NumPy

>>> import numpy as np
>>> np.sqrt(25)
5.0
>>> np.sqrt([2, 3, 4])
array([1.41421356, 1.73205081, 2.        ])

docs

Negative

For negative reals, it’ll return nan, so np.emath.sqrt() is available for that case.

>>> a = np.array([4, -1, np.inf])
>>> np.sqrt(a)
<stdin>:1: RuntimeWarning: invalid value encountered in sqrt
array([ 2., nan, inf])
>>> np.emath.sqrt(a)
array([ 2.+0.j,  0.+1.j, inf+0.j])

Another option, of course, is to convert to complex first:

>>> a = a.astype(complex)
>>> np.sqrt(a)
array([ 2.+0.j,  0.+1.j, inf+0.j])

SymPy

Depending on your goal, it might be a good idea to delay the calculation of square roots for as long as possible. SymPy might help.

SymPy is a Python library for symbolic mathematics.

import sympy
sympy.sqrt(2)
# => sqrt(2)

This doesn’t seem very useful at first.

But sympy can give more information than floats or Decimals:

sympy.sqrt(8) / sympy.sqrt(27)
# => 2*sqrt(6)/9

Also, no precision is lost. (√2)² is still an integer:

s = sympy.sqrt(2)
s**2
# => 2
type(s**2)
#=> <class 'sympy.core.numbers.Integer'>

In comparison, floats and Decimals would return a number which is very close to 2 but not equal to 2:

(2**0.5)**2
# => 2.0000000000000004

from decimal import Decimal
(Decimal('2')**Decimal('0.5'))**Decimal('2')
# => Decimal('1.999999999999999999999999999')

Sympy also understands more complex examples like the Gaussian integral:

from sympy import Symbol, integrate, pi, sqrt, exp, oo
x = Symbol('x')
integrate(exp(-x**2), (x, -oo, oo))
# => sqrt(pi)
integrate(exp(-x**2), (x, -oo, oo)) == sqrt(pi)
# => True

Finally, if a decimal representation is desired, it’s possible to ask for more digits than will ever be needed:

sympy.N(sympy.sqrt(2), 1_000_000)
# => 1.4142135623730950488016...........2044193016904841204

Binary search

Disclaimer: this is for a more specialised use-case. This method might not be practical in all circumstances.

Benefits:

• can find integer values (i.e. which integer is the root?)
• no need to convert to float, so better precision (can be done that well too)

I personally implemented this one for a crypto CTF challenge (RSA cube root attack),where I needed a precise integer value.

The general idea can be extended to any other root.

def int_squareroot(d: int) -> tuple[int, bool]:
    """Try calculating integer squareroot and return if it's exact"""
    left, right = 1, (d+1)//2
    while left<right-1:
        x = (left+right)//2
        if x**2 > d:
            left, right = left, x
        else:
            left, right = x, right
    return left, left**2==d

EDIT:

As @wjandrea have also pointed out, **this example code can NOT compute **. This is a side-effect of the fact that it does not convert anything into floats, so no precision is lost. If the root is an integer, you get that back. If it’s not, you get the biggest number whose square is smaller than your number. I updated the code so that it also returns a bool indicating if the value is correct or not, and also fixed an issue causing it to loop infinitely (also pointed out by @wjandrea). This implementation of the general method still works kindof weird for smaller numbers, but above 10 I had no problems with.

Overcoming the issues and limits of this method/implementation:

For smaller numbers, you can just use all the other methods from other answers. They generally use floats, which might be a loss of precision, but for small integers that should mean no problem at all. All of those methods that use floats have the same (or nearly the same) limit from this.

If you still want to use this method and get float results, it should be trivial to convert this to use floats too. Note that that will reintroduce precision loss, this method’s unique benefit over the others, and in that case you can also just use any of the other answers. I think the newton’s method version converges a bit faster, but I’m not sure.

For larger numbers, where loss of precision with floats come into play, this method can give results closer to the actual answer (depending on how big is the input). If you want to work with non-integers in this range, you can use other types, for example fixed precision numbers in this method too.

