Mathematics desk
< December 19	<< Nov \| December \| Jan >>	December 21 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 20

apeirogon and zerogon

Yesterday's Wiktionary "Word of the day" was "apeirogon" - a polygon with an infinite number of sides and vertices. Wikipedia also has an article about that: apeirogon.

The Wiktionary definition for apeirogon says that "zerogon" is a synonym...yet the definition it provides for zerogon defines it as a polygon with zero edges and vertices.

Seems to me that wiktionary is incorrect - zero is not infinity (well, DUH!)
Seems to me that the reason someone there thought these were synonyms is that (allegedly) a circle could be described as either a zerogon (it has no vertices) or as an apeirogon (a regular apeirogon is arguably a circle). I was under the impression that polygons had to have straight edges - so I don't think a circle is a zerogon...but then IANAM.
If wiktionary is correct, or even if the terms are loosely related, why no mention of zerogon in our apeirogon article?
If zerogon is a real term, shouldn't we have an article about it and an entry in Template:Polygons - which already has entries for weird shit like the monogon and digon?

Either way, something needs fixing - I'm just not quite sure which. SteveBaker (talk) 14:13, 20 December 2015 (UTC)[reply]

In the case of a regular polygon, they seem synonymous to me, as the sides on the regular apeirogon are all infinitely short, which really makes them points, not sides. But for the general case, you could have an apeirogon with a finite number of measurable sides and an infinite number of (infinitely short) sides, elsewhere. That's not a zerogon. StuRat (talk) 18:16, 20 December 2015 (UTC)[reply]

"Zerogon"? I guess I can accept "0-gon", but this looks really weird. What's it supposed to be in Ancient Greek? "Medenagon"? Anyway, digons are way above monogons and 0-gons in legitimacy, being proper abstract polytopes. A regular apeirogon may also be construed as a line subdivided equally into an infinite number of congruent segments. Each interior angle is thus pi, as expected.

A 0-gon would have to have a boundary and no sides, so drawing it on a surface would have to give a single point with the interior being the rest of the surface. This very OR-ish. I have not seen references to 0-gons in respectable sources at all. (Coxeter at least sort of mentions monogons.) Further, it has an unbounded area unless the surface you draw it on is closed, like a sphere.

To me, though, the biggest problem with polygons with 0 or 1 sides is that they have no 1-polytopes as elements. Double sharp (talk) 18:30, 20 December 2015 (UTC)[reply]

BTW a monogon can be easily constructed on a conic surface. Take a circular sector, choose two points on radii, equally distant from the vertex, and join them with a line. When you wrap the figure to make a cone, the line segment becomes a legitimate monogon. --CiaPan (talk) 07:27, 22 December 2015 (UTC)[reply]

Usually I see apeirogon defined as a half-plane bounded by a line divided into equal segments, or else (in the hyperbolic plane) a polygon inscribed in a horocycle. —Tamfang (talk) 10:01, 22 December 2015 (UTC)[reply]

Sounds like a joke. A null polytope or nullitope is an abstract polytope with no vertices, so maybe its a null polygon or nulligon? Tom Ruen (talk) 18:39, 20 December 2015 (UTC)[reply]

Is there any concept besides Division by zero that does not exist, but gets analysed throughly?

Is there any concept besides division by zero that does not exist, yet has a Wikipedia article?--Jubilujj 2015 (talk) 19:11, 20 December 2015 (UTC)[reply]

Bottom type, Russell's paradox, Least interesting number, Penrose triangle, Cantor's diagonal argument, Fermat's Last Theorem, an enormous amount of mathematics is about things that don't exist. Dmcq (talk) 20:47, 20 December 2015 (UTC)[reply]

Add the Set of all sets, a.k.a. Universal set, closely related to Russel's paradox. --CiaPan (talk) 07:05, 22 December 2015 (UTC)[reply]

Division by zero is in fact possible in some natural and useful structures, which the article is at pains to explain. --Trovatore (talk) 20:57, 20 December 2015 (UTC)[reply]

Once upon a time the idea that a number might exist, but not be the ratio of two integers, seemed totally irrational. Later, a number that was the square root of a negative number was a similarly imaginary concept. In both cases, analyzing these concepts led to important extensions of mathematics by giving rise to the understanding that they were possible after all. --76.69.45.64 (talk) 00:03, 21 December 2015 (UTC)[reply]

I am really confused by your question. Are there any concept besides "Division by zero" that does not exists? Dude, there are no concepts that does not exists because by definition, if it is a concept then that concept exists! Concepts exists. You are like asking. Are there any imaginations that are not imagined by any humans? 175.45.116.66 (talk) 05:13, 21 December 2015 (UTC)[reply]

Cubic function formula proof

Anyone know How do we solve the cubic $ax^{3}+bx^{2}+cx+d=0$ step-by-step, without skipping any logical steps?

Most proofs I've seen so far always do this, and I end up not believing a word. I just wonder if it's easy to solve just like the quadratic function $ax^{2}+bx+c=0$ .

Please find me honest answers because I'm slow. For example, no-one explains why it's allowed to erase $bx^{2}$ somehow out of the equation. יהודה שמחה ולדמן (talk) 19:56, 20 December 2015 (UTC)[reply]

See cubic equation#Reduction to a depressed cubic Sławomir
Biały 20:10, 20 December 2015 (UTC)[reply]

There is nothing new under the sun. You want a formula for solving the cubic equation like a formula for solving the quadratic equation? Here it is. [Formula for cubic equation]. 175.45.116.66 (talk) 00:55, 21 December 2015 (UTC)[reply]

Multiplicatively perfect numbers

A number n is multiplicatively perfect if the product of the divisors of n is equal to n². Is it true that the multiplicatively perfect numbers are just 1, the cubes of primes, and the product of two distinct primes? 172.56.30.117 (talk) 23:54, 20 December 2015 (UTC)[reply]

This seems like a homework problem. Hint: if p and q are primes, what are all the divisors of pq? And what are all the divisors of p³? --76.69.45.64 (talk) 00:06, 21 December 2015 (UTC)[reply]

More generally, suppose you want to find numbers n so that the product of the divisors is n^k. If n = p^aq^b... then the exponent of p in the product of divisors is (1/2)a(a+1)(b+1)..., the exponent of b is (a+1)(1/2)a(b+1)..., ans so on. So the problem amounts to solving the simultaneous equations (1/2)a(a+1)(b+1)... = ka, (a+1)(1/2)b(b+1)... = kb, etc. But with cancellation these reduce to the same equation (a+1)(b+1)... = 2k. In other words look at every way of factoring 2k and subtract 1 from the exponents to get the possible values for a, b, .... So for k=3 you get numbers of the form p⁵ and p²q; for k=18 you get numbers of the form p³⁵, p¹⁷q, p¹¹q², p⁸q³, p⁵q⁵, p⁸qr, p⁵q²r, p³q²r², and p²q²rs. A corollary is that the product of the factors of n is always a power of n unless unless n is a perfect square. Also he product of the factors of n² is always an odd power of n. --RDBury (talk) 06:38, 21 December 2015 (UTC)[reply]