Python 2 is_integer not returning properly

Question

from fractions import Fraction

counter = 0;
a = int(raw_input())
b = int(raw_input())

if 1 <= a <= 10 ** 8:
    if a <= b <= 10 ** 8:
        for i in range(a, b+1):
            if float(i**Fraction(1,3)).is_integer() == True:
                counter += 1
                print(i)

print(str(counter))
print (str(float(64**Fraction(1,3)).is_integer()))

This code returns false which results in the if statement not being run at all. Furthermore, the cube root of 64 is 4, therefore, the result should be an integer. However, in the range between 1 and 100 inclusive, 1, 8, and 27 return true for this case. Any help would be appreciated as to why the cube root of 64 is not returning true.

You're running into the issue that floating point numbers are not very accurate - see https://docs.python.org/2/tutorial/floatingpoint.html — Peter Gibson, May 24 '18 at 01:55
Possible duplicate of [Is cube root integer?](https://stackoverflow.com/questions/23621833/is-cube-root-integer) — James, May 24 '18 at 02:00
Possible duplicate of [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — Barmar, May 24 '18 at 02:04

Naveen · Answer 1 · 2018-05-24T02:09:09.613

0

You have used float for the cube root value. The following code gives you the value of float(i**Fraction(1,3)) = 3.9999999

Try rounding the value first before converting it to float.

if float(round(i**Fraction(1,3),2)).is_integer() == True:

edited May 24 '18 at 02:09

answered May 24 '18 at 02:01

Naveen

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I wouldn't suggest rounding first - this would return true for any value – Peter Gibson May 24 '18 at 02:04
Thanks Peter, I missed the floating point in the round function. – Naveen May 24 '18 at 02:10
`float(round(0.5)).is_integer()` returns True though – Peter Gibson May 24 '18 at 02:12
1

`float(round(0.5,2)).is_integer()` returns FALSE. – Naveen May 24 '18 at 02:16
That's better, but still worth mentioning the precision tradeoff – Peter Gibson May 24 '18 at 02:26
Note that `(a^3+1)^(1/3)=a+1/(3*a^2)+...` so that rounding to two digits will not be enough around `a^3=1000` and above, as `1001^(1/3)=10,0033322228..` confirms. For cubes close to the upper bound `10^8` one would need 6 digits after the dot. – Lutz Lehmann May 24 '18 at 06:14

Python 2 is_integer not returning properly

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