cfrac {32}{3}}+{\cfrac {64/3}{2050-1-{\cfrac {1}{2050-{\cfrac {1}{2050-\ddots }}}}}}={\cfrac {32}{3}}+{\cfrac {64}{6150-3-{\cfrac {9}{6150-{\cfrac {9}{6150-\ddots }}}}}},$

which is now [10;1,2, 10,2,1,20,1,2, 10,2,1,20,1,2, ...] evaluated at the third term and every six terms thereafter.

## Lucas sequence method

the Lucas sequence of the first kind Un(P,Q) is defined by the recurrence relations:

$U_{n}(P,Q)={\begin{cases}0&{\text{if }}n=0\\1&{\text{if }}n=1\\P\cdot U_{n-1}(P,Q)-Q\cdot U_{n-2}(P,Q)&{\text{Otherwise}}\end{cases}}$

and the characteristic equation of it is:

$x^{2}-P\cdot x+Q=0$

it has the discriminant $D=P^{2}-4Q$  and the roots:

${\begin{matrix}x_{1}={\frac {P+{\sqrt {D}}}{2}},&x_{2}={\frac {P-{\sqrt {D}}}{2}}\end{matrix}}$

all that yield the following positive value:

$\lim _{n\to \infty }{\frac {U_{n+1}}{U_{n}}}=x_{1}$

so when we want ${\sqrt {a}}$ , we can choose $P=2$  and $Q=1-a$ , and then calculate $x_{1}=1+{\sqrt {a}}$  using $U_{n+1}$  and $U_{n}$ for large value of $n$ . The most effective way to calculate $U_{n+1}$  and $U_{n}$ is:

${\begin{bmatrix}U_{n}\\U_{n+1}\end{bmatrix}}={\begin{bmatrix}0&1\\-Q&P\end{bmatrix}}\cdot {\begin{bmatrix}U_{n-1}\\U_{n}\end{bmatrix}}={\begin{bmatrix}0&1\\-Q&P\end{bmatrix}}^{n}\cdot {\begin{bmatrix}U_{0}\\U_{1}\end{bmatrix}}$

Summary:

${\begin{bmatrix}0&1\\a-1&2\end{bmatrix}}^{n}\cdot {\begin{bmatrix}0\\1\end{bmatrix}}={\begin{bmatrix}U_{n}\\U_{n+1}\end{bmatrix}}$

then when $n\to \infty$ :

${\sqrt {a}}={\frac {U_{n+1}}{U_{n}}}-1$

## Using Pell's equation

Pell's equation (also known as Brahmagupta equation since he was the first to give a solution to this particular equation[citation needed]) and its variants yield a method for efficiently finding continued fraction convergents of square roots of integers. However, it can be complicated to execute, and usually not every convergent is generated. The ideas behind the method are as follows:

• If (p, q) is a solution (where p and q are integers) to the equation $p^{2}=S\cdot q^{2}\pm 1\!$ , then ${\frac {p}{q}}$  is a continued fraction convergent of ${\sqrt {S}}$ , and as such, is an excellent rational approximation to it.
• If (pa, qa) and (pb, qb) are solutions, then so is:
$p=p_{a}p_{b}+S\cdot q_{a}q_{b}\,\!$
$q=p_{a}q_{b}+p_{b}q_{a}\,\!$
(compare to the multiplication of quadratic integers)
• More generally, if (p1, q1) is a solution, then it is possible to generate a sequence of solutions (pn, qn) satisfying:
$p_{m+n}=p_{m}p_{n}+S\cdot q_{m}q_{n}\,\!$
$q_{m+n}=p_{m}q_{n}+p_{n}q_{m}\,\!$

The method is as follows:

• Find positive integers p1 and q1 such that $p_{1}^{2}=S\cdot q_{1}^{2}\pm 1$ . This is the hard part; It can be done either by guessing, or by using fairly sophisticated techniques.
• To generate a long list of convergents, iterate:
$p_{n+1}=p_{1}p_{n}+S\cdot q_{1}q_{n}\,\!$
$q_{n+1}=p_{1}q_{n}+p_{n}q_{1}\,\!$
• To find the larger convergents quickly, iterate:
$p_{2n}=p_{n}^{2}+S\cdot q_{n}^{2}\,\!$
$q_{2n}=2p_{n}q_{n}\,\!$
Notice that the corresponding sequence of fractions coincides with the one given by the Hero's method starting with $\textstyle {\frac {p_{1}}{q_{1}}}$ .
• In either case, ${\frac {p_{n}}{q_{n}}}$  is a rational approximation satisfying
$\left|{\frac {p_{n}}{q_{n}}}-{\sqrt {S}}\right|<{\frac {1}{q_{n}^{2}\cdot {\sqrt {S}}}}.$

## Approximations that depend on the floating point representation

A number is represented in a floating point format as $m\times b^{p}$  which is also called scientific notation. Its square root is ${\sqrt {m}}\times b^{p/2}$  and similar formulae would apply for cube roots and logarithms. On the face of it, this is no improvement in simplicity, but suppose that only an approximation is required: then just $b^{p/2}$  is good to an order of magnitude. Next, recognise that some powers, p, will be odd, thus for 3141.59 = 3.14159 × 103 rather than deal with fractional powers of the base, multiply the mantissa by the base and subtract one from the power to make it even. The adjusted representation will become the equivalent of 31.4159 × 102 so that the square root will be 31.4159 × 10.

