Who Needs Floating-Point Anyway?

Suppose we truly want floating point arithmetic free of rounding error. Can we do this for all combinations of floating point numbers? Well. Let's for now just consider finite arithmetic; that is, neither infinity nor NaN. In this case, all we need to do is define the smallest positive floating point number as "1", and shift the decimal point accordingly. According to the IEEE-754 standard, for double-precision this number is

2^-1074

which in binary looks like

0 00000000000 0000000000000000000000000000000000000000000000000001

To treat this as "1" we must also look at what the largest number is. Checking IEEE-754, we have

2¹⁰²³ × (1 + (1 − 2⁻⁵²)), or in bits,

0 11111111110 1111111111111111111111111111111111111111111111111111

Putting this together, we need to represent a number as large as

2¹⁰²³ × (1 + (1 − 2⁻⁵²)) / (2⁻¹⁰⁷⁴).

Which, we can calculate with the useful bc (with sufficiently high scale) as

36385714125121573300846800698456749842842774431060269030973563199251\
83520276313187422051044619975257814616895952553597550412366074125973\
05594915359190782200698392412987448013052928786408355279308639946743\
57611588999020693594474762898847930291552594690170203187215045688094\
95566077392257613796983034261186022501993558219960112146924922314987\
24661213715671558623030843303146025660694164325513330061947744775142\
60351201969859368060220131234488198148976536169638305696900504838830\
71976087551424621650897680388272858267735217700412928884785446308400\
63729813907563445195499310977439636039716323348918368319786868700433\
55177324550146752512

So, to fit everything inside a fixed-point value, we need the base-2 logarithm of this:

$ bc -l
scale=60;
l(2^1023 * (1 + (1 - 2^-52)) * 2^1074)/l(2)
$ 2097.999999999999999839828674809254106809409981385123854002550379

Which is just a tad under 2098 bits. But remember, we need negative numbers as well so we need 2099 bits to get the sign for twos-complement. Then, once we have some value in fixed-point one can multiply it by 2^-1074 to get the correct value. What purpose does this have? For example, the GNU MPFR library has arbitrary-precision floating point values. Knowing that we need 2099 bits means we can use any value with at least 2099 bits of precision to do any sequence of floating point operations with no error, as long as the results are correctly rounded. MPFR is carefully designed to be correctly rounded, so that's covered for us too!

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