Is 3*x+x always exact?

Question

Assuming strict IEEE 754 (no excess precision) and round to nearest even mode, is 3*x+x always == 4*x (and thus exact in absence of overflow) and w

Answer 1

In the following, math shown in code format is computed with IEEE 754 in round-to-nearest mode, and math not in code format is exact.

Let p be the number of bits in the significand.

Let f be the factor 2ⁿ-1 for a positive integer n and be exactly representable (n ≤ p).

Let U(x) be the ULP of x. For normal values, U(x) ≤ 2^1-px.

Let t be f*x. If f*x is subnormal, then it is exactly fx. If it is normal, then t = fx+e for some |e| ≤ ½U(fx) ≤ 2^-px. Note that if |e| is exactly half an ULP, then it must equal the lowest bit of x that is set (since otherwise e would have more than one bit set and could not be half of an ULP).

Substituting for f, t = (2ⁿ-1)x+e.

t+x = (2ⁿ-1)x+e+x = 2ⁿx+e.

Consider t+x. By IEEE-754 requirements of round-to-nearest, this must be within ½ an ULP of t+x, which we know to be 2ⁿx+e. Clearly 2ⁿx is representable (barring overflow), and |e| ≤ ½U(fx) ≤ ½U(2ⁿx). Therefore t+x must be 2ⁿx unless |e| is exactly half an ULP and the low bit of x’s significand is odd (since an even low bit wins the tie and gives 2ⁿx).

If n is 1, then f is 1, and e is 0. If 2 ≤ n, then |e| ≤ 1/4 U(2ⁿx) < ½U(2ⁿx). So a case where |e| is half an ULP and x’s low bit is odd does not occur.

Therefore t+x must be 2ⁿx. (Overflow and NaN left as an exercise for the reader.)

Additionally, I tested exhaustively for IEEE-754 32-bit binary floating-point.