A simple method can be written to find this value.

public class FloatEqualsZero {
    public static void main(String [] args) {
        float x = 1;
        while(x != 0 && -x != 0) {
            x *= 0.1;
            System.out.println(x);
        }
    }
}

This outputs the following:

0.1
0.01
9.999999E-4
9.999999E-5
9.999999E-6
9.999999E-7
...
1.0E-37
1.0E-38
1.0E-39
1.0E-40
1.0E-41
1.0E-42
1.0E-43
9.8E-45
1.4E-45
0.0

This (and similar tests) show that (x == 0) really only is true when x is 0.0f or -0.0f

how come a primitive float value can be -0.0?

floating point numbers are stored in memory using the IEEE 754 standard meaning that there could be rounding errors. You could never be able to store a floating point number of infinite precision with finite resources.

You should never test if a floating point number == to some other, i.e. never write code like this:

if (a == b)

where a and b are floats. Due to rounding errors those two numbers might be stored as different values in memory.

You should define a precision you want to work with:

private final static double EPSILON = 0.00001;

and then test against the precision you need

if (Math.abs(a - b) < epsilon)

So in your case if you want to test that a floating point number equals to zero in the given precision:

if (Math.abs(a) < epsilon)

And if you want to format the numbers when outputting them in the GUI you may take a look at the following article and the NumberFormat class.

easily by some checking you can do it First if(floatVal==0) sop("it is zero") Second float b=floatVal; if(b/floatVal!=1) sop("it is zero"); But you should be also aware about the divide by zero error here.