### Points and lines

4th November, 2014

Last week, we Leeds mathematicians were treated to a talk by ~~Professor Green~~ Professor Green with the delightfully elementary title of *Points and Lines*. It reported on joint work of Ben Green and Terence Tao. Here I’m going to talk about the background to their work. As always, any errors are mine.

The idea is that we will be presented with some *points*, specifically a finite collection of points in the plane. We want to predict the answer to questions of the form: how many *lines* are there through exactly \(n\) many of these points? We want to do this while knowing as little as possible about the details of what we’ll be given. Actually, on a first pass, these questions are pretty trivial:

- There are guaranteed to be infinitely many lines through zero points since when you’re positioning a line, a finite set of points is very easy to miss.
- Similarly, there will be infinitely many lines which hit exactly one point, since once you’ve decided which to hit, you can easily tweak that line to miss the rest.
- Must there be any lines which hit exactly two points? Not necessarily. If all the points lie on a single straight line (and if there are more than two of them), then any line either goes through 0, 1, or all of them.
- Arranging the points around a circle illustrates that there also need not be any lines which hit exactly 3 points. The same argument works for n= 4,5,6…

So far, so straightforward: we can predict the answer exactly for \(n=0,1\) and not at all otherwise. But let’s think again: in the case of \(n=3\), there’s obviously nothing special about a circle. There are numerous other curves which would do just as well: a parabola, hyperbola, squircle, catenary, … all of these and many others share the property than any straight line hits them at most twice.

However, for the case \(n=2\), there is something undeniably special about demanding that all our points lie along a single straight line. So the question arises: is this configuration the only obstacle? Must it be the case that *either* all the points lie along a line, *or* we will be able to find a line hitting exactly two of them? This problem was first posed, so far as we know, by J.J. Sylvester in the back pages of the Educational Times in 1893, under the unassuming title of *mathematical question 11851*. (The Royal Society’s *Sylvester Medal* is depicted above, of which Ben Green is the most recent winner, for another famous result of his jointly with Tao.)

The question was later re-asked by Paul Erdős and then solved in 1940 by Tibor Gallai, who established that the answer was *yes*: if a set of points do not all lie on a line, then there must be some line through exactly two of them.

A later proof by Leroy Kelly is, as Ben put it, a `classic of mathematics’ and actually induced an audible gasp from the audience. So let’s see it:

We assume that our points don’t all lie along a single line. That means we can definitely find two of them which lie on a line **L**, along with some other point **X** lying off **L**, as shown. In fact, there are likely to be numerous such configurations; we choose the one where the perpendicular distance from **X** to **L** is the least possible.

The claim is then that **L** contains *only* those two points. Suppose not. Then there must be two points **A & B** both lying on **L** and on the same side of the perpendicular line **XL**. Now, let **M = XA**. This line also contains two points (obviously), but the distance from **B** to **M** is less than that from **X** to **L**, which is a contradiction. **QED**

Clever, right? Anyway, for reasons unexplained, lines which pass through exactly two of our specified collection of points are known as *ordinary* lines. The Sylvester-Gallai theorem above therefore says that (so long as our points are not colinear) an ordinary line will always exist. But the next question is: how many must there be? In most cases there will be lots: if you try this for a random scattering of \(m\) points, then in all likelihood, whenever you draw a line through two points it will miss all the others. Thus there are around \(\frac{1}{2}m^2\) ordinary lines.

But it is not hard to come up with sets with fewer. For instance, take \(m-1\) points all lying along a line, and add in one extra point \(X\) off it. Then the only ordinary lines are those which connect \(X\) to another point, giving \(m-1\) many.

A cunning construction by Károly Böröczky brings this down even further: a set of \(m\) points with only \(\frac{1}{2}m\) many ordinary lines.[1]

Now we come to the theorem of Green and Tao. In their paper, they have shown that Böröczky’s examples are as good as it gets. For large enough values of \(m\), they prove that every non-colinear set of \(m\) points must have at least \(\frac{1}{2}m\) many ordinary lines.

How large is `large enough’? The bound they establish is in the vicinity \(10^{10^{10}}\), but the true answer may be as low as 14. For smaller sets, it is possible to do better. The left-hand picture here shows a set of 7 points with only 3 ordinary lines.

And how is it proved? Well, I’m not going to go into details, but the key step is the unexpected (to me) connection of the property `has few ordinary lines’ with the property `can be covered with a single cubic curve’.

