# A knotty problem **URL:** \ **Category:** Factual Questions\ **Created:** [July 30, 2017, 4:59pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448 "2017-07-30T16:59:01Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [July 30, 2017, 4:59pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/1 "2017-07-30T16:59:01Z")
Recently Ethan Segal on _Starts With A Bang_ had this to say, in making an analogy with topological defects (bolding mine): > [@](#): > > imagine that you tied a knot in the string: just a loop, criss-cross, tuck, and pull.\<…\>Now, you can do some interesting things with that knot-containing line. You can tie another knot in it the exact same way, and now you’ll have two topological defects that both add. But if you tie a knot in the opposite direction, i.e., you made the same loop, but you criss-crossed the ends the opposite way before tucking and pulling, you make a knot that’s the topological opposite of the original knot. **If you very carefully brought both the original knot and this new, oppositely-tied knot together, you’d find that they could undo each other** , Really? Can you actually get two opposite thumb knots to cancel each other, without passing the free ends of the string through any of the loops? I haven’t found a demonstration yet. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [July 30, 2017, 5:16pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/2 "2017-07-30T17:16:22Z")
In knot theory we learn that the unknot cannot be written as the composition of two non-trivial knots, so it sounds like Segal may be lying to us. ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [July 30, 2017, 5:19pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/3 "2017-07-30T17:19:20Z")
Absolutely not (no pun intended). There’s a simple theorem in knot theory that says that no knot possesses an inverse knot, such that the two will cancel out when tied in the same string. Here’s an elementary proof, from the book _Knots_ by Alexi Sossinsky: By contradiction, assume two knots A and B are inverse knots. Now imagine an infinitely long string with alternating knots tied like this: ABABABAB… By grouping them like this, they all cancel and the string can be completely unknotted: (AB)(AB)(AB)(AB)… But by grouping them like this, they all cancel except the first one, and string remains with knot A in it: A(BA)(BA)(BA)(BA)… A string cannot be both knotted and unknotted, so the original assumption that inverse knots exist must be false. There is a bit of handwaving in this argument because of the assumption that an infinite string of knots can exist, but this can be made mathematically rigorous. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 30, 2017, 9:24pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/4 "2017-07-30T21:24:11Z")
You also have to assume that inverse knots would be bidirectional inverses: AB is not necessarily equal to BA, and so AB = identity would not necessarily imply that BA = identity. Though there’s probably some rigorous way to address that, too. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 30, 2017, 9:32pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/5 "2017-07-30T21:32:16Z")
On thinking about it some more, I guess it depends on what you mean by “tying another knot”. If you start with any knot, and then feed the end of the rope back along parallel to itself, you can turn the knot into a knot-on-a-bight, which is equivalent to no knot at all. ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [July 30, 2017, 10:19pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/6 "2017-07-30T22:19:36Z")
> [@Chronos](#): > > You also have to assume that inverse knots would be bidirectional inverses: AB is not necessarily equal to BA, and so AB = identity would not necessarily imply that BA = identity. Though there’s probably some rigorous way to address that, too. It’s rather simple to prove that knots are commutative – for any two knots, AB = BA. A knot can always be shrunk, “slid” along the string, through the other knot, and to the other side. There’s a nice picture in the Sossinsky book but I haven’t found one online. Hopefully it’s fairly easy to visualize this. ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [July 30, 2017, 10:22pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/7 "2017-07-30T22:22:15Z")
> [@Chronos](#): > > On thinking about it some more, I guess it depends on what you mean by “tying another knot”. If you start with any knot, and then feed the end of the rope back along parallel to itself, you can turn the knot into a knot-on-a-bight, which is equivalent to no knot at all. In knot theory, and as I read it, in the OP’s quote from Segal, two knots are _composed_ by tying them separately, in nonintersecting areas of the string, then manipulating the string to produce one knot from the two separate knots. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [July 30, 2017, 10:32pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/8 "2017-07-30T22:32:10Z")
