如何判断数字是否可以精确表示为32位IEEE浮点数？

Question

This is a task from my textbook

Given a hypothetical computer with a word length of 32 bits divided the following way:
1 bit for the sign bit
9 bit biased exponent
23 bit mantissa part

Is 2^-1 + 2^-29 a machine number on this computer? How might i go about determining this?

Answer 1

abs((-1) - (-29)) is larger than the mantissa width so no, that would require too many significant figures (mantissa bits).

If you wrote them out in a non-exponential binary-point notation, like 0.10000...0001 , the gap between the 2 ones would be larger than the mantissa. (It's like a decimal point, but it's binary place values so we call it a binary point.)

ie 2 ^-29 is less than 1 ulp for 2 ^-1 . (ULP = unit in the last place = low bit of the mantissa.)

Binary floats can represent a fixed number of (binary) "significant figures" equal to the mantissa width, regardless of the exponent. ( https://en.wikipedia.org/wiki/Significant_figures )

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Note that the stored width of the mantissa is only 23 bits, but there's an implicit leading 1 (for normalized numbers). https://en.wikipedia.org/wiki/Single-precision_floating-point_format

So 1 + 2 ^-23 is exactly representable. ie 1 ulp relative to 1.0 = 2^-23 . (Related: https://en.wikipedia.org/wiki/Machine_epsilon like C FLT_EPSILON is half a ulp of 1.0, so 2^-24 for binary32 floats. It's the largest value you can add to 1.0 and still have the result round to 1.0 , ie the largest rounding error you can get when adding to 1.0 )

Note that many modern documents use "significand" instead of "mantissa" because it's more mathematically correct.

Answer 2

Is 2^-1 + 2^-29 a machine number on this computer?
How might i go about determining this?

Assuming the hypothetical format is using base 2 (that was not specified), then 2 ^-1 and 2 ^-29 are both exactly representable being small powers of 2. Let us call them a_1 and a_29 and assume a_1 > a_29 > 0

"1 bit for the sign bit, 9 bit biased exponent, 23 bit mantissa part" is akin to binary32 . Should it follow all those rules (including sub-normals), then any subtraction between different values will not result in 0. This is important for the next step.

Set sum/difference acc = a_1 +/- a_29 and acc_1 = acc - a_1 . If acc_1 == +/-a_29 , then acc was exactly represented, else acc was not exactly the sum/difference a_1 +/- a_29 , just a rounded result.

In the case of 2 ^-1 and 2 ^-29 , the difference is not exactly representable as the "23 bit mantissa part" is too narrow to encode exactly a relative difference of 2 ^{-1 - 29} . We would need about "27-28 bit mantissa part" to do so.