[1] For the cognoscenti: it achieves this by moving the action to the real projective plane, arranging \(m\) points symmetrically around the circle, and placing another \(m\) on the line at infinity at positions corresponding to the directions determined pairs of points from the first set. Then the only ordinary lines are the \(m\) many tangents to the circle at points. Finally, since no-one mentioned anything about lines at infinity being permitted, the whole set-up needs to be transported back to the plain old Euclidean plane, which can be achieved via a projective transformation. This however will distort the circle into an ellipse. See page 11 of their paper for a picture. )

Categories: Geometry, Maths | Comments (0) | Permalink

### Lecture: Gödel, Incompleteness & Unprovable Theorems

23rd July, 2014

If you’ve got a spare hour[1] at some stage here is a video of a talk I gave last month on Gödel, Incompleteness & Unprovable Theorems.

[1] The joy of the pause button is that the hour doesn’t have to form a single contiguous block.

Categories: Elwes Elsewhere, Logic, Maths | Comments (0) | Permalink

### Faro Shuffles

27th February, 2014

If you shuffle a deck of cards perfectly eight times, you get back exactly to where you started. Here’s a video of the magician Adam West doing it:

In this post, we’ll see why this works. First let’s tighten up our terminology: a “perfect shuffle” here is really what magicians call a *Faro Out-Shuffle:* you split the deck in half, and then interleave the two halves. To make things easier, let’s work with a deck of ten cards, initially in numerical order.

Step one: split 1,2,3,4,5 from 6,7,8,9,10.

Step two, interleave the two: 1,6,2,7,3,8,4,9,5,10.

Notice that the outermost cards (1 and 10) remain in place. (That’s why this is an *out-shuffle* — interleaving the other way produces an *in-shuffle*.)

At the same time, notice that the card initially in 4th position is now in 7th, and vice versa. We can indicate this in *cycle notation* by putting the two together in brackets: (4 7).

What about the rest? The card in second position to start with is now in third place, while the card from third position is now in fifth, the card from fifth is now in ninth,… Continuing this line of thought produces the cycle (2 3 5 9 8 6).

So the shuffle is totally described by the collection of cycles (1) (10) (4 7) (2 3 5 9 8 6). (I’ve included the two trivial cycles of length one.)

Now, notice that performing any cycle of length two twice brings us back to where we started: in this case 4 goes to 7 and then back to 4. Similarly, performing a cycle of length 6 six times takes use back to the starting position. In fact then, performing this ten card shuffle six times brings everything back to where it started.

Ok. What about a full 52 card deck? Well, for cards initially in the top half of the deck (i.e. numbered 1-26), the card in \(n\)th position is moved to \((2n -1)\)th position after one shuffle, (meaning \(1 \to 1, \ 2 \to 3,\ 3 \to 5, \ldots, 26 \to 51\)).

For cards in the second half (numbered 27-52), the card in \(n\)th position is moved to \((2n-52)\)th position (meaning \(27 \to 2, \ 28 \to 4, \ldots, 55 \to 54, \ 56 \to 56 \)).

Repeatedly applying the first rule, we can see that 2 goes to 3 which goes to 5 which goes to 9 which goes to 17 which goes to 33. Now applying the second rule, 33 goes to 14 which goes to 27 (by the first rule) which goes back to 2 (by the second rule). Thus we have the cycle:

(2 3 5 9 17 33 14 27).

The same reasoning tells us the other cycles which make up the shuffle, and altogether they turn out to be

(2 3 5 9 17 33 14 27) (4 7 13 25 49 46 40 28) (6 11 21 41 30 8 15 29)

(10 19 37 22 43 34 16 31) (12 23 45 38 24 47 42 32) (20 39 26 51 50 48 44 36)

(18 35) (1) (52)

which is to say six cycles of length eight, one of length two, and the two trivial cycles of length 1. Each of these cycles will take us back to the starting position when applied eight times… as the video above shows!

**Question:** what if we do a *Faro in-shuffle* instead? This means interleaving the other way. On ten cards, after one shuffle the deck reads 6,1,7,2,8,3,9,4,10,5. How many applications of this rule are needed to bring us back to the starting position in this case?

Categories: Maths, puzzles | Comments (2) | Permalink

### Total Numbers of Permutations

29th January, 2014

I’m teaching a third year Combinatorics module this term, which has led me to revisit some old friends: permutations and combinations. I thought I knew them well, but have already learned something new!

