Python: how to retain accuracy of square of sqrt()?

Question

I thought this is a popular question but I have not been able to find an answer:

When square root of an integer is squared, I expect the result is an integer but I get the following:

from math import sqrt
from decimal import Decimal 

a = 5
b = sqrt(Decimal(a))
c = 2 * b ** 2
print(a, c)
if c.is_integer():
    print('c is integer')
    do something ...
else:
    print('c is NOT integer')

and the result:

5 10.000000000000002
c is NOT integer

I also tried without decimal.Decimal() but it did not work either. What is the best way to get around this? The above code is simplified for the question but I would like a generic solution if possible. I am using Python 3.5.

Answer 1

If you want an 'accurate' representation of sqrt(5) you'll have to acquaint yourself with a CAS or computer algebra system such as sympy.

>>> import sympy
>>> s = sympy.sqrt(5)
>>> s
sqrt(5)
>>> s*s
5
>>> type(s*s)
<class 'sympy.core.numbers.Integer'>
>>> float(s)
2.23606797749979

sqrt(5) is a so called irrational number. This means it cannot be written as a fraction. With their normal arithmetic computers can only store fractions and not even all of those, so on that road you are out of luck.

Here is a translation of your snippet to sympy, demonstrating the function simplify:

import sympy

def check_int(x, name='input'):
    print('ah', name, '=', x)
    sx = sympy.simplify(x)
    if sx != x:
        print('let me think...')
        print("that's", sx)
    if isinstance(sx, (int, sympy.Integer)):
        print(name, 'is integer')
    else:
        print(name, 'is NOT integer')
    print()

s = sympy.sqrt(5)
t = sympy.sqrt(3)
check_int(5)
check_int(s)
check_int(s**2)
check_int((s+1)**2)
check_int((s+1)**2-2*s)
check_int((s-t)**2)

Output:

ah input = 5
input is integer

ah input = sqrt(5)
input is NOT integer

ah input = 5
input is integer

ah input = (1 + sqrt(5))**2
input is NOT integer

ah input = -2*sqrt(5) + (1 + sqrt(5))**2
let me think...
that's 6
input is integer

ah input = (-sqrt(3) + sqrt(5))**2
let me think...
that's -2*sqrt(15) + 8
input is NOT integer

Answer 2

You should get familiar with how real numbers are stored in computers. Long story short, they either have a finite precision (and you just deal with it) or you need to do symbolic calculations (which is not used in everyday software and still has its limitations).

If in your specific problem domain you really need an accurate square root (and later accurate square of your square root), you could probably invent your own type SquareRoot and overload necessary operators. Please note that that won't work for whole math, like roots of third degree (x^(1/3)) or anything else.