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In this case, the function in the example representing a Doom value is a definite value in the set resulting from f(d, o), where d is the current date, and omni
by esoter 6y ago
In this case, the function in the example representing a Doom value is a definite value in the set resulting from f(d, o), where d is the current date, and omniscience is o. Where the date of Doom game creation is t, when d ~< t, Doom is 0. Otherwise, the result from f(d, o) is either a set containing a single possible integer solution if o is high or a set of an unknown number of possible integer solutions if o is low, with the length of that set also being dependent on proximity of d to t.
Doom is a real number under any circumstance as there is only one value it can be of the set produced by f(d, o) and that value is a real number.
Doom cannot always be represented by a fraction, because it is an indeterminate constant when d is higher than t and o is not high.
- core-questions 6y ago> Imaginary number is irrational I don't think that's the case. `i` is not irrational, it's just perpendicular to 1.
- esoter 6y agoYes, i is neither rational nor irrational. I just edited to correct/remove that. It’s possible that I cannot define an irrational number as an real number member in a sometimes unknown location within a result set of a function, where the size of the set being exactly one results in the possibility of the number being known exactly. But it has a lot in common with irrational numbers, aside from it looking to meet the qualifications of being irrational. If you assume total omniscience, then the result of the function would always result in a real, rational number, and that is the number I’m trying to define as irrational, because the answer as defined is always an integer, but depending on the values of the parameters to the function, which number it is would be unknown, and that cannot be defined as a fraction.