#1

but i want to know if the logic behind it is correct, i didnt think about the round() function , should I be using that instead? thanks in advance.

``````def is_int(x):
if x == 0:
return True
elif x / int(x) == 1:
return True
else:
return False``````

#2

I tried your code, and I got the same answer as my code did:

``````def is_int(x):
if x - round(x) > 0:
return False
else:
if x < 0 and x - round(x) < 0:
return False
else:
return True``````

So I guess you did it right, you just used a different way to reach your goal.
Sorry In advance for the bad response...I'm new at helping people

#3

Your logic is correct but naturally there is more than one way to determine if a number is an integer (in general, not just in python).

Your approach works, but required the exception for 0. You could have done something like
if X!=int(X):
return False

else:
return True

or something like

if X>int(x) or X<int(x):
return false

else:
return True

Or like you mentioned the round function. If you use the round function on X and that still is equal to X return True. All are valid ways of checking for integers.

#4

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