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Python

Enzo Cnop
Enzo Cnop
5,157 Points

Simple way to ask python to check if a number is/is not/ a float.

For the code challenge after creating a number game, I was asked to create a function that would return True if the number entered is even, and False, if the number entered is odd. The code that let me pass the challenge is below.

def even_odd(number):
  if number % 2 == 0 :
    return True
  else:
    return False

My question: Why can't I write this code? (please ignore indentation, i'm not sure how to format code on this website yet.)

def even_odd(number):
  if number % 2 == float :
    return False
  else:
    return True

Surely, python recognizes what a 'float' is, so why can't it check whether or not a number is a float this way? I don't even understand how the code that let me pass the challenge works. Say 5 is the number entered. 6 % 2 = 3, by my understanding of this logic, the code should return false, because the resulting integer is not equal to 0. How does this work. Please help me understand.

2 Answers

Hey Enzo,

Many languages work solely with integer values unless specified, so 5 / 3 = 1 (1.67, but it returns the integer value). In this case, you could probably get away with it, since Python will actually return a float when performing functions with integers. You could use:

isinstance(number, float)

But in the case of even/odd, the modulus operator (%) returns the remainder, so it is not returning the value like a division operator. The example you cited above, 6 / 3 = 2, but 6 % 3 = 0, since three divides completely into 6. If you did 6 % 5, you would get 1, as there is a remainder of 1 when six is divided by 5. In this challenge, any even number divided by 2 will equal 0; thus, it is even. If it isn't 0, then the number is odd. I hope this helps!

Selemawit Gebremedhin
Selemawit Gebremedhin
1,331 Points

num = int (input ("Enter number: "))

def is_odd (num):

if num % 2 != 0:

    return True

else:

    return False

I don't know why it is giving me error