python, square and cubic

Question

Hello,

I need to build a program that prompts the user to enter a number, and from that number, I need to validate whether it is cubic and square. I'm having some difficulties writing it because, despite the SQRT() command as well as MATH.POW(x, 1.0/3) helping me find the square or cube of the number, I don't know how to validate to say whether the number is square and cubic... I'm missing the validation part and I'm unable to do it; I can't find enough documentation.
Thank you

Configuration: Mac OS X (10.10.2) / Chrome 42.0.2311.90

Anonymous user · Accepted Answer

Hi,A number is a square if its square root is an integer? The same for cubic?

dsy73 · Answer

You can check if the returned value is an integer by using X.is_integer() assuming your float is named X.

Anonymous user · Answer

You can also compare 6.0 to 6. Here, they are equal. But for example, let's take the square root of 2, which is about 1.4. In int, it would give either 1 or 2, but that's different from 1.4, so 2 is not a square number.

fiddy · Answer

Hello,

It's best not to use sqrt() which returns a float. Generally speaking, when you want to work with integers, it's better to stick with integers to avoid rounding errors.

Here you can for example do:
def isSquare(nb):
x = 1
while True:
tmp = x
x = (x**2 + nb)/(2*x)
if abs(x - tmp) < 2: break
return nb==x**2

--

Google is your friend

fiddy · Answer

Hello,

It's best not to use sqrt() which returns a float. Generally speaking, when you want to work with integers, it's better to stick with integers to avoid rounding errors.

Here you can for example do:
def isSquare(nb):
x = 1
while True:
tmp = x
x = (x**2 + nb)/(2*x)
if abs(x - tmp) < 2: break
return nb==x**2

--

Google is your friend

Python, square and cubic

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