4 ms·
It is false - real numbers fulfill the trichotomy property, which is precisely the lack of such a relationship: every two real number is either less than, equal
by kaba0 3y ago
It is false - real numbers fulfill the trichotomy property, which is precisely the lack of such a relationship: every two real number is either less than, equal or greater than.
But the numerical context can still be correct: (edit: ~~imaginary~~) complex numbers for example don’t have such a property.
- fsckboy 3y agoi'm learning about this right here as I read, but do you mean complex numbers rather than imaginary?
- kaba0 3y agoYep, that’s what I meant, sorry!
- TeMPOraL 3y agoSame question. Or more generally, vectors. They don't have a total order, because as we define "less than"/"greater than" in terms of magnitude (length), this means for any vector V (other than 0), there's an infinitely many vectors that are not equal to V, but whose length is equal to length of V. Is this is what ≹ is talking about?
- shemii 3y agoAs I've written in another comment here, a great example of a number-y field which is not totally ordered is Games ⊂ Surreal Numbers ⊂ ℝ. There you have certain "numbers" which can can confused with (read incomparable) whole intervals of numbers. Games are really cool :)
- SetTheorist 3y agoYou've got the subset relationships backwards: Reals are a subset of the field of Surreal Numbers which is a subset of the group of Games. (Probably better phrased as embeddings, rather than subsets, but the point remains...) Note that Games do _not_ form a field: there is no general multiplication operation between arbitrary games.