How do I calculate arbitrary precision square root in Python?

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The result is:
1.4142135623730951
1.41421356237309504880168872420969807856967187537694807317667973799073247846210703 => reference
SQRT OVER

only 16 decimals long and the last digit is also wrong (1 instead of zero)

What I have tried:

I tried this code:

Python

from math import sqrt

def main():
    n = '2.0'
    root2th = sqrt(float(n))
    print(root2th)
    print('1.41421356237309504880168872420969807856967187537694807317667973799073247846210703', '=> reference')

if __name__ == '__main__':
    main()
    print('SQRT OVER')

Posted 1-Jan-23 4:51am

giocip

Updated 1-Jan-23 5:06am

v3

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0x01AA 1-Jan-23 10:56am

Read this: 9.4. decimal — Decimal fixed point and floating point arithmetic — Python 2.7.18 documentation[^]

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0x01AA · Accepted Answer · 2023-01-01T05:06:00

While I have no expirience in Python, a short google lead me to this:

from math import sqrt
# loading decimal library
from decimal import *

def main():
    n = '2.0'
    a = Decimal(2)
    root2th = sqrt(float(n))
    print(root2th)
    print(a.sqrt())
    print('1.41421356237309504880168872420969807856967187537694807317667973799073247846210703', '=> reference')

if __name__ == '__main__':
    main()
    print('SQRT OVER')

The result is:

1.4142135623730951
1.414213562373095048801688724
1.41421356237309504880168872420969807856967187537694807317667973799073247846210703 => reference
SQRT OVER

[Edit]
Arbitrary presicion, see line getcontext().prec = 80

from math import sqrt
# loading decimal library
from decimal import *

def main():
    n = '2.0'
    getcontext().prec = 80
    a = Decimal(2)
    root2th = sqrt(float(n))
    print(root2th)
    print(a.sqrt())
    print('1.41421356237309504880168872420969807856967187537694807317667973799073247846210703', '=> reference')

if __name__ == '__main__':
    main()
    print('SQRT OVER')