Why does Python give the “wrong” answer?

x = 16

sqrt = x**(.5)  #returns 4
sqrt = x**(1/2) #returns 1

Yes, I know import math and use sqrt. But I’m looking for an answer to the above.

sqrt=x**(1/2) is doing integer division. 1/2 == 0.

So you’re computing x^(1/2) in the first instance, x⁽⁰⁾ in the second.

So it’s not wrong, it’s the right answer to a different question.

You have to write: sqrt = x**(1/2.0), otherwise an integer division is performed and the expression 1/2 returns 0.

This behavior is “normal” in Python 2.x, whereas in Python 3.x 1/2 evaluates to 0.5. If you want your Python 2.x code to behave like 3.x w.r.t. division write from __future__ import division – then 1/2 will evaluate to 0.5 and for backwards compatibility, 1//2 will evaluate to 0.

And for the record, the preferred way to calculate a square root is this:

import math
math.sqrt(x)

import math
math.sqrt( x )

It is a trivial addition to the answer chain. However since the Subject is very common google hit, this deserves to be added, I believe.

/ performs an integer division in Python 2:

>>> 1/2
0

If one of the numbers is a float, it works as expected:

>>> 1.0/2
0.5
>>> 16**(1.0/2)
4.0

What you’re seeing is integer division. To get floating point division by default,

from __future__ import division

Or, you could convert 1 or 2 of 1/2 into a floating point value.

sqrt = x**(1.0/2)

This might be a little late to answer but most simple and accurate way to compute square root is newton’s method.

You have a number which you want to compute its square root (num) and you have a guess of its square root (estimate). Estimate can be any number bigger than 0, but a number that makes sense shortens the recursive call depth significantly.

new_estimate = (estimate + num / estimate) / 2

This line computes a more accurate estimate with those 2 parameters. You can pass new_estimate value to the function and compute another new_estimate which is more accurate than the previous one or you can make a recursive function definition like this.

def newtons_method(num, estimate):
    # Computing a new_estimate
    new_estimate = (estimate + num / estimate) / 2
    print(new_estimate)
    # Base Case: Comparing our estimate with built-in functions value
    if new_estimate == math.sqrt(num):
        return True
    else:
        return newtons_method(num, new_estimate)

For example we need to find 30’s square root. We know that the result is between 5 and 6.

newtons_method(30,5)

number is 30 and estimate is 5. The result from each recursive calls are:

5.5
5.477272727272727
5.4772255752546215
5.477225575051661

The last result is the most accurate computation of the square root of number. It is the same value as the built-in function math.sqrt().

Perhaps a simple way to remember: add a dot after the numerator (or denominator)

16 ** (1. / 2)   # 4
289 ** (1. / 2)  # 17
27 ** (1. / 3)   # 3

You can use NumPy to calculate square roots of arrays:

 import numpy as np
 np.sqrt([1, 4, 9])

I hope the below mentioned code will answer your question.

def root(x,a):
    y = 1 / a
    y = float(y)
    print y
    z = x ** y
    print z

base = input("Please input the base value:")
power = float(input("Please input the root value:"))


root(base,power)