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July 2

Bases of Fermat quotients

In the article Fermat quotient, p is referred to as the base of the Fermat quotient? What is the reason for this? To me it would seem more natural to refer to a as the base of the Fermat quotient and to p simply as the exponent of the Fermat quotient or something like that. -- Toshio Yamaguchi (tlk−ctb) 13:48, 2 July 2012 (UTC)[reply]

I imagine it's something like octal or the binary numeral system. Expressing a number as a fermat quotient base p would be like expressing a number in base 8 or base 2.--Wikimedes (talk) 06:24, 8 July 2012 (UTC)[reply]

Material conditional and functional completeness

I don't understand the distinction that the functional completeness article seems to make regarding the material conditional operator. The "Minimal functionally complete operator sets" section of this says that only NAND and NOR are minimally complete, by which I think it means to say that with only the axioms of {False, NAND} one can build the full suite of boolean logical operators; and one can do the same starting with {False, NOR}. But one can do likewise with {False,→} too (that is, one can build not, then or, then nor, etc., from just those two). I don't understand what distinction that article is drawing which precludes the material conditional operator from also being complete. -- Finlay McWalterჷTalk 16:42, 2 July 2012 (UTC)[reply]

For NAND and NOR you don't need a special symbol for False: False = A nand A = A nor A. However for → you do need a special symbol for False. --87.68.251.173 (talk) 09:43, 3 July 2012 (UTC)[reply]

Name for set including Moebius strip and Klein Bottle?

I know that the Moebius strip and Klein Bottle are non-orientable surfaces, but is there a name for the small class(?) of objects there are in where a Sn (sphere of dimension n) is tracing through the n+2 dimension so that it is matched up to its inverted self? (Yes, I know that that sentence contains terms that might not be exact). Also is there a name for the next step up from a Klein Bottle (S3 tracing through the 5th(?) dimension) for a non-orientatable surface?Naraht (talk) 18:20, 2 July 2012 (UTC)[reply]

I'm not sure exactly what you're after. Perhaps something to do with holonomy? Sławomir Biały (talk) 19:24, 2 July 2012 (UTC)[reply]

I get the feeling that the user may be asking about homotopy groups. — Fly by Night (talk) 19:35, 2 July 2012 (UTC)[reply]

I don't think either. This list would have (presumably) *one* entity at each dimension, with the Moebius strip being the movement of S1 embedded in 3 dimension, the Klein Bottle with the movement of S2 in 4 dimensions and so on.Naraht (talk) 20:33, 2 July 2012 (UTC)[reply]

I'm not sure what you mean by "movement of S1 embedded in 3 dimensions". The boundary of the Mobius strip is S1. How is this "moving"? Maybe you mean that the Mobius strip is a kind of mapping cylinder for the identification of antipodal points of S1. But then that doesn't correlate with the Klein bottle being the "movement of S2 in 4 dimensions" since the mapping cylinder for the identification of antipodal points of S2 would be a 3-manifold rather than a surface. Sławomir Biały (talk) 12:37, 3 July 2012 (UTC)[reply]

Think of the Mobius strip as a line segment (which is actually the same as S1) moving through 3 dimensions until it gets back where it started, except backwards. Think of a Klein Bottle as a Circle moving through 4 dimentions until it gets back where it started, except backwards.Naraht (talk) 18:20, 3 July 2012 (UTC)[reply]

Ah, by S1 you mean

B^{1}

, the 1-ball, not the sphere! Sławomir Biały (talk) 18:37, 3 July 2012 (UTC)[reply]

The answer is that the Mobius strip and (solid) Klein bottle are homeomorphic to non-orientable vector bundles over

S^{1}

(the circle). There is one in each dimension, obtained by twisting the trivial bundle

S^{1}\times B^{n}

by the Mobius bundle. Sławomir Biały (talk) 18:41, 3 July 2012 (UTC)[reply]

OK, that makes sense. The 1-D sphere would just be two points. Does the "next dimension up" from a (solid) Klein bottle have a name?

S^{1}

is the circle.

S^{0}

is two points (see FbN's post below). Thus

S^{1}\times B^{1}

is the cylinder,

S^{1}\times B^{2}

is the solid torus, and so on. You get the thing your after by twisting these so-called "trivial" bundles (because they can be expressed as a product) by the Mobius bundle. Sławomir Biały (talk) 19:32, 3 July 2012 (UTC)[reply]

Not sure about a name. It would just be a sum of three Mobius bundles. Sławomir Biały (talk) 19:13, 3 July 2012 (UTC)[reply]

There's a big difference between a sphere and a ball. The definitions are:

S^{n}:=\{(x_{1},\ldots ,x_{n+1})\in \mathbb {R} ^{n+1}:x_{1}^{2}+\cdots +x_{n+1}^{2}=1\}\,.

B^{n}:=\{(x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}:x_{1}^{2}+\cdots +x_{n}^{2}\leq 1\}\,.

The 1-sphere is a circle in the plane while the 1-ball is a closed line segment in the line. For example, the skin of an orange is a 2-sphere while the orange itself is a 3-ball. (Notice that one is "hollow" while the other is "solid".) — Fly by Night (talk) 19:27, 3 July 2012 (UTC)[reply]

OK, I didn't realize that S^n and B^n weren't in the same dimension the way that I thought. (S^2 is equivalent to the surface of B^3) Still sort of surprised there is no name for the next one up...Naraht (talk) 16:50, 5 July 2012 (UTC)[reply]

Almost. What you called the surface, is usually called the boundary. We use the symbol