Recent content by CrankFan

I'm not certain about what the author of the book wants. It's mentioned in section (7.1) that the great circle associated with the point P means the unique great circle of points \frac{\pi}{2} distant from P. If question 1(a) asked: Determine the equation of the great circle associated with...

I had a suspicion that that is the case but didn't want to jump to any conclusions. Yes. We're talking about the unit sphere: S^2, sorry for not making that clearer. I scanned the page from the book and posted it here so that people can see the exact question...

Hi. This question is from the book Geometry by Brannam, Esplen, Gray and it comes from the chapter 7, section 1 review problems which cover introduction to spherical geometry. Determine the equation of the great circle that passes through the point (1 / sqrt(6), -1 / sqrt(6), 2 / sqrt(6))...

I guess I said that because I wasn't thinking clearly. Thanks for the correction.

Yes. Any set A is equinumerous to the set A_s = \{ \{x\} \in P(A) | x \in A \} For any set A Let f_A denote the bijective map from A to A_s. If |P(A)| = |P(B)| then there is a bijection g from P(A) to P(B) and the restriction of g to the set A_s, is a bijection from A_s to B_s. Thus a...

{ \cos^{-1} \theta } \over { \sin^{-1} \theta } ...I wouldn't say that I find it really annoying. It's just one of those notational things that could lead to confusion.

I agree with Matt that your use of P(X) was confusing. As such I didn't get too far into the proof in the direction you opted and didn't read the back and forth too much but I think Matt gave a hint on how to do it. Here's an idea for how to prove it in the other direction |X| >= |Y|...

Could that be Equivalents of The Axiom of Choice? There is a second edition/volume? here here

If I understand you correctly, you want to find solutions, in integers, for the equation y^2 = x^2 + s where s is your "starting point". finding solutions for x and y boilds down to factoring s since that equation is equivalent to (y - x)(y + x) = s. The bad news is that in general there isn't...

If "for all n in N, P(n)" has a counter example: k then "not P(k)" is a theorem of ZF.

Formal undecidability is with respect to a specific formal theory, not all of them. For example if you add CH to the axioms of ZF you get a formal theory in which CH is decidable. What you're asking about only works if the statement in question is roughly of the form for all n \in \mathbb{N}...

http://www.math.uchicago.edu/~mileti/teaching/math278/settheory.pdf

You seem to be confused. There is no bijection from the empty set to any set that isn't the empty set, including the natural numbers.

I think you mean recursive here, instead of finite. For example, consider the first order induction schema (of first order PA).

Model Theory here: http://www.math.psu.edu/simpson/notes/