Absolute rounding error and subnormal numbers

Question

is there an x in subnormal numbers in IEEE Single format such that:

|round()−|/||>/2.

if there is not please explain.

Note: round's mode is to the nearest

= machine epsilon = in this case 2^-23

Answer 1

Let x be 2⁻¹⁰⁰⁰. The two IEEE-754 binary32 (“single precision” floating-point) values nearest x are 0 and 2⁻¹⁴⁹. So round(x) is 0, so |round(x)−x| / |x| = |0−2⁻¹⁰⁰⁰| / |2⁻¹⁰⁰⁰| = 1, which is greater than 2⁻²³.

For another example, let x be 2⁻¹⁴⁹+2⁻¹⁵⁰ = 1½•2⁻¹⁴⁹. The two binary32 values nearest it are 2⁻¹⁴⁹ and 2⁻¹⁴⁸ (which equal 1•2⁻¹⁴⁹ and 2•2⁻¹⁴⁹). These are equidistant from x, and the rule for breaking ties is to choose the number with the even low bit, which is 2⁻¹⁴⁸. Then |round(x)−x| / |x| = |2⁻¹⁴⁸ − (2⁻¹⁴⁹+2⁻¹⁵⁰)| / |2⁻¹⁴⁹+2⁻¹⁵⁰| = |2•2⁻¹⁴⁹ − 1½•2⁻¹⁴⁹| / |1½•2⁻¹⁴⁹| = |2−1½| / |1½| = ½ / 1½ = ⅓, which is greater than 2⁻²³.