Pedantic aside: I guess you mean non-associative multiplication? That is, ax(bxc) versus (axb)xc. Since rounding happens at different times, they can give diffe

Pedantic aside: I guess you mean non-associative multiplication? That is, ax(bxc) versus (axb)xc. Since rounding happens at different times, they can give different results. Floating point multiplication is indeed non-commutative as well, but that has to do with NaN handling, not imprecision.