> Doesn't the risk of lack of precision apply to both floating point and fixed-point? I'd think the reciprocal would need to be exactly represented in the type,

> Doesn't the risk of lack of precision apply to both floating point and fixed-point? I'd think the reciprocal would need to be exactly represented in the type, or to have more bits than the original type to produce a correct result in the last bit (or two bits?). Yes, but this was historically okay on an x87 FPU which had more precise representation than the common external formats.