Edit 2, on other answers:

Currently, and afaik, the only other answer that has similar or better precision for large numbers than this implementation is the one that suggest SymPy, by Eric Duminil. That version is also easier to use, and work for any kind of number, the only downside is that it requires SymPy. My implementation is free from any huge dependencies if that is what you are looking for.

Python’s fractions module and its class, Fraction, implement arithmetic with rational numbers. The Fraction class doesn’t implement a square root operation, because most square roots are irrational numbers. However, it can be used to approximate a square root with arbitrary accuracy, because a Fraction‘s numerator and denominator are arbitrary-precision integers.

The following method takes a positive number x and a number of iterations, and returns upper and lower bounds for the square root of x.

from fractions import Fraction

def sqrt(x, n):
    x = x if isinstance(x, Fraction) else Fraction(x)
    upper = x + 1
    for i in range(0, n):
        upper = (upper + x/upper) / 2
    lower = x / upper
    if lower > upper:
        raise ValueError("Sanity check failed")
    return (lower, upper)

See the reference below for details on this operation’s implementation. It also shows how to implement other operations with upper and lower bounds (although there is apparently at least one error with the log operation there).

• Daumas, M., Lester, D., Muñoz, C., "Verified Real Number Calculations: A Library for Interval Arithmetic", arXiv:0708.3721 [cs.MS], 2007.

Alternatively, using Python’s math.isqrt, we can calculate a square root to arbitrary precision:

• Square root of i within 1/2ⁿ of the correct value, where i is an integer:Fraction(math.isqrt(i * 2**(n*2)), 2**n).
• Square root of i within 1/10ⁿ of the correct value, where i is an integer:Fraction(math.isqrt(i * 10**(n*2)), 10**n).
• Square root of x within 1/2ⁿ of the correct value, where x is a multiple of 1/2ⁿ:Fraction(math.isqrt(x * 2**(n)), 2**n).
• Square root of x within 1/10ⁿ of the correct value, where x is a multiple of 1/10ⁿ:Fraction(math.isqrt(x * 10**(n)), 10**n).

In the foregoing, i or x must be 0 or greater.

Newton’s method

Most simple and accurate way to compute square root is Newton’s method.

You have a number which you want to compute its square root (num) and you have a guess of its square root (estimate). Estimate can be any number bigger than 0, but a number that makes sense shortens the recursive call depth significantly.

new_estimate = (estimate + num/estimate) / 2

This line computes a more accurate estimate with those 2 parameters. You can pass new_estimate value to the function and compute another new_estimate which is more accurate than the previous one or you can make a recursive function definition like this.

def newtons_method(num, estimate):
    # Computing a new_estimate
    new_estimate = (estimate + num/estimate) / 2
    print(new_estimate)
    # Base Case: Comparing our estimate with built-in functions value
    if new_estimate == math.sqrt(num):
        return True
    else:
        return newtons_method(num, new_estimate)

For example we need to find 30’s square root. We know that the result is between 5 and 6.

newtons_method(30,5)

number is 30 and estimate is 5. The result from each recursive calls are:

5.5
5.477272727272727
5.4772255752546215
5.477225575051661

The last result is the most accurate computation of the square root of number. It is the same value as the built-in function math.sqrt().

This answer was originally posted by gunesevitan, but is now deleted.

Arbitrary precision square root

This variation uses string manipulations to convert a string which represents a decimal floating-point number to an int, calls math.isqrt to do the actual square root extraction, and then formats the result as a decimal string. math.isqrt rounds down, so all produced digits are correct.

The input string, num, must use plain float format: ‘e’ notation is not supported. The num string can be a plain integer, and leading zeroes are ignored.

