Why does the math
module return the wrong result?
A = 12345678917
print \'A =\',A
B = sqrt(A**2)
print \'B =\',int(B)
>
The documentation for the math module states "It provides access to the mathematical functions defined by the C standard." It also states "Except when explicitly noted otherwise, all return values are floats."
Those together mean that the parameter to the square root function is a float value. In most systems that means a floating point value that fits into 8 bytes, which is called "double" in the C language. Your code converts your integer value into such a value before calculating the square root, then returns such a value.
However, the 8-byte floating point value can store at most 15 to 17 significant decimal digits. That is what you are getting in your results.
If you want better precision in your square roots, use a function that is guaranteed to give full precision for an integer argument. Just do a web search and you will find several. Those usually do a variation of the Newton-Raphson method to iterate and eventually end at the correct answer. Be aware that this is significantly slower that the math module's sqrt function.
Here is a routine that I modified from the internet. I can't cite the source right now. This version also works for non-integer arguments but just returns the integer part of the square root.
def isqrt(x):
"""Return the integer part of the square root of x, even for very
large values."""
if x < 0:
raise ValueError('square root not defined for negative numbers')
n = int(x)
if n == 0:
return 0
a, b = divmod(n.bit_length(), 2)
x = (1 << (a+b)) - 1
while True:
y = (x + n//x) // 2
if y >= x:
return x
x = y