  # Xerxes

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Meson
• Birthday May 19

## Profile Information

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UK
• College Major/Degree
PhD@the best
• Favorite Area of Science
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• Biography
I am a weird person
• Occupation
Trainee alcoholic

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• ### shagufta

1. And what, pray, is this this called "reality"? Surely not snooker balls on trampolines?
2. You have a problem with mathematics? How else would you describe the Theory of General Relativity?
3. Actually, curvature is defined as the second derivative of the metric field. So if the metric field is constant the curvature field is zero - elementary calculus. Non-zero otherwise
4. I ask again - where is your argument that says this is the case? I have one, although to say it is hand waving would be to flatter it!
5. Why? Or Rather how? Your argument is...........
6. As this question was posted in the Linear Algebra and Group Theory section it seems reasonable to assume it was referring to the theorem that the order of any group is an integer multiple of that of any of it's subgroups (proof is easy - try it; hint - use the definition of cosets) Whether it applies in so-called daily life, I have no idea. Should we care?
7. I wonder if it might be helpful to point out the difference between the ordinal numbers and the cardinal numbers, since we seem to restricting the discussion to the natural numbers $$\mathbb{N}$$ So an ordinal number roughly speaking describes the position of an element in an ordered set. In contrast a cardinal number describes the size of a set, ordered or otherwise. Notice these are quite different concepts. Now, by construction the natural numbers are ordered (there is a theorem that any set can be ordered - the proof is hellacious and not relevant here). So it is fairly easy to see that, for any subset of $$\mathbb{N}$$ (it's ordered recall) if there exists a largest ordinal $$n$$ then this corresponds to the cardinality of our subset and it must be finite. Otherwise the gloves are off. The largest non-finite ordinal, by an arbitrary convention is denoted as $$\omega$$. This is still an ordinal., and can in no way denote the cardinality of a non finite subset of $$\mathbb{N}$$ e.g $$\mathbb{N}$$ itself. For this we use the arbitrary symbol $$\aleph_0$$. Any help?
8. In fact in science "points of view" have no value. None. Whatever. The so-called "curvature of spacetime" arises because in the presence of mass/energy the most useful/appropriate coordinates to describe the geometry of the 4-dimensional spacetime manifold are curvilinear, not rectilinear. "Spacetime curvature" is an unhelpful pop-science term. It is easy to see that in the (theoretical) absence of a gravitational source, or if you prefer, at infinite distance from one, the rectilinear (i.e. quasi-Cartesian) coordinates will suffice. Yes, as far as is known - but note that, in this context, gravitation is considered a form of mass/energy. That is gravitation causes itself!! For this reason, mathematicians call the General Theory of Relativity non-lineaar
9. Hi Function. If you were submitting your thesis to a UK university board, there are no "rules". There are, however, conventions, not only in theses but generally. 1. You can refer to yourself however you like - but John Maynard Octavious Smith, Jr. would be considered pretentious 2. In attributions or thanks, it is usually sufficient to use the suffix title only - Prof., Dr., etc. But be careful - in a clinical context, do not refer to someone as Mr. unless they are a surgeon. 3. "Trailing" qualifications (MD, FRCS, PhD, BSc etc) are not normally used in this context 4. If in doubt, ask the people concerned what they would prefer. That seems the simplest course
10. Good God, did I really say this? It appears that I did - I cannot imagine what I was thinking, as it is quite clearly nuts. Apologies
11. No, I don't think this is quite correct. If $n \in \mathbb{N}$ and $\mathbb{N}$ is infinite but countable i.e has cardinality $\aleph_0$, then the vector spaces $\mathbb{R}^n$ and $\mathbb{C}^n$ are necessarily infinite dimensional vector spaces.
12. Then if you know better than those who try to guide you, you do not need to ask the question, right? Just a reminder: the image of your first function is $h(x)$. Then you claim it's preimage is $h^{-1}(x)$. So $h(x)=x$. Agreed? And so on.......
13. No, this forum doesn't quite work like that. First if this is homework, you are in the wrong sub-forum - try "homework". Second, nobody will do your homework for you, although they may give hints if you show what you have tried. If it is not homework, you are wasting everybody's time with a pointless question
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