First a quick refresher course:

**Combinations:** say \(C(n,k)\) is the the number of subsets of size \(k\) from a collection of size \(n\). So if you select 3 pieces of fruit from a total of 6 (all different), and don’t care about the order, then the number of selections you might make is \(C(6,3)\). The formula^{[1]} here is \(C(n,k) = \frac{n!}{k!(n-k)!}\) and \(C(6,3)=20\).

**Permutations:** If you do care about the order of selection (maybe you want to arrange your fruit into military columns so that “strawberry, apple, pear” is different from “apple, pear, strawberry”), then the number you need is \(P(n,k)\), the number of permutations of \(k\) objects from \(n\). Its formula is \(P(n,k) = \frac{n!}{(n-k)!}\) and \(P(6,3)=120\).

**Total numbers of combinations:** A set of size \(n\) has \(2^n\) subsets in total. To say the same thing another way, \(C(n,0)+C(n,1)+…+C(n,n)= 2^n\). So the total number of collections of fruit you could take from the box of 6 is \(2^6=64\) (this includes both the empty-plate option and the take-everything option).

All of the above I have known for many years. But I recently discovered something new (to me):

**Total numbers of permutations:** if you pick between \(0\) and \(n\) objects, in order, from a set of size \(n\), how many choices can you make? To ask the same thing another way, what is \(P(n,0)+P(n,1)+…+P(n,n)\)? And the answer turns out to be \( \lfloor n! \cdot e \rfloor \), that is the number you get if you round \(n! \cdot e\) down to a whole number. So the number of military columns of fruit (including the empty column) you could make from your six are \( \lfloor 6! \cdot e \rfloor = 1957\).

I think this is very cool, and the proof of this is not hard. It’s been written out very nicely here by Michael Lugo. (When I first saw this result I assumed that it would be a consequence of Stirling’s approximation – but that’s not required.)

[1] These formulae all use factorial \(! \) notation where \(7!=7\times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)

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### Infinite Series – A Health and Safety Warning

18th January, 2014

There’s been a bit of a rumpus recently about this video from *Numberphile* which purports to show that \[1+2+3+\ldots = -\frac{1}{12}\]

It got posted on Slate with an accompanying blogpost by Phil Plait, which then got shot to pieces by various incensed bloggers, and Phil now has a much better follow-up post.

I’m not going to rehearse the story again – the short version is that the equation above is a really bad and misleading way of communicating a really interesting and surprising result. But I thought I’d share another titbit which has a similar moral, namely:

Infinite series often behave in very weird ways.

This in turn has two corollaries, both amply illustrated in the story above:

Unexpected and cool things can happen.

but at the same time

It’s very easy to start talking nonsense if you’re notextremelycareful.

The video is about the series \(1+2+3+4+5+\ldots \) This is what we call a *divergent* series, meaning that as you add up more and more terms, the series grows without limit. An example of a *convergent* series is \(0.9+0.09+0.009+0.0009+\ldots\) Here, as you add up more and more terms, you get ever closer to the fixed value 1. This is the *limit* of the series. Convergent series are generally friendlier beasts than divergent series, but the purpose of this post is to illustrate that even convergent series can do shocking things.

Humans are not used to infinite series. But we are used to adding up finite collections of numbers. If we have three numbers \(A,B,C\) we know that the order in which we add them up does not matter: \(A+B+C=B+A+C=A+C+B\) etc.. This sort of thinking is so natural and automatic that it is all too easy to transfer it to the realm of infinite series, where they may not hold – not even (necessarily) for convergent series.

This post is about the following series: \[S: \ \ 1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \frac{1}{5} – \frac{1}{6} + \frac{1}{7} – \frac{1}{8} + \frac{1}{9} \ldots\]

The first obvious question is: does \(S\) converge? And the answer is yes. It turns out to converge to a number near 0.69. (More precisely, but unexpectedly and coolly, the limit is^{[1]} the natural logarithm of 2 or \( \ln 2 \). But you don’t have to know what that means to follow the rest of the post.)

