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August 6[edit]

Is there any advantage to a floating point number system for computer that is other than binary or base ten (BCD)[edit]

Floating point number system on computers are binary that is base two. The only other floating point number system is BCD or based ten.

But binary floating point cannot represent 1/10 accurately while both binary and BCD/base ten cannot represent 1/3 accurately.

So why not have a floating point number system using base 720720 which has many many divisors because it is Superior highly composite number? 96.66.16.169 (talk) 16:00, 6 August 2017 (UTC)[reply]

It is not efficient to perform computations and store information with such a huge base. At this point, you might as well store the rational number exactly (in the form

(p,q)

representing

p/q

).--Jasper Deng (talk) 18:14, 6 August 2017 (UTC)[reply]

Many calculator programs do store results as rationals. This avoids rounding error for simple calculations such as 1.01 * 100 - 101 not being 0. It does not remove rounding errors for more complex operations though, e.g. (1/(√5 - 2) - 2)² - 5 not 0.

It should be noted that BCD is not exactly the same a decimal; it uses two digits like binary but each group of four represents 1 decimal digit. Computers can easily use octal and hexadecimal using a similar scheme. You could have system that uses a group of 20 binary digits to store 1 base-720720 digit, but you'd still run into rounding errors when, say, the denomitor is 17. --RDBury (talk) 20:10, 6 August 2017 (UTC)[reply]

It is more efficient to use a power of 2. Bubba73 ^{You talkin' to me?} 23:19, 6 August 2017 (UTC)[reply]

Densely packed decimal used in IEEE decimal floating point uses base 1000 to hold the digits which is almost as efficient - efficiency isn't a major consideration for that. Dmcq (talk) 09:18, 9 August 2017 (UTC)[reply]

A floating point representation on a ternary computer might have some advantages. For example, repeated rounding would be stable (i.e. rounding once to the final precision produces the same result as with any number of intermediate-precision rounding steps); no tie-breaking rule is needed from a number that is already represented using a finite number of ternary digits. This stability does not extend to common scenarios though: rounding numbers before addition can still lead to a different result than rounding the result. —Quondum 02:55, 13 August 2017 (UTC)[reply]