In JAVA , why does Imaginary number ( taking square root of a negative number ) and division by zero give no error ?

Question

In JAVA ,
The code : System.out.println(Math.sqrt(-16.0)); will print : NaN .
Similarly if we compile : System.out.println(0.0/0.0) ; , it will give the same output .
Now change the numerator and compile : System.out.println(17.0/0.0); , it will give the output : Infinity .
The same thing gives an error when we perform the operations on an Integer .
The best part is trying to compile : System.out.println(Math.sqrt(-0.0)); , it will give -0.0 as the output .

How can such weird outputs be explained logically ?

Answer 1

Might such output can be explained logically, here are some points which can help :
● domain for sqrt funcn is R+ U {0}, so taking sqrt of negative number will give NAN.
● 0/0 is indeterminate form, as upper 0 want to take value to 0 & lower zero want to take value to ○○, so it gives NAN here also.
● see syntax for Math.sqrt() , we pass double type in it (not integer)
● output for 17.0/0.0 is quite understable & for -0.0 it will be taken as 0.0 only (return type of Math.sqrt() function is also double)
//for java

Answer 2

It's up to each programming language how to implement divisions by zero etc. Some might throw an error, others may get results like NaN or infinity. There's not the one correct way that is valid for all programming languages