Wikipedia:Reference desk/Archives/Mathematics/2017 September 18

= September 18 =

Recent result on infinite cardinals
To quote from the aleph number talk page:

Just ran across this article this weekend: https://www.scientificamerican.com/article/mathematicians-measure-infinities-and-find-theyre-equal/ which seems to imply that all infinite sets have equal cardinality. It's been way to long since I studied any set theory for me to tie that into aleph numbers, but it's probably worth a mention. 38.109.137.2 (talk) 15:12, 18 September 2017 (UTC)

The title of that article is clumsily worded. No, it doesn't say at all that all infinite sets have the same cardinality. It says that two particular infinite cardinalities, ones important enough to have been given names ( p {\displaystyle {\mathfrak {p}}} {\mathfrak {p}} and t {\displaystyle {\mathfrak {t}}} \mathfrak{t}), turn out to be provably equal, which was contrary to the previous expectation. This isn't the right place to go into detail. I'd love to say more at the mathematics reference desk, if you'd care to ask a question there. --Trovatore (talk) 16:57, 18 September 2017 (UTC)

I lack the access to the journal to read the original article, and Google isn't exactly good for searching for the definitions of p and t. I understand the aleph zero and aleph one concepts (cardinality of integers and reals) and kind of get aleph two (cardinality of the set of all ranges on the real number line). Upon further reading of the articles referencing the paper I noted that while it was specifically talking about the p and t infinities, it never really defined them, and I was assuming that they corresponded to aleph numbers. Having looked a bit more into aleph numbers, I note that there is appears to be an assumption that there is no "aleph 1.5"; that is all infinite cardinalties have to fall into an integer numbered aleph. So what, roughly, are p and t? Are there any repercussions to the P versus NP problem from this? 38.109.137.2 (talk) 20:04, 18 September 2017 (UTC)
 * Ah, before we get there we have to clear up a misconception. You write, I understand the aleph zero and aleph one concepts (cardinality of integers and reals).  But in fact &alefsym;1 is not necessarily the cardinality of the reals.  The cardinality of the reals is 2&alefsym;0, or 𝖈.
 * The assertion that these two cardinalities, &alefsym;1 and 𝖈, are equal, is called the continuum hypothesis. It is unknown whether it is true or false.  What is known is that it can neither be proved nor disproved in the most common axiomatization of set theory, ZFC. --Trovatore (talk) 20:17, 18 September 2017 (UTC)
 * OK, no problems there? Cool.  I suppose I should go ahead and answer the actual question, then.
 * First you need to understand the structure P(&omega;)/Fin. Basically this is the set of all sets of natural numbers, except that you ignore finite sets.  So you consider the even numbers, {0, 2, 4, 6, 8, 10, ...}, to be the same as the same set, except leaving out 2 and 4 and adding 3 and 5: {0, 3, 5, 6, 8, 10, ...}.  All finite sets are the same as the empty set &mdash; they're the "zero" of this structure.
 * And you have the natural ordering on this structure as "almost inclusion" &mdash; this is the subset order, except again you ignore finite differences. So A&le;B if and only if A&sube;B with possibly finitely many exceptions (that is, A∖B is a finite set).
 * So now 𝖕 is the smallest cardinality of a collection S of elements of P(&omega;)/Fin for which any finite subcollection has a nonzero lower bound, but S itself does not. (A nonzero lower bound for such a collection is called a pseudointersection, which is what the 𝖕 stands for; the property that any finite subcollection has a pseudointersection is called the strong finite intersection property.)
 * On the other hand, 𝖙 is the smallest cardinality of a chain of nonzero elements of P(&omega;)/Fin that has no nonzero lower bound. (The symbol 𝖙 stands for "tower".)
 * It's easy to see that 𝖕&le;𝖙, because any chain as in the definition of 𝖙 is automatically a collection as in the definition of 𝖕.
 * However, until the recent result, apparently the people most familiar with the subject thought that it was probably consistent with ZFC that 𝖕 is strictly less than 𝖙. I am not one of those people that is, I am not an expert on this subject so I can't help you with why they thought that.
 * It is now known that ZFC proves that 𝖕 and 𝖙 are equal.
 * By the way, I looked up the definitions and translated it into the stuff about P(&omega;)/Fin in my head; it seems clearer to me that way. It is possible that I have made a mistake &mdash; if anyone sees one please let me know. --Trovatore (talk) 06:11, 19 September 2017 (UTC)

Triaugmented triangular prism as die
Can a Triaugmented triangular prism be used as a fair 14-sided die? If not, can it at least be used as a fair 7-sided die by printing each number on two faces? Neon Merlin  20:37, 18 September 2017 (UTC)


