Can A Three-String Knot be Constructed?

  • Thread starter Thread starter spill
  • Start date Start date
AI Thread Summary
The discussion centers on the possibility of constructing a three-string knot that collapses into an unknot when any one of the strings is removed. It is clarified that if a knot cannot self-intersect, removing a string will result in a non-knot curve unless the arc is closed. The concept of Brunnian links is introduced as a solution, where the removal of any component results in the remaining components becoming an unlink. The Borromean rings are cited as a specific example of such links. The conversation concludes with a confirmation that Brunnian links are indeed the sought-after solution.
spill
Messages
4
Reaction score
0
This is a problem that's been troubling me recently. I'm neither a boy scout nor a knot theoretician, so I'm not sure how to progress.

Is it possible for three pieces of string to be tied together without the knot being topologically equivalent to a knot tying two of the pieces together with the third string tied either around one of the other pieces or around the first knot itself?

Oh dear, that wasn't very clear. Let's try again.

Is it possible to construct a knot from three pieces of string where removing any of the pieces will cause the knot to collapse into an unknot?

Help appreciated.
 
Mathematics news on Phys.org
Since a knot cannot self-intersect, removing an arc from a knot will make it a curve that is not a knot. If you mean remove an arc and close the curve, then there are many obvious choices of making an unkot from a nontrivial knot. If you meant 3 links and the resulting 2 links collapsing into the unlink, then the Borromean rings is an example of one such link. Such links are called http://www.cs.ubc.ca/nest/imager/contributions/scharein/brunnian/brunnian.html and can have any number of components.
I've probably misunderstood your question, though. :)
 
Last edited by a moderator:
Brunnian links - exactly what I was looking for. Thanks!
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top