Java's Math.IEEERemainder function states:
The remainder value is mathematically equal to f1 - f2 × n, where n is the mathematical integer closest to the exact mathematical value of the quotient f1/f2, and if two mathematical integers are equally close to f1/f2, then n is the integer that is even
For the following:
double f1 = 0.1;
double f2 = 0.04;
System.out.println(Math.IEEEremainder(f1, f2));
The output is -0.019999999999999997
However, 0.1/0.04 = 2.5 which is equidistant from both the integers 2 and 3. Shouldn't we pick n = 2 here, resulting in 0.1 - 0.04*2 = 0.02, instead of -0.02 ?
See: Is floating point math broken?
You would think that 0.1 / 0.04 would return exactly 2.5, but that's not true. According to this article, 0.1 cannot be accurately represented using IEEE 754, and is actually represented as 0.100000000000000005551....
In this case, the quotient is slightly higher due to that minuscule offset, which results in a value of 3 for n, as it's no longer equidistant between 2 and 3.
Computing it results in the following:
0.1 - 0.04 * 3 = 0.1 - 0.12 = -0.02 ~= -0.019999999999999997