If the integer part of the adjusted mantissa is taken, there can only be the values 1 to 99, and that could be used as an index into a table of 99 pre-computed square roots to complete the estimate. A computer using base sixteen would require a larger table, but one using base two would require only three entries: the possible bits of the integer part of the adjusted mantissa are 01 (the power being even so there was no shift, remembering that a normalised floating point number always has a non-zero high-order digit) or if the power was odd, 10 or 11, these being the first two bits of the original mantissa. Thus, 6.25 = 110.01 in binary, normalised to 1.1001 × 22 an even power so the paired bits of the mantissa are 01, while .625 = 0.101 in binary normalises to 1.01 × 2−1 an odd power so the adjustment is to 10.1 × 2−2 and the paired bits are 10. Notice that the low order bit of the power is echoed in the high order bit of the pairwise mantissa. An even power has its low-order bit zero and the adjusted mantissa will start with 0, whereas for an odd power that bit is one and the adjusted mantissa will start with 1. Thus, when the power is halved, it is as if its low order bit is shifted out to become the first bit of the pairwise mantissa.

A table with only three entries could be enlarged by incorporating additional bits of the mantissa. However, with computers, rather than calculate an interpolation into a table, it is often better to find some simpler calculation giving equivalent results. Everything now depends on the exact details of the format of the representation, plus what operations are available to access and manipulate the parts of the number. For example, Fortran offers an EXPONENT(x) function to obtain the power. Effort expended in devising a good initial approximation is to be recouped by thereby avoiding the additional iterations of the refinement process that would have been needed for a poor approximation. Since these are few (one iteration requires a divide, an add, and a halving) the constraint is severe.

Many computers follow the IEEE (or sufficiently similar) representation, and a very rapid approximation to the square root can be obtained for starting Newton's method. The technique that follows is based on the fact that the floating point format (in base two) approximates the base-2 logarithm. That is $\log _{2}(m\times 2^{p})=p+\log _{2}(m)$

So for a 32-bit single precision floating point number in IEEE format (where notably, the power has a bias of 127 added for the represented form) you can get the approximate logarithm by interpreting its binary representation as a 32-bit integer, scaling it by $2^{-23}$ , and removing a bias of 127, i.e.

$x_{\text{int}}\cdot 2^{-23}-127\approx \log _{2}(x).$

For example, 1.0 is represented by a hexadecimal number 0x3F800000, which would represent $1065353216=127\cdot 2^{23}$  if taken as an integer. Using the formula above you get $1065353216\cdot 2^{-23}-127=0$ , as expected from $\log _{2}(1.0)$ . In a similar fashion you get 0.5 from 1.5 (0x3FC00000).

To get the square root, divide the logarithm by 2 and convert the value back. The following program demonstrates the idea. Note that the exponent's lowest bit is intentionally allowed to propagate into the mantissa. One way to justify the steps in this program is to assume $b$  is the exponent bias and $n$  is the number of explicitly stored bits in the mantissa and then show that

$(((x_{\text{int}}/2^{n}-b)/2)+b)\cdot 2^{n}=(x_{\text{int}}-2^{n})/2+((b+1)/2)\cdot 2^{n}.$

/* Assumes that float is in the IEEE 754 single precision floating point format
* and that int is 32 bits. */
float sqrt_approx(float z)
{
int val_int = *(int*)&z; /* Same bits, but as an int */
/*
* To justify the following code, prove that
*
* ((((val_int / 2^m) - b) / 2) + b) * 2^m = ((val_int - 2^m) / 2) + ((b + 1) / 2) * 2^m)
*
* where
*
* b = exponent bias
* m = number of mantissa bits
*
* .
*/

val_int -= 1 << 23; /* Subtract 2^m. */
val_int >>= 1; /* Divide by 2. */
val_int += 1 << 29; /* Add ((b + 1) / 2) * 2^m. */

return *(float*)&val_int; /* Interpret again as float */
}


The three mathematical operations forming the core of the above function can be expressed in a single line. An additional adjustment can be added to reduce the maximum relative error. So, the three operations, not including the cast, can be rewritten as

val_int = (1 << 29) + (val_int >> 1) - (1 << 22) + a;


where a is a bias for adjusting the approximation errors. For example, with a = 0 the results are accurate for even powers of 2 (e.g., 1.0), but for other numbers the results will be slightly too big (e.g.,1.5 for 2.0 instead of 1.414... with 6% error). With a = -0x4B0D2, the maximum relative error is minimized to ±3.5%.

If the approximation is to be used for an initial guess for Newton's method to the equation $(1/x^{2})-S=0$ , then the reciprocal form shown in the following section is preferred.

### Reciprocal of the square root

A variant of the above routine is included below, which can be used to compute the reciprocal of the square root, i.e., $x^{-{1 \over 2}}$  instead, was written by Greg Walsh. The integer-shift approximation produced a relative error of less than 4%, and the error dropped further to 0.15% with one iteration of Newton's method on the following line. In computer graphics it is a very efficient way to normalize a vector.

float invSqrt(float x)
{
float xhalf = 0.5f*x;
union
{
float x;
int i;
} u;
u.x = x;
u.i = 0x5f375a86 - (u.i >> 1);
/* The next line can be repeated any number of times to increase accuracy */
u.x = u.x * (1.5f - xhalf * u.x * u.x);
return u.x;
}


Some VLSI hardware implements inverse square root using a second degree polynomial estimation followed by a Goldschmidt iteration.

## Negative or complex square

If S < 0, then its principal square root is

${\sqrt {S}}={\sqrt {\vert S\vert }}\,\,i\,.$

If S = a+bi where a and b are real and b ≠ 0, then its principal square root is

${\sqrt {S}}={\sqrt {\frac {\vert S\vert +a}{2}}}\,+\,\operatorname {sgn}(b){\sqrt {\frac {\vert S\vert -a}{2}}}\,\,i\,.$

This can be verified by squaring the root. Here

$\vert S\vert ={\sqrt {a^{2}+b^{2}}}$

is the modulus of S. The principal square root of a complex number is defined to be the root with the non-negative real part.