Floating point comparison precision

Question

Given 3 IEEE-754 floats a, b, c that are not +/-INF and not NaN and a < b, is it safe to assume that a - c < b - c? Or, can you give an example when this is incorrect?

Answer 1

Suppose a is approximately 0.00000000000000001, b is approximately 0.00000000000000002, and c is 1. Then a − c and b − c will both equal −1.

(That's assuming double-precision, aka 64-bit, values. For higher-precision values, you'll need to add some more zeroes.)

---

Edited to add explanation:

If we ignore denormalized values and not-a-number values and infinities and so on, and just focus on IEEE 754 double-precision floating-point value for the sake of having something concrete to look at, then — in terms of the binary representation, a floating-point value consists of a sign bit s (0 for positive, 1 for negative), an eleven-bit exponent e (with an offset of 1023, such that e =0 means 2 ⁻¹⁰²³ and e =1023 means 2 ⁰ , ie 1), and a 52-bit fixed-point significand m (representing 52 places past the binary point, so it ranges from [0,1) with finite precision). The actual value of the representation is therefore (−1) ^s × (1 + m ) × 2 ^{e −1023} .

Because the significand is fixed-point, and has a fixed number of bits, the precision is very finite. A value like 1.00000000000000001 and a value like 1.00000000000000002 are identical for very many places past the decimal — more places than a double-precision significand can hold.

When you perform addition or subtraction between a very large number and a very small number (relative to each other: in our example, 1 is "very large"; alternatively, we could have used 1 as the very small value and chosen a very large value of 10000000000000000), the resulting exponent is going to be determined almost entirely by the very large number, and the significand of the very small number has to get scaled appropriately. In our case, it gets divided by about 10 ¹⁷ ; so it simply disappears. The significand doesn't hold enough bits to be able to distinguish that.