I was working with numbers of 200 digits in python. When finding the square root of a number using math.sqrt(n) I am getting a wrong answer.
In[1]: n=9999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999998292000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000726067
In[2]: x=int(math.sqrt(n))
In[3]: x
Out[1]: 10000000000000000159028911097599180468360808563945281389781327
557747838772170381060813469985856815104L
In[4]: x*x
Out[2]: 1000000000000000031805782219519836346574107361670094060730052612580
0264077231077619856175974095677538298443892851483731336069235827852
3336313169161345893842466001164011496325176947445331439002442530816L
In[5]: math.sqrt(n)
Out[3]: 1e+100
The value of x is coming larger than expected since x*x (201 digits) is larger than n (200 digits). What is happening here? Is there some concept I am getting wrong here? How else can I find the root of very large numbers?
math.sqrt returns an IEEE-754 64-bit result, which is roughly 17 digits. There are other libraries that will work with high-precision values. In addition to the decimal and mpmath libraries mentioned above, I maintain the gmpy2 library (https://code.google.com/p/gmpy/).
>>> import gmpy2
>>> n=gmpy2.mpz(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)
>>> gmpy2.get_context().precision=2048
>>> x=gmpy2.sqrt(n)
>>> x*x
mpfr('99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0',2048)
>>>
The gmpy2 library can also return integer square roots (isqrt) or quickly check if an integer is an exact square (is_square).