 * I see no reason to suppose it would be a fair die. You can use a normal 6 sided dice to get 1 to 7 fairly by throwing it once to get a then again to get b, if they are both 6 throw them again, then calculate 6a+b, divide by 7 and add 1 to the remainder to give the result 1 to 7. Only the one case in 36 needs another throw and it is unlikely you'll have to throw too many times. Dmcq (talk) 22:01, 18 September 2017 (UTC)


 * A fair die needs to be isohedral. It is not enough that all faces are the same shape; they also need to all lie in the same symmetry orbit. The triaugmented triangular prism is not isohedral. Double sharp (talk) 04:38, 19 September 2017 (UTC)


 * You can get a fair die for 1 to seven by having a seven sided prism with seven faced pyramids at each end then each number should be on one of each of the triangles at the end and a rectangle on the prism. It should be easy to see this will choose 1 to 7 fairly. Dmcq (talk) 22:50, 19 September 2017 (UTC)


 * Provided the bases are regular heptagons, yes.  Or use an octahedron and discard one number, when it comes up. StuRat (talk) 23:41, 19 September 2017 (UTC)

Symmetry axes of an ellipsoid
Ellipsoid says


 * $${x^2 \over a^2}+{y^2 \over b^2}+{z^2 \over c^2}=1,$$
 * where $a, b, c$ are positive real numbers.
 * where $(a, 0, 0)$ are positive real numbers.


 * The points $(0, b, 0)$, $(0, 0, c)$ and ᙭᙭᙭ lie on the surface. The line segments from the origin to these points are called the semi-principal axes of the ellipsoid.

This seems to imply that there are also other axes. Can we say that every chord through the center is an axis of twofold symmetry? (This appears to me to be true for ellpses.) References if possible, please. Loraof (talk) 22:17, 18 September 2017 (UTC)


 * No, I don't believe so, and I don't believe this applies to a (planar) ellipse, either. Just draw a highly eccentric ellipse aligned with the X and Y axes and a 45 degree angle line through the center. You will see it is not symmetrical about that line.  Same is true in 3D.  Here's a rough pic to use to visualize the 2D case: .  Chord CW isn't perpendicular to the ellipse boundary, as you see, so if one half of the ellipse is mirrored about it, it's tangent would be discontinuous with the original half.  StuRat (talk) 22:33, 18 September 2017 (UTC)


 * You're talking about reflectional symmetry. I'm talking about twofold rotational symmetry—if you rotate an ellipse by 180° it occupies the same position as it originally did. [added subsequently:] But rotational symmetry in 2D is about a point, not a line, so I guess it doesn't work for the ellipse wrt a chord.


 * So my question remains, slightly clarified: Does the ellipsoid have twofold rotational symmetry? Loraof (talk) 02:32, 19 September 2017 (UTC)


 * No, the three semi-principal axes are the only axes of twofold rotational symmetry, if the ellipsoid is not a sphere. For example, x=y=z is not an axis of symmetry of $$\frac{x^2}4+y^2+z^2 = 1$$. $$(2,0,0)$$ lies on the ellipsoid, but rotating the point 180 degrees about x=y=z will take it off the ellipsoid to $$(-\tfrac23,\tfrac43,\tfrac43)$$. C0617470r (talk) 07:15, 19 September 2017 (UTC)


 * The semi-principal axes are actually principal semi-axes. Bo Jacoby (talk) 09:27, 19 September 2017 (UTC).

Thanks. So why are the three axes of rotational symmetry called (at least in the Wikipedia article Ellipsoid) "the principal axes" instead of "the (only) axes"? Loraof (talk) 16:48, 19 September 2017 (UTC)


 * An ellipsoid (as for any solid) has an infinite number of axes passing through the centre. The three axes of the ellipsoid that have rotational symmetry (of infinite order or of order 2) are called the principal axes.  No other axis has rotational symmetry of any order.  If rotated about any other axis, the ellipse would "wobble" and there would be stresses on the axis of rotation. The ellipsoid also has three planes of reflective symmetry.  Each of these planes contains two of the principal axes.   D b f i r s   18:04, 19 September 2017 (UTC)


 * So an axis of any solid object is by definition any chord through the center? If so, does the "center" of a solid object here mean the centroid? Loraof (talk) 03:08, 20 September 2017 (UTC)


 * I suppose any straight line could be an axis, but usually, in rotational dynamics, we consider rotation about an axis through the centre of mass (or center on your side of the pond). For an object of uniform density, the centroid coincides with the centre of mass.  The centre of gravity is also the same point in a uniform gravitational field.  In other contexts, an axis might implicitly mean an axis of symmetry, which I guess is what you were thinking of.    D b f i r s   16:48, 20 September 2017 (UTC)


 * The term principal axes for an ellipsoid is analogous to the major and minor axes of an ellipse. It's just that an ellipsoid has three axes so it isn't worth the effort to distinguish between them. Semi-principal axes (or principal semiaxes) is analogous to semi-major and semi-minor axes. C0617470r (talk) 19:03, 20 September 2017 (UTC)