4 ms·
In this case it can, but only conditionally. At all times, with sufficient knowledge and defined state you could instantaneously determine the Doom value, so i
by esoter 6y ago
In this case it can, but only conditionally.
At all times, with sufficient knowledge and defined state you could instantaneously determine the Doom value, so it is a real number.
With the example, prior to creation of the game Doom, Doom value would be 0. Around the point of creation, the value would probably be 1, because prior to that point Doom did not exist, and then it would.
After creation, there could be a limit given the available hardware, languages, etc. and code for Doom in those languages. A large number of variations of the code in various languages could still produce Doom. Given sufficient knowledge, there could be a finite value of Doom. But practically if we had to feed an actual set of numbers that could equal Doom into a machine given what we could ascertain, it would be a finite set.
In our example, since there exists a defined value for Doom, it is real.
Since that defined value can only conditionally be known, and otherwise is a set of unknown length, it is conditionally not a constant that is representable by a fraction.
- AlDante2 6y agoNeither 0 nor 1 are irrational numbers. So Doom is not necessarily an irrational number. There is a maximum length of all Doom programs in all languages. Therefore, if I understand your definition correctly, Doom is always finite and hence not irrational. That the length of the set is unknown is irrelevant - we don't know how many books there are in the world, but we know that it is a finite number. Not knowing what fraction represents Doom is very different from saying there is no such fraction.