Now, let’s reorder the elements of \(S\). Notice that the numbers on the bottom alternate between being odd and even. Let’s mix things up so that they go ‘odd even even odd even even’ instead, but being careful to keep the sign of every element the same at it was before:

\[ 1 – \frac{1}{2} – \frac{1}{4} + \frac{1}{3} – \frac{1}{6} – \frac{1}{8} + \frac{1}{5} – \frac{1}{10} – \frac{1}{12} + \ldots \]

To reiterate, we have not added or removed any elements here, just reordered the elements of \(S\). But if you look carefully at this new series, you’ll see that the positive terms are all followed by the negative of their halves – I’ll group these together in brackets to make it clearer:

\[ \left(1 – \frac{1}{2} \right)- \frac{1}{4} + \left( \frac{1}{3} – \frac{1}{6} \right) – \frac{1}{8} + \left( \frac{1}{5} – \frac{1}{10} \right) – \frac{1}{12} + \ldots \]

Now let’s compute the bits in the brackets and give a name to the new series:

\[ T: \ \ \frac{1}{2} – \frac{1}{4} + \frac{1}{6} – \frac{1}{8} + \frac{1}{10} – \frac{1}{12} + \ldots \]

But what’s this? If you compare a term of \(T\) to the term in the corresponding position in the original series \(S\), you’ll see that it’s exactly half. And indeed, the limit of \(T\) is exactly half the limit of \(S\) – around 0.34 (more preciely, \(\frac{1}{2} \ln 2 \)).

So just by rearranging the terms of \(S\) we have completely changed its limit! That might seem surprising, but Riemann’s rearrangement theorem tells us that this is the tip of the iceberg. It is actually possible to rearrange the terms of \(S\) so that it converges to any number you care to name – or so that the series diverges!

So, be very careful when handling infinite series! Here endeth the lessen. But to continue with the mathematics a little: this happens because \(S\) is what is termed *conditionally convergent*, that means that if you replace all its negative terms with their positives:

\[H: \ \ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} \ldots\]

you get a series which doesn’t converge. \(H\) is called the harmonic series; that it diverges was one of the first truly surprising facts that mathematicians discovered about infinite series. *Absolutely convergent series* (ones where the corresponding entirely positive series converges, as 0.9 + 0.09 + 0.009 +… does) are much better behaved, and Riemann’s theorem doesn’t apply there.

**Finally, a shameless plug:** All this material is covered in my book Maths 1001.

[1] This is the case \(x=1\) of the series expansion for \( \ln (1+x)\) which is given by the *Mercator series:* \(1 – x + \frac{x^2}{2} – \frac{x^3}{3} + \ldots \)

Categories: Maths | Comments (3) | Permalink

### Flipping Classrooms and Hegartymaths

17th October, 2013

As the years have rolled by, the blackboard may have been replaced by a whiteboard, and then by a smartboard, but otherwise the formula has remained by and large the same: theory in the classroom followed by exercises for homework.

The idea of “flipping” the classroom, is to reverse this process. The insight is that in the age of youtube, today’s students can perfectly well take in the lectury bit at home — with the added advantage of being able to pause, rewind, and rewatch in their own time. This leaves lesson-time free for practice, providing the teacher with more time to go around talking to students individually or in groups, meaning more opportunities to help those who are struggling or to supply extra challenges to those who need stretching.

I have no experience of this method myself — all the same it immediately appeals to me as a way in which technology may really be able to improve the teaching and learning experience, rather than just adding bells and whistles. One person who is convinced by this new approach is Colin Hegarty, an old friend of mine from university, who I’m delighted to say has returned to the world of maths after a few years in finance, and is now expertly flipping classrooms in North London.

Even if you don’t have the good fortune to be one of Mr Hegarty’s students, you can still peruse the 611 (!) videos that he and his colleague Brian Arnold have created for this purpose, all of which are freely available on Hegartymaths, or on youtube. Judging by the 800,000 odd views their videos have gathered, it’s not only their own pupils who are benefiting from these experiments in flipping.

As a sample I’m embedding one in which Colin talks through the Chinese postman problem:

Categories: Education, Technology | Comments (3) | Permalink

### Interviewed by Kevin Houston

7th October, 2013

Categories: Elwes Elsewhere | Comments (0) | Permalink

### Constructible Numbers

13th September, 2013

**This blog-post is an extract from my book Maths in 100 Key Breakthroughs**

### Constructible Numbers

**A sure route to mathematical fame is to resolve a problem that has stood open for centuries, defying the greatest minds of previous generations. In 1837, Pierre Wantzel’s seminal analysis of constructible numbers was enough to settle not just one, but an entire slew of the most famous problems in the subject, namely those relating to ruler-and-compass constructions.**

As with so much in the history of mathematics, the topic had its origins in the empire of ancient Greece. The geometers of that period were interested not only in contemplating shapes in the abstract, but also in creating them physically. Initially, this was for artistic and architectural purposes, but later for the sheer challenge it posed. In time, mathematicians came to understand that the obstacles theyencountered in these ruler-and-compass constructions brought with them a great deal of mathematical insight. Nowhere was this more true than in the ancient enigma of squaring the circle, and what that revealed about the number \(\pi\).

