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MHB Collection of Most Bizarre Theorems and Conjectures

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Hi, MHB.

I have decided set up a thread on collection of most bizarre statements of mathematical problems. I have thought of some, and I hope other MHB members would add some - it would be fun! The statement of the problem you are going to post should be

1) Very weirdly related to the reality
2) Must seem unapproachable in terms of mathematical language at a glance
3) Very tough to solve even if described mathematically
4) A little famous in mathematical community.

Here are my choices :

Theorem (Toeplitz, partially solved by Stromquist) : Every simple closed curve that you can draw by hand will pass through the corners of some square.

Theorem (Brouwer, solved by Hadamard & Brouwer) : You have two identical pieces of paper with the same picture printed on them. Put one flat on a table and the other one you crumple up without tearing it and place it on top of the first one, then there is some point in the picture on the crumpled-up page that is directly above the same point on the bottom page.

Balarka
.
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Theorem (Toeplitz, partially solved by Stromquist) : Every simple closed curve that you can draw by hand will pass through the corners of some square.

.
Interesting ! , what do you mean by partially solved ?
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Interesting ! , what do you mean by partially solved ?
Stromquist's theorem says settles the matter for nice curves, but several simple and closed curves has fractal-like behavior, so surely, this is not enough.

Here's a picture of a particularly nasty curve with an inscribed square :

inscribed_square.gif

Have fun twiddling with this nice little conjecture! Who knows, you might even solve it!

NOTE : The conjecture has been proved for triangles, and shown that the vertices of all the triangles that is similar to the given is dense in any nice simple closed curve.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Here's another :

Theroem (Plücker) : The number of real bitangents of any quartic curve is either 28, 16 or smaller than 9.

A construction of 28 by Trott (7 given, others can be derived from symmetry) :

 
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ModusPonens

Well-known member
Jun 26, 2012
45
This is one of the most fascinating ideas for a topic I've seen! So far living to the expectation of what it could be. Thanks. :)

Here is my modest contribution, which possibly doesn't fit all criteria:

A Banach Space X with dim (X) = $\infty$ doesn't have a countable vector space basis.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
ModusPones said:
Here is my modest contribution, which possibly doesn't fit all criteria
Although your problem doesn't satisfy (1), I certainly do accept theorems realted to countability, or decidability.

Here are some others;

Theorem (Jaffe & Ruberman) : Any sextic surface on P3 have at most 65 double points.

A construction by Barth :



Theorem (Banach & Tarski) : Given a solid ball in 3‑dimensional space, there exists a decomposition of the ball into a finite number of non-overlapping pieces which can then be put back together in a different way to yield two identical copies of the original ball.

Theorem (Poincaré, solved by Perelman) : In a compact 3-dimensional surface without boundary, if every loop can be tightened to a certain point in the surface, then the surface is homeomorphic (i.e., similar) to a 3-sphere.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Most of the theorems seem to be connected to Algebraic topology here is one that I see interesting

Brouwer fixed-point theorem: Every continuous mapping \(\displaystyle f : D^n \to D^n\) has a fixed pint.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Most of the theorems seem to be connected to Algebraic topology here is one that I see interesting

Brouwer fixed-point theorem: Every continuous mapping \(\displaystyle f : D^n \to D^n\) has a fixed pint.
Heh :D, a similar version of this was given by me in the first post to satisfy condition (1) :

myself said:
Theorem (Brouwer, solved by Hadamard & Brouwer) : You have two identical pieces of paper with the same picture printed on them. Put one flat on a table and the other one you crumple up without tearing it and place it on top of the first one, then there is some point in the picture on the crumpled-up page that is directly above the same point on the bottom page.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
The hairy ball theorem: you cannot comb a spherical dog in such a way that its fur lies smoothly flat at each point.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
The hairy ball theorem: you cannot comb a spherical dog in such a way that its fur lies smoothly flat at each point.
I definitely missed this one - very nice!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
A coast line has infinite length. -Benoit Mandelbrot
Its dimension is something between that of a curve and a surface.
A specific example is the Koch curve with dimension $\log_3 4 \approx 1.26186$.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
A specific example is the Koch curve with dimension $\log_3 4 \approx 1.26186$.
Indeed a very good example. This post made me think of a conjecture :

The Hausdorff dimension of a certain Weierstrass function \(\displaystyle \sum_{k=1}^{\infty} \frac{\sin (2^k x)}{\sqrt{2}^k}\) is exactly 3/2.

Also, thinking of fractals, the Banach-Tarski paradox holds for a 3-banach tree (and the construction is particularly of interest).
 

