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?
Using the decimal module:
import decimal
D = decimal.Decimal
n = D(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)
with decimal.localcontext() as ctx:
ctx.prec = 300
x = n.sqrt()
print(x)
print(x*x)
print(n-x*x)
yields
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999145.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999983754999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998612677
99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0E-100
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