How Can I Put Type() With If Statement In Python?

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(

Solution 1:

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")

Solution 2:

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).

Solution 3:

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")