Python | sympy.is_comparable() method

With the help of sympy.is_comparable() method, we can compute the real number with precision and return True if computed value is real with precision by using sympy.is_comparable() method.

Syntax : sympy.is_comparable() Return : Return boolean value if real number is computed.

Example #1 : In this example we can see that by using sympy.is_comparable() method, we are able to evaluate the expression and if computed value is real with precision it will return True else False. Python3 1=1

# import sympy
from sympy import * 

x, y, z = symbols('x y z')

# Use sympy.is_comparable() method
gfg = (3 * x + 2 * tan(y + I * pi) + log(2 * x)).is_comparable
  
print(gfg)

Output :

False

Example #2 : Python3 1=1

# import sympy
from sympy import * 

x, y, z = symbols('x y z')

# Use sympy.is_comparable() method
gfg = (pi / 2).is_comparable
  
print(gfg)

Output :

True