With IEEE-754, 0 < ABS(const) < 1, is (x / const) * const guaranteed to return distinct results for distinct values of X?

Question

Assume I do this operation:

(X / const) * const

with double-precision arguments as defined by IEEE 754-2008, division first, then multiplication.

const is in the range 0 < ABS(const) < 1.

Assuming that the operation succeeds (no overflows occur), are distinct arguments of X to this operation guaranteed to return distinct results?

In other words, are there any X1, X2 and 0 < ABS(const) < 1 so that X1 <> X2, but (X1 / const) * const = (X2 / const) * const?

Answer 1

Yes.

public class TestDoubleDivision
{
    public static void main(String[] args)
    {
        final Random random = new Random();
        int i = 0;
        while (i < 10)
        {
            final double c = random.nextDouble();
            final double x1 = 10.0 * random.nextDouble();
            final double x2 = nextDouble(x1);

            if (x1 / c * c == x2 / c * c)
            {
                System.out.printf("x1 = %.20f, x2 = %.20f, c = %.20f\n", x1, x2, c);
                i++;
            }
        }
    }


    private static double nextDouble(double d1)
    {
        return Double.longBitsToDouble(Double.doubleToLongBits(d1) + 1);
    }
}

prints

x1 = 5.77383813703796800000, x2 = 5.77383813703796900000, c = 0.15897456707659440000
x1 = 2.97635611350670850000, x2 = 2.97635611350670900000, c = 0.15347615678619309000
x1 = 7.98634439050267450000, x2 = 7.98634439050267500000, c = 0.83202322046715640000
x1 = 0.11618686267768408000, x2 = 0.11618686267768409000, c = 0.09302449134082225000
x1 = 0.98646731978098480000, x2 = 0.98646731978098490000, c = 0.40549842805620606000
x1 = 3.95828649870362700000, x2 = 3.95828649870362750000, c = 0.75526917984495820000
x1 = 1.65404856207794440000, x2 = 1.65404856207794460000, c = 0.14500102367827516000
x1 = 5.72713430182017500000, x2 = 5.72713430182017550000, c = 0.68241935505532810000
x1 = 3.71143195248990980000, x2 = 3.71143195248991000000, c = 0.21294683305890750000
x1 = 5.66441726170857800000, x2 = 5.66441726170857900000, c = 0.69355199625947250000

Answer 2

(I just wanted to add something to starblue's answer -- it's too long to fit into a comment.)

I find it easier to see what's going on -- and I'm hoping you will too -- when I can see the full binary value of a double. I put starblue's examples in a C program and converted the output to binary (using my conversion program at http://www.exploringbinary.com/converting-floating-point-numbers-to-binary-strings-in-c/ ). Here's the output, plus the result of the calculation:

x1 = 101.1100011000011010010000011001001011111001110000111
x2 = 101.11000110000110100100000110010010111110011100001111
c  = 0.00101000101100101000111010100110011111010101111100001
r  = 100100.01010001101110101101000101101100011111011010101

x1 = 10.111110011111001001111001011010001100001011001111011
x2 = 10.1111100111110010011110010110100011000010110011111
c  = 0.0010011101001010001101101010001000011100110010101011
r  = 10011.011001001001100010101001001110011100011111011111

x1 = 111.1111110010000001000100001110001111001101010100101
x2 = 111.11111100100000010001000011100011110011010101001011
c  = 0.11010100111111110111100101001001011010110100010111011
r  = 1001.100110010100010010100101110100000100000110000011

x1 = 0.0001110110111110011011000001011101101100111011010101
x2 = 0.00011101101111100110110000010111011011001110110101010001
c  = 0.0001011111010000011100111111110000001001001011101001
r  = 1.00111111101111011111001110101010100101010101010101

x1 = 0.1111110010001001000111110100110100001000001101111111
x2 = 0.11111100100010010001111101001101000010000011011111111
c  = 0.01100111110011101011111010110111000101001011000000111
r  = 10.011011101100011101000000101000110110101011010011111

x1 = 11.1111010101010010010000111001010000100001011000111
x2 = 11.111101010101001001000011100101000010000101100011101
c  = 0.110000010101100101010010001010110001110001011111111
r  = 101.00111101101010110100110000011111101001010010101111

x1 = 1.1010011101101111101110100000000000011110110111110111
x2 = 1.1010011101101111101110100000000000011110110111111
c  = 0.00100101000111101100100101111110100101011010111111001
r  = 1011.011010000011101100001011000110000010011111110001

x1 = 101.10111010001001010111100100111110000111100001000011
x2 = 101.101110100010010101111001001111100001111000010001
c  = 0.101011101011001100001000111011000001111010111011011
r  = 1000.0110010001110100001001010000000101111000011111011

x1 = 11.101101100010000001100111100010010100011000001001111
x2 = 11.10110110001000000110011110001001010001100000101
c  = 0.0011011010000011101011110000001111000110010101111111
r  = 10001.01101101110011010100011111101110101011001010001

x1 = 101.10101010000101110011111111101001111011111010101111
x2 = 101.1010101000010111001111111110100111101111101011
c  = 0.1011000110001100100111111010011000000010100011
r  = 1000.0010101011010001010101111000111101110100001000001

(BTW, the "* const" part of the expression is unnecessary: the division by const alone shows that X1 / const == X2 / const.)

You can really see what's going on when you compare the double values with the true, arbitrary precision values. Take the first example, for instance:

x1/c = x2/c (double) = 100100.01010001101110101101000101101100011111011010101

x1/c (true)          = 100100.01010001101110101101000101101100011111011010100 1011...

x2/c (true)          = 100100.01010001101110101101000101101100011111011010101 0111...

I put a space between significant bits 53 and 54, where the rounding occurs in a double. x1/c rounds up, and x2/c rounds down (truncates), becoming the same value.