Also, the string should form a topological circle, so there are no loose ends to reeve back through and undo all the knots. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 30, 2017, 10:39pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/9 "2017-07-30T22:39:03Z")
Well, while you’re tying a knot, the ends have to be loose in some sense. But you’re right that it wouldn’t be a proper composition. And I also see what you mean about knots being commutative. Knot theory is an area I’ve never given all that much deep thought. ---
**Author:** ![Senegoid](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senegoid/32/6606_2.png) [@Senegoid](https://boards.straightdope.com/u/Senegoid)\ **Post date:** [July 31, 2017, 2:06am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/10 "2017-07-31T02:06:23Z")
I’m really skeptical about any “proof” that involves an infinite string of knots as **markn+** cites above, since infinite series in general simply do NOT behave like finite series. Would the same be true of an infinite series (vs a finite series) of knots? Proof that the number 1 can have no additive inverse: Consider the series: 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 . . . . by which we mean, of course: (+1) + (-1) + (+1) + (-1) + (+1) + (-1) + (+1) + (-1) . . . . This can be grouped this way: ((+1) + (-1)) + ((+1) + (-1)) + ((+1) + (-1)) + ((+1) + (-1)) . . . . = 0 or: (+1) + ((-1) + (+1)) + ((-1) + (+1)) + ((-1) + (+1)) + ((-1) + (+1)) . . . . = +1 The fact that the same series can be grouped to sum to 0 or +1 seems to be a contradiction. If we start this proof by assuming that (+1) has an additive inverse (which I didn’t say specifically, but suppose I did), then I’ve just proved by contradiction that it can’t be so. Does the “rigorous proof” that **markn+** mentions have a similar failing, or does it get around it? ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [July 31, 2017, 2:18am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/11 "2017-07-31T02:18:08Z")
The connected sum of knots in **markn+**'s proof is not like adding numbers. An infinite series like 1-1+1-1-1+1… means to consider the sequence of partial sums 1, 1-1, 1-1+1, etc. But, in his proof, the infinite connected sum A#B#A#… really means to consider an infinite string of knots all at once. The result is still a well-defined knot. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [July 31, 2017, 2:38am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/12 "2017-07-31T02:38:07Z")
I wonder why Siegel bullshitted his readers with that false fact? You would think that with a non-obvious statement like that he would have at least taken a piece of string and tried it himself. ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [July 31, 2017, 2:42am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/13 "2017-07-31T02:42:43Z")
> [@Chronos](#): > > a knot-on-a-bight, which is equivalent to no knot at all. Which interestingly is how you tie a [bowline on a bight](https://s-media-cache-ak0.pinimg.com/736x/00/b5/4c/00b54c3f4d07444de10812f8b118209f.jpg) in the middle of a rope without needing to bother with the free ends. ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [July 31, 2017, 2:44am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/14 "2017-07-31T02:44:22Z")
> [@markn\_1](#): > > In knot theory, and as I read it, in the OP’s quote from Segal, two knots are _composed_ by tying them separately, in nonintersecting areas of the string, then manipulating the string to produce one knot from the two separate knots. Or rather, no knot, in the belief that a knot plus it’s mirror-image equals no knot overall- which apparently isn’t the case. ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [July 31, 2017, 2:32pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/15 "2017-07-31T14:32:20Z")
