https://epoch.ai/frontiermath/open-problems

Crossposted from https://piefed.social/c/science/p/1406456/first-shape-found-that-cant-pass-through-itself

In 1876, Peter Guthrie Tait set out to measure what he called the “beknottedness” of knots.

Tait had an idea for how to determine if two knots are different. First, lay a knot flat on a table and find a spot where the string crosses over itself. Cut the string, swap the positions of the strands, and glue everything back together. This is called a crossing change. If you do this enough times, you’ll be left with an unknotted circle. Tait’s beknottedness is the minimum number of crossing changes that this process requires. Today, it’s known as a knot’s “unknotting number.”

If two knots have different unknotting numbers, then they must be different. But Tait found that his unknotting numbers generated more questions than they answered.

If Tait missed something, so did every mathematician who followed him. Over the past 150 years, many knot theorists have been baffled by the unknotting number. They know it can provide a powerful description of a knot. “It’s the most fundamental [measure] of all, arguably,” said Susan Hermiller (opens a new tab) of the University of Nebraska. But it’s often impossibly hard to compute a knot’s unknotting number, and it’s not always clear how that number corresponds to the knot’s complexity.

To untangle this mystery, mathematicians in the early 20th century devised a straightforward conjecture about how the unknotting number changes when you combine knots. If they could prove it, they would have a way to compute the unknotting number for any knot — giving mathematicians a simple, concrete way to measure knot complexity.

Researchers searched for nearly a century, finding little evidence either for or against the conjecture.

Then, in a paper posted in June, Hermiller and her longtime collaborator Mark Brittenham (opens a new tab) uncovered a pair of knots that, when combined, form a knot that is easier to untie than the conjecture predicts. In doing so, they disproved the conjecture (opens a new tab) — and used their counterexample to find infinitely many other pairs of knots that also disprove it.

The result demonstrates that “the unknotting number is chaotic and unpredictable and really exciting to study,” she added. The paper is “like waving a flag that says, we don’t understand this.”

I'm not very advanced in math or coding but I do enjoy some of the low level challenges. Anyone else do these? Link

@math
Hi! I am currently working on an open Diophantine equation listed in a publication by Dr Bogdan Grechuk: y^3 + xy = x^4 + 4. The question if any integer solutions exist eludes brute force attempts even with scripts in systems such as SageMath. Empirical work suggests there may be no solutions. I have been trying to prove by various approaches that no integer solutions exist. Dr Gerchuk gave critical feedback on my first two approaches but I am hoping to get more informal feedback or even collaboration on this before I send it to him a third time.
Here is the Overleaf link to the project and I am open to also share a link to edit and possibly coauthor if anyone is interested in helping get this case closed or to lay it to rest.
https://www.overleaf.com/read/ncqwpfxzkqfr#5621d2

Part (b). So, this is from the introduction of a discreet math book, How to Prove it, idk where to even start with this, and figure that since it's part of the 1st question in the intro that I should know how to do it. But, alas.

It's based on a fairly common interview question but spiced up a bit.

It's based on this but I spiced it up a bit: https://medium.com/@shelvia1039/brain-teaser-2-tiger-and-sheep-9293acd97012

https://goatmatrix.net/c/MatrixFun/F71YGTSzZg

Project website of the guys who proofed the Einstein tile tiles the plane non-periodically.

Random thought on magic squares:

If I view the smallest possible non-trivial magic square

2 7 6
9 5 1
4 3 8

since its rows and diagnoals sum up to 2+5+8 = 2+7+6 = 4+5+6 = 2+9+4 = … = 15

in the article as a 3x3 Matrix, its determinant is Δ = -360 . Its inverse:

-37/360 19/180 23/360
17/90 1/45 -13/90
-7/360 -11/180 53/360

note how this is a magic square, rows and diagonals sum up to 1/15.

https://matrix.reshish.com/inverse.php

Now if you are really bored (I can not do this): proof that for any non trivial magic squares the inverse …

• exists (i.e. every non-trivial magic square has an inverse)
• is a magic square.

Two students who discovered a seemingly impossible proof to the Pythagorean theorem in 2022 have wowed the math community again with nine completely new solutions to the problem.

While still in high school, Ne'Kiya Jackson and Calcea Johnson from Louisiana used trigonometry to prove the 2,000-year-old Pythagorean theorem, which states that the sum of the squares of a right triangle's two shorter sides are equal to the square of the triangle's longest side (the hypotenuse). Mathematicians had long thought that using trigonometry to prove the theorem was unworkable, given that the fundamental formulas for trigonometry are based on the assumption that the theorem is true.

Jackson and Johnson came up with their "impossible" proof in answer to a bonus question in a school math contest. They presented their work at an American Mathematical Society meeting in 2023, but the proof hadn't been thoroughly scrutinized at that point. Now, a new paper published Monday (Oct. 28) in the journal American Mathematical Monthlyshows their solution held up to peer review. Not only that, but the two students also outlined nine more proofs to the Pythagorean theorem using trigonometry.

Calculator: https://www.omnicalculator.com/everyday-life/dilution-ratio

If I type in the dilution ratio and final volume it calculates the concentrate amount and water amount but I don't know how it does that and want to find out how it does that

Then I am stuck. I think the provided answer contains an error. But even if they are right, why does this last step equal f(x,y) + g(y) ????

(For the sake of intuition, 1/√0=0)

I'm thinking re the latest vid of @mindyourdecisions

No need to view his vid. Here's the problem –

Brian has some boxes of paper clips. Some boxes hold 10 clips and some boxes hold 100. He has some paper clips left over. He has 3 more boxes with 100 paper clips than he has boxes with 10 paper clips. He has 2 fewer paper clips left over than he has numbers of boxes with 100 paper clips. What number of paper clips could he have?

• let x1 be the number of boxes with 10 clips
• x2 be the number of boxes with 100 clips
• n be the number of leftover clips

I thought of 100x2 = 10x1 + 300

Is that equation right? Something tells me I shouldn't equate 100x2 to 10x1 plus 300. Something tells me I shouldn't make an equation re number of clips as it isn't explicit in the problem. I'm confused.

Hi,

I found online a nice (and seemed easy) math problem.

Rocket A travel from Mars to Earth in 200 days
Rocket B travel from Earth to Mars in 150 days, but take off 30 days later

When they cross each other, which one is the closet to the earth ?

So they give a "flat" answer, without giving any explanation on how they reach this conclusion.

What would be your simplest Mathematical solution for this ?

Thanks.

So this is bugging me for a while and I'm just do dumb to get how I solve this, but here's the situation:
Given I take a local backup of my system daily and have a retention policy that keeps a backup of the past 7 days each, a backup of the past 4 weeks each and a backup of the past 6 month each. That's either 17 backups or less if you consider some backups being counted as a daily and weekly or as a weekly and monthly. But that's not that important.
The interesting part is, that I also take a remote backup of my local backup daily, which has the same retention policy, so it's cascading. Here there is obviously a huge overlap of backups, but I can't wrap my head around, how I calculate this.
Is anybody willing and/or interested to solve this for and with me?

This year's Abel Prize has just been awarded to Michael Talagrand. I didn't knew about his work, but it seems really interesting and he made an effort to make it really accessible both to read and access.

Isn't it just "composite"?

Every arrow in category can be composed, the set(or class or whatnot..) of that is composite.