The digits argument specifies the number of decimal places in the result string, i.e., the number of digits after the decimal point.

from math import isqrt

def str_sqrt(num, digits):
    """ Arbitrary precision square root

        num arg must be a string
        Return a string with `digits` after
        the decimal point

        Written by PM 2Ring 2022.01.26
    """

    int_part , _, frac_part = num.partition('.')
    num = int_part + frac_part

    # Determine the required precision
    width = 2 * digits - len(frac_part)

    # Truncate or pad with zeroes
    num = num[:width] if width < 0 else num + '0' * width
    s = str(isqrt(int(num)))

    if digits:
        # Pad, if necessary
        s = '0' * (1 + digits - len(s)) + s
        s = f"{s[:-digits]}.{s[-digits:]}"
    return s

Test

print(str_sqrt("2.0", 30))

Output

1.414213562373095048801688724209

For small numbers of digits, it’s faster to use decimal.Decimal.sqrt. Around 32 digits or so, str_sqrt is roughly the same speed as Decimal.sqrt. But at 128 digits, str_sqrt is 2.2× faster than Decimal.sqrt, at 512 digits, it’s 4.3× faster, at 8192 digits, it’s 7.4× faster.

Here’s a live version running on the SageMathCell server.

find square-root of a number

while True:
    num = int(input("Enter a number:n>>"))
    for i in range(2, num):
        if num % i == 0:
            if i*i == num:
                print("Square root of", num, "==>", i)
                break
    else:
        kd = (num**0.5)  # (num**(1/2))
        print("Square root of", num, "==>", kd)

OUTPUT:-

Enter a number: 24
Square root of 24 ==> 4.898979485566356
Enter a number: 36
Square root of 36 ==> 6
Enter a number: 49
Square root of 49 ==> 7

✔ Output CLICK BELOW & SEE ✔

One way to calculate square roots is by using logarithm

let n be a number and d be the square root of that number:

Let us prove mathematically:

n(1/2) = d

apply log2 on both the sides of the equation

log2(n(1/2)) = log2(d)

Now it looks like this:

log2(d) = 1/2 * log2(n)

Or

d = 2(1/2 * log2(n))

So we can write programmatically in python as:

from math import log2
d = 2**(0.5*log2(n))

Square root of 9

2**(0.5*log2(9))
#3.0

Square root of 0.000000000000000009012345362

2**(0.5*log2(0.000000000000000009012345362))
#3.002056855224434e-09

A correctly rounded, arbitrary-precision square root of any int, float, or Fraction, without looping or converting to Decimal

(Copied mostly from my original answer here)

Note: requires python 3.8+ as I’m using math.isqrt under the hood to work the magic.

def sqrt(x: Union[int, float, Fraction], precision: int = 53) -> Fraction:
    a, b = x.as_integer_ratio()
    d = a.bit_length() - b.bit_length()
    s = max(precision - 1 - (d-(b<<(d>0 and d) > a<<(d<0 and -d)) >> 1), 0)
    a <<= s << 1
    n0 = math.isqrt(a // b)
    n1 = n0 + 1
    return Fraction(n1 if n0 * n1 * b < a else n0, 1 << s)

More precisely, the following are guaranteed to hold:

• |√(x) - sqrt(x, precision=p)| < 0.5ulpₚ(√(x))
• sqrt(float(x), precision=53) == math.sqrt(x)
• sqrt(x * x) == x if x is an integer, avoiding this problem
• More generally, sqrt(x, precision=p) will be correctly rounded to precision max(p, floor(x).bit_length())
• For the mathematically curious: d-(b<<(d>0 and d) > a<<(d<0 and -d)) >> 1 is equal to ⌊log₂(√(a/b))⌋ for a > 0.

If you need the result as a float instead of a Fraction, just do float(sqrt(x)) (although this may lose precision or overflow if the final result is too big for a float).

Note: if you are only operating on integers, there is a slightly simpler function, which is equivalent to the above for any integer or Fraction with a denominator of 1:

def sqrt_of_int(x: int, precision: int = 53) -> Fraction:
    s = max(precision - (x.bit_length() + 1 >> 1), 0)
    x <<= s << 1
    n0 = math.isqrt(x)
    n1 = n0 + 1
    return Fraction(n1 if n0 * n1 < x else n0, 1 << s)

An alternative implementation of the original function could then be as follows:

def sqrt(x: Union[int, float, Fraction], precision: int = 53) -> Fraction:
    a, b = x.as_integer_ratio()
    return sqrt_of_int(a, precision) / sqrt_of_int(b, precision)

This version has the advantage that sqrt(a/b)*sqrt(b/a) == 1, which the original does not guarantee. On the other hand, I’m not sure if the rounding guarantees hold anymore.