DreamWeaver

Well-known member
Sep 16, 2013
337
What a great idea for a thread, Mathbalarka!!! Well played... (Muscle)


This one doesn't quite fit your criteria, but I think/hope it's odd enough to fit in well here.


The Hairy Ball Theorem (no giggles, please! :rolleyes:)

When considering a tennis ball, or other hairy ball, if one attempts to comb all of the hairs in a particular direction, this attempt will be successful in the general sense, except for a tufty patch at one (arbitrary) pole, and a bald patch at another.

Now, considering the weather systems (ie wind currents) of planet Earth to be analogous to hairs on a ball being swept in arbitrary directions, the Hairy Ball Theorem tells us that - at all times - there must be at least two 'weather singularities', ie cyclones, hurricanes, etc, somewhere, on our beloved planet... AT. ALL. TIMES... :eek:

Not bad weather prediction for a hairy ball... (Heidy)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
There exists no retraction between a disk and a circle.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Theorem (Riemann) : If an infinite series is conditionally convergent, then a permutation exists between the terms which, after summed, converges or diverges.

A consequence of this can be made so that (1) is satisfied : The product of all real numbers without 0 converges to -1. (*)

(*) Note how the order is not specified at this point!

PS : The theorem is called Riemann rearrangement theorem.

Balarka
.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Theorem (Riemann) : If an infinite series is conditionally convergent, then a permutation exists between the terms which, after summed, converges or diverges.
Even more bizarre than that, if an infinite series is conditionally convergent, then the terms can be rearranged to make the series converge to any sum that you care to choose.

For example, the series $1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \ldots$ converges (conditionally) to $\ln2$. But by rearranging the terms you can make it converge to $\ln3$, or to $-999$, or to $\pi^{17}$, or to anything else you choose.

The proof is a lot easier than you might expect. Notice that the odd-numbered terms are all positive, and form a divergent series. The even-numbered terms are all negative, and also form a divergent series. To rearrange the series so that it converges to the limit $l$, start by taking the positive terms $1+\frac13 + \frac15 + \ldots$, stopping as soon as their sum exceeds $l$. Then take sufficiently many of the negative terms $-\frac12 - \frac14 - \frac16 - \ldots$ until the sum dips down below $l$. Then take some more of the positive terms until the sum goes back up above $l$, and so on. That way, you force the rearranged series to converge to $l$.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Opalg said:
Even more bizarre than that, if an infinite series is conditionally convergent, then the terms can be rearranged to make the series converge to any sum that you care to choose.
Indeed. $S = 1 - 1 + 1 - 1 + 1 - 1 + ...$ can be made to converge to pretty much anything in $\mathbb{Z}$, and of course 1/2. This is what I at least have off the top of my head.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Great thread! Keep 'em coming... (heart)
 

Shobhit

Member
Nov 12, 2013
23
The Riemann hypothesis (in the language of Integrals and Series ):

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$

where $\gamma$ denotes the Euler Mascheroni Constant.

Reference: V. V. Volchkov, On an equality equivalent to the Riemann hypothesis
 
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mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Yes, I knew that form of RH, it's particularly nice. (although not much of a way to approach the conjecture :p)

Okay, here's one from number theory :

Theorem (Tanaka, as a disproof of Polya's conjecture) : The smallest number n for which most of the number smaller than n has even number of prime factors is n = 906150257.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Conjecture (Jaeger) : Every bridgeless graph has a cycle-continuous mapping to the Petersen graph.

And, oh, how did I forgot these two :

Theorem (Proved by Appel & Haken) The regions of a map can be colored using at most four colors so that no two adjacent regions have the same color.

Conjecture (McKay, Radziszowski & Exoo) R(5, 5) = 43 where R(a, b) is the least number of vertices of a graph red or blue colored for which a subgraph of a number of vertices or a subgraph of b vertices exists with edges painted entirely red or entirely blue, respectively.
 

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

Humble conjecture: "mente oscura".

[tex]Let \ n \in{N} \ / \ P(n)=n^2-n+41[/tex]

[tex]Let \ p \in{N} \ / p>1 \ / \ p|P(n)[/tex]

Conjecture:

[tex]p \geq{41}[/tex]

Regards.
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573

mente oscura

Well-known member
Nov 29, 2013
172

DreamWeaver

Well-known member
Sep 16, 2013
337
Just a thought, mind, but since there's been plenty of abstract ideas/theorems on here already, does anyone have any nice visual or geometric ones to share...???

(Hug)