Re:Non-computability of solutions of certain equations on di...
by nine_k on Saturday August 03, 2024 @10:21PM
(#41150752)
An exact comparison is undecidable. You can only say if the numbers sufficiently differ. This is equivalent to saying that x > 0 + ε or x < 0 - ε, where ε > 0. If you represent numbers in binary, ε = 2^(-n), where n is the maximum number of bits you allocate to represent a number. Obviously you have n < ∞ in any finite machine.
[ Parent
| on HN ]
There are no comments on this comment yet.