> [@Senegoid](#): > > I’m really skeptical about any “proof” that involves an infinite string of knots as **markn+** cites above, since infinite series in general simply do NOT behave like finite series. You are right to be skeptical. In fact, Sossinsky raises the exact objection to his proof that you raise (the analogy with a sum of alternating +1 and -1). He says > [@Sossinsky](#): > > The error in reasoning is the same in both cases: one cannot manipulate _infinite_ sums or compositions (they must be defined beforehand) as one manipulates finite sums or compositions. > > In the case of knots, however, a slight modification of the argument makes it totally rigorous. Just place copies of the knots a and b one by one in an infinite series of boxes that become smaller and smaller and converge to a point – which by the way, rigorously defines the infinite composition C – and replace the questionable arithmetic operations by correctly defined topological manipulations. I will skip the technical details. The reader will have to take my word that this clever proof using infinite compositions is more than a brilliant sophism (like Achilles and the tortoise); indeed it is the basis of a rigorous mathematical argument. ---
**Author:** ![markn\_1](https://avatars.discourse-cdn.com/v4/letter/m/f9ae1b/32.png) [@markn\_1](https://boards.straightdope.com/u/markn_1)\ **Post date:** [July 31, 2017, 2:35pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/16 "2017-07-31T14:35:27Z")
> [@Lumpy](#): > > Or rather, no knot, in the belief that a knot plus it’s mirror-image equals no knot overall- which apparently isn’t the case. The absence of a knot is also considered to be a knot, called the _unknot_, just as zero is considered to be a number. **DPRK** mentioned back in post #2 that the unknot cannot be the composition of two nontrivial knots (that is, neither being the unknot), which is just another way of saying that no knot has an inverse. ---
**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\ **Post date:** [August 2, 2017, 3:25am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/17 "2017-08-02T03:25:19Z")
> [@Lumpy](#): > > Recently Ethan Segal on _Starts With A Bang_ had this to say, in making an analogy with topological defects (bolding mine): Segal isn’t talking about a useful knot like a thumbknot. Where he says “knot”, replace it with “loop”. He didn’t make that extremely clear, but he did introduce loops and then start talking about knots. I was recently fitting a chain to a chain saw. Due to the way I dropped the chain, I saw eight small loops. (Ignoring the loop of the O shape.) But, I was sure that that factory wouldn’t sell me a knotted chain, that I’d have to cut apart and join back together. Well it turns out that if you start with the circle of chain, O shape, then you can get an even number of loops in it, and there must be an equal number of clockwise up loops and anticlockwise up loops. Just remember you can only cancel any loop with a loop of the opposite direction… When you cancel enough loops, you have a chain you can fit to the chain saw. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [August 2, 2017, 1:56pm UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/18 "2017-08-02T13:56:09Z")
What you wrote is clear and makes sense, and is something most people can visualize and have hands-on experience with. Yet Segal (obviously not the venerable Irving Segal!) wrote “…criss-cross, tuck, and pull,” which is highly misleading, especially in combination with the word “knot.” Also, he said to take a piece of string, not a ribbon or a chain. If you have a simple loop on a string, a 180-degree twist undoes it without requiring an anti-loop. Therefore you may be giving him too much credit, not that he should be getting any credit for sloppy writing in the first place. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 3, 2017, 4:19am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/19 "2017-08-03T04:19:52Z")
The infinite sum reasoning in the knot-theoretical proof given above is perfectly valid and has been well-understood for over half a century. This is known as the Mazur swindle; look it up for more information. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [August 3, 2017, 8:58am UTC](https://boards.straightdope.com/t/a-knotty-problem/792448/20 "2017-08-03T08:58:07Z")
For what it’s worth, **Senegoid** ’s reasoning about Grandi’s series IS fundamentally isomorphic to the reasoning in the Mazur swindle; they were correct to draw that analogy. However, the distinction is that there IS a natural infinite summation operator with the expected properties for knots (in the way Sossinsky describes above in post 15), while this very reasoning in the context of Grandi’s series shows that such an infinite summation operation can NOT exist for signed integers or such. There IS a natural infinite summation operation on unsigned cardinalities, and in this way one could prove that any two cardinalities whose sum is zero are themselves zero, but this is obvious on its own terms anyway (if a union of two sets is empty, both the sets were themselves empty, clearly). [Next page](https://boards.straightdope.com/t/a-knotty-problem/792448.md?page=2)