**Classical problems**

Greek geometers decided on a set of simple rules for building shapes, using only the simplest possible tools: a ruler and pair of compasses. The ruler is unmarked, so it can only be used for drawing straight lines, not for measuring length (therefore these are sometimes called straight-edge-and-compass constructions). The compass is used to draw circles, but it may only be set to a length that has already been constructed.

Today’s schoolchildren still learn how to use these devices to divide a segment of straight line into two equal halves and to bisect a given angle. These were two of the very first ruler-and-compass constructions. A more sophisticated technique allows a line to be trisected, that is, divided into three equal parts. What of trisecting an angle, though? Various approximate methods were discovered, which were accurate enough for most practical purposes, but no one could find a method which worked exactly. This proved a mystery, and gave the first hint that there was real depth beneath this question. But what does it mean if one task can be carried out by ruler and compass and another cannot?

The most famous of the ruler-and-compass problems, and indeed one of the most celebrated questions in mathematics, is that of squaring the circle. The question is this: given a circle, is it possible to create, by ruler and compass, a square which has exactly the same area? At the heart of this question lies the number \(\pi\) (see page 54). The problem ultimately reduces to this: given a line 1 unit long, is it possible to construct by ruler and compass another line exactly \(\pi\) units long?

Another classical problem was that of doubling the cube. This problem had its origins in a legend from around 430 BC. To overcome a terrible plague, the citizens of the island of Delos sought help from the Oracle of Apollo. They were instructed to build a new altar exactly twice the size as the original. At first they thought it should be easy: it could be done by doubling the length of each side. But that process leads to the volume of the altar increasing by a factor of 8 (since that is the number of smaller cubes that can fit inside the new one). To produce a cube whose volume is double that of the original, the sides need to be increased by a factor of \( \sqrt[3]{2}\) (that is the cube root of 2, just as 2 is itself the cube root of 8). The question of doubling the cube therefore reduces to this: given a line segment 1 unit long, is it possible to construct another exactly \( \sqrt[3]{2}\) units long?

**Wantzel’s deconstruction**

Working in the turbulent setting of France in the early 19th century, Pierre Wantzel turned these ancient questions over in his mind. He recognized that the form of many ruler-and-compass questions is the same. The key to them was this: given a line 1 unit long, which other lengths can be constructed? And which cannot? If a line of length \(x\) can be constructed, then Wantzel deemed \(x\) a *constructible* number. Setting aside the geometrical origins of these problems, he devoted himself to studying the algebra of constructible numbers. Some things were obvious: for example, if \(a\) and \(b\) are constructible, then so must be \(a + b\), \(a – b\), \(a \times b\), and \(a \div b\). But these operations do not exhaust the range of constructible numbers; Wantzel realized that it is also possible to construct square roots, such as \(\sqrt{a}\).

His great triumph came in 1837, when he showed that everything constructible by ruler and compass must boil down to some combination of addition, subtraction, multiplication, division and square roots. Since \(\sqrt[3]{2}\) is a cube root, and cannot be obtained via these algebraic operations, it followed immediately that the Delians’ ambition to double the cube was unattainable. A similar line of thought revealed the impossibility of trisecting an angle.

As for the greatest problem of all, squaring the circle, the final piece didn’t fall into place until 1882, when Ferdinand von Lindemann proved that \(\pi\) is a transcendental number (see page 197). Then Wantzel’s work immediately implied the non-constructibility of \(\pi\), and the impossibility of squaring the circle was finally established.

Categories: Bookery, Geometry, Number theory | Comments (0) | Permalink

### Maths in 100 Breakthroughs

12th September, 2013

*Maths in 100 Key Breakthroughs*is published by Quercus and is now available as a softback or e-book. You can buy it from the publisher here, or in the usual other places.

As the title suggests, its hundred chapters, ordered chronologically, each deal with a major mathematical development (e.g. Aristotle’s analysis of logical syllogisms circa 350BC, the discovery of transcendental numbers in 1844, and the creation of Weaire-Phelan foam in 1993). My hope is that it should be accessible, attractive, and entertaining to people with little or no background in the subject – jargon and technical notation are kept to a minimum, and each chapter is accompanied by a beautiful full-page colour illustration.

