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Any help clearing up a point of confusion would be very much appreciated so I can get to grips with rational geometry...
You repeatedly make it clear that the set of real numbers is very complicated and are to be avoided. You then start using vectors which (to me at the moment..) seem to be clearly over the scalar field of real numbers. help(!) :)...
mediteight 2 years ago
Almost all the vectors, and points, I use are meant to have rational coordinates, unless otherwise stated.
Sometimes we need to go to certain quadratic extension fields (like adjoining sqrt(2) to the rationals) but the full complexity of the `real numbers' is a million miles away...
njwildberger 2 years ago
Many thanks for your time - I haven't come across quadratic extension fields so that sounds very likely to be the cause of my confusion. Can I just check - are you saying your vector space used in these examples is over a field that is a quadratic extension field? (Would it be possible to have a definition (or a pointer to one) of a quadratic extension field?
thanks again - Paul
mediteight 2 years ago
All the discussion about rotations and reflections is purely over the rational numbers.
One thing that takes some getting used to is that there is no rational point on the unit circle also on the line x=y, i.e. at spread 1/2 or angle 45 degs. But there are plenty of other rational points to make up for that!
I will be introducing quadratic extension fields later on. It is a more advanced concept.
njwildberger 2 years ago
I see - many thanks. It was exactly the rotation of spread 1/2 that was making me think you were using the the reals.
It seemed slightly disturbing for such a simple rotation to not result in a vector still within the vector space.
As a final check (promise!) am i right that:
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mediteight 2 years ago
...
1) the vector space is over the rationals
2) in the paramaterisation of the circle, if we restrict t to the rationals then the result of rotating any vector, by the rotation corresponding to a certain value of t, will result in a vector still having rational coordinates.
Is this true? If it is, is the proof straightforward?
mediteight 2 years ago