I have TypeError: a float is required when I try putting type() with if statement
example:
from math import *
x=input("Enter a Number: ")
if type (sqrt(x)) == int:
print (sqrt(x))
else:
print("There is no integer square root")
obviously I can't use x= float(x)
there are 2 problems here :
1st
x=input("Enter a Number: ")
will be a string, not a number, using int(x) should fix this
2nd
if type (sqrt(x)) == int:
sqrt() always return a float, you can use float.is_integer(), like this : sqrt(x)).is_integer() to check if the square root is a integer.
final code :
from math import *
x=int(input("Enter a Number: "))
if sqrt(x).is_integer():
print (sqrt(x))
else:
print("There is no integer square root")
sqrt(x).is_integer() rather than float.is_integer(sqrt(x)). There's no need for the unbound method here.You have to convert your input string to float before passing it to sqrt(). As far as Python is concerned, sqrt("5.4") makes no more sense than sqrt("Bob"). And x=float(x) is a fine way to do that conversion.
There's no question of types here: input() will always be string, and sqrt() will always be float (never int).
It seems that instead of an int, you just want a whole number. To check if something is a whole number, do a modulus division by 1.
from math import *
x = float(input("Enter a Number: "))
if sqrt(x) % 1 == 0:
print(sqrt(x))
else:
print("There is no integer square root")
x= float(x)” – why not?float.is_integermethod to test if a float is an int.isinstance.TypeError: a float is required. That's becausexis a string. You can't take a square root of a string. Do note that even if you dosqrt(float(x))your if-statement will never evaluate to true becausemath.sqrtalways returns a float.