My major concern was to avoid wrenching these breakthroughs out of context and artificially presenting them as stand-alone events. After all, mathematicians typically make advances by contemplating the insights of previous generations and answering questions posed by earlier thinkers. Without Kelvin’s conjecture (and perhaps without the work of Pappus and Thomas Hales on related geometrical questions) the discovery of Weaire-Phelan foam would have been less exciting. Equally, it often takes time and further insight for the significance of a breakthrough to become apparent: it was some years after their initial discovery that the deep importance of transcendental numbers was recognised.

So I hope that the book not only presents some wonderful discoveries, but also tells the back-stories, gives some sense of what the characters involved thought they were up to, and discuss why their work matters to us today.

Categories: Bookery | Comments (8) | Permalink

### The revenge of the Perko pair

14th August, 2013

**Perko pair**is a wonderful mathematical story, and one I have told on numerous occasions, so let me do so again:

### The Perko Pair I: the story so far

In the late 19th century, mathematicians and physicists started producing tables of knots. The idea here is that some knots are genuinely different from each other, while others can be deformed to match each other without cutting or gluing, making them essentially the same thing from a topologist’s point of view. The trouble is whether or not two knots are really the same is extremely hard to tell on first sight…. as we shall see.

The tables begin with the unknot, the only knot with no crossings. Then comes the trefoil or overhand knot, which (so long as we don’t distinguish between it and its mirror image) is the only knot with three crossings. There’s similarly one knot with four crossings, then two with five, and so on. In the late 19th century Peter Guthrie Tait and Charles Little got as far as listing the knots with 10 crossings.

However the early knot tabulators (unsurprisingly) made a few errors, and it was not until 1976 that Dale Rolfsen put together a comprehensive list of the knots with up to 10 crossings, based on earlier efforts by John Conway, in turn building on the work of James Alexander in the 1920s. Of the ten-crossing knots they counted 166 separate varieties. (Today’s mathematicians have made it as far as 16 crossings.)

The surprise was that Rolfsen and Conway had also made a mistake! Even armed with 20th century techniques of algebraic topology, a duplication had slipped through. In 1973^{[1]} Kenneth Perko had been studying a 19th century table of Little, and had realised that two ten-crossing knots (subsequently labelled by Rolfsen as 10_{161} and 10_{162}) were actually the same thing in disguise.

This story has a very clear moral: telling knots apart is difficult. *Really* difficult. Even after 75 years of brainpower, mathematicians were still coming badly unstuck, and who knows how long it might have taken without Perko’s alertness. And of course, this just for knots with a paltry ten crossings. Imagine trying to decipher knots with thousands…

Now for the sequel:

### The Perko Pair II: Perko strikes again

The above story has been told countless times, and is usually accompanied by a picture of the offending pair of knots, something like this:

This error has infected, most likely among many other places, Wolfram Mathworld and Mathematics 1001 by myself. And guess who pointed this error out…

The explanation of the mistake is that two updated versions of Rolfsen’s 1976 table are now in circulation. In both, 10_{162} has been deleted as it should be. But one version (occurring for instance in the latest editions of his book) keeps his original numbering up to 10_{166} with a space between 10_{161} and 10_{163}. In the other, see here for example, Rolfsen’s 10_{163} has been renumbered as 10_{162}, and 10_{164} as 10_{163}, etc., thus counting up to 10_{165}.

So the knot numbered 10_{162} in the picture above was actually Rolfsen’s original 10_{163}, and thus **not** equivalent to 10_{161}! For the avoidance of doubt, here is the real Perko pair, with drawings provided by the man himself. Accept no imitations!

While we’re at it, let’s also put paid to the to the idea that Ken is an ‘amateur’ mathematician. Before going on to a career in law, he says “at Princeton I studied under the world’s top knot theory topologists (Fox, Milnor, Neuwirth, Stallings, Trotter and Tucker)”. I apologise for suggesting otherwise. At least I didn’t pronounce him dead unlike certain other popular maths authors… Also, he adds “That stuff about rope on the living room floor is pure internet nonsense. I did it with diagrams on a yellow legal pad. Ropes wouldn’t work anyway since the two knots are non-isotopic mirror images of each other.”

[1] Note the corrected date, usually given as 1974.

Categories: Topology | Comments (20) | Permalink

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