Several years ago, I encountered KnoTiles, which are puzzle pieces that can be used to make mathematical knots. Inspired by KnoTiles, I made puzzle pieces that can be assembled to make mathematical braids. I call the puzzle pieces *BraidTiles.* (Here’s a printable pdf, which I recommend printing in color on heavy cardstock.)

Notice that the twists come in two varieties—left strand over right strand and right strand over left strand. If you replace one twist with the other, you will get a different braid!

]]>Enjoy!

[Update 3/20: I changed a few of the clues.]

]]>However, because it is not yet π Day, I thought I’d introduce you to another number that appears in the book—one that is not as well known but that has a very surprising property. The number is *i ^{i}*; that is, At a glance, this looks like the most imaginary number possible—an imaginary number raised to an imaginary power. But in fact, as Leonhard Euler pointed out to Christian Goldbach in a 1746 letter,

Euler proved that for any angle expressed in radians,

If we take radians (that is, ), then Euler’s formula yields

Therefore,

which is clearly real! Moreover, like π, *i ^{i}* is a transcendental number (and hence an irrational number). This fact was proved in 1929 by the 23-year-old Russian Aleksander Gelfond.

(Actually, as Euler pointed out, *i ^{i}* does not have a single value; rather, it takes on infinitely many real values. The angle

Happy π Day Eve! ( Day? Day?)

]]>So, here is my first creation. It has a mathematical theme. I’ve enjoyed solving crossword puzzles off and on over the years, but I would by no means call myself an aficionado. Thus, I tried my best to write typical crossword-like clues. If anyone would like to offer feedback, post it in the comments below or send me comments by email.

Here’s a printable pdf of the puzzle and a pdf of the solution.

]]>The key to the trick is that there is a hole at the bottom of the post that leads to a pipe that runs up through the post and down through the base of the cup. As the cup fills, so does the pipe. When the liquid gets to the top, it flows down the pipe and gravity does the rest. The pipe acts as a siphon much as the plumbing does in a modern toilet.

Here’s a video of me using the cup and me using a Pythagorean cup that I made.

To make the cup, I pushed a bendy straw through the bottom of a styrofoam cup. It was bent over on the inside so that the opening of the straw was at the base of the cup. (I also put a bead of clue around the hole in the cup so that the liquid wouldn’t leak out around the straw.) It worked beautifully!

If I were a talented potter or had any skills in designing objects for a 3D printer, I’d make a mug with the tube running through the handle. Here’s what I have in mind. If any of you make one, send me a photo—or better, a video—or better still, the real thing!

As for the name “Pythagorean cup”: Do any math/physics/Greek historians have information about the history of this cup? Based on nothing but a gut feeling, I highly doubt this was created by Pythagoras. But I’d definitely be interested in learning the source of this attribution.

Update: On January 1, I received a printable STL file from Pablo Untroib from Argentina. I haven’t had a chance to print it yet. But Pablo did, and it is fantastic. See his video below. He said that when it is sitting on a flat table, it may not leak, but when you pick up the cup, it starts pouring out. A great prank cup!

Further update: On January 4, I received an email from blogger James Stanley who also created a Pythagorean cup—one with the slightly modified design shown below. He wrote about it on his blog. It looks fantastic—I’ll have to print that one too!]

]]>In the logic section of my Discrete Mathematics class (our “intro-to-proofs” class), the students learned about the converse of a conditional statement: the converse of “if A, then B” is “if B, then A.” Most notably, a conditional statement is not logically equivalent to its converse. “If I am over six feet tall, then I am over five feet tall” is a true statement (in my case, the hypothesis is false and the conclusion is true). But the converse, “If I am over five feet tall, then I am over six feet tall” is false.

On my exam, I had a page with the following instructions and the following problem:

*For each of the following problems, determine if such an example exists. If not, state IMPOSSIBLE and give a brief explanation. If so, give an explicit example that satisfies the conditions.*

*A statement that begins “If x≥1, then . . .” and its converse such that the statement is***true**and its converse is**false**.

The answer that I was looking for was something like:

(True) statement: If *x*≥1, then *x*≥0.

(False) converse: If *x*≥0, then *x*≥1.

I class we had discussed that typically when you encounter a conditional like “if *x*≥1, then *x*≥0″ in mathematics, there is an implied universal quantifier. So, mentally, we read it as “for all real numbers *x*, if *x*≥1, then *x*≥0.” My first statement (if *x*≥1, then *x*≥0) is definitely true for all *x*, but the second statement (if *x*≥0, then *x*≥1) is not true for all *x* (for instance, it is not true when *x=*0).

Not everyone answered in this way. Interestingly, I discovered that it was not the case that the more prepared the student, the more likely she or he would get the problem correct (my intended answer). Here’s what I mean:

**Least prepared student:**This student might get it totally wrong, leave the problem blank, not know what the converse of an if-then statement is, and so on.**More prepared:**If a student knew what a converse is, then she or he might get the problem correct without realizing that there were any subtleties.**Yet more prepared:**In class, we talked about the fact that we can only assign truth values to statements. If there is a variable in a sentence, then it is not a statement (it is a predicate). For instance, “the integer*n*is even” is not a statement. It is not true or false unless we know the value of*n*. So, some students wrote on my exam that it was IMPOSSIBLE to solve because we don’t know the value of*x—*it is a predicate, not a statement*.*This was not the answer I was looking for, but I gave them full credit because they were, technically speaking, correct. (In a sense, their answer was more correct than my intended answer.)**Most prepared:**These students would have understood the previous argument, but would also have recalled our discussion of the implied universal quantifier and would have remembered that we had homework problems on this, and thus they would have given the answer I was looking for.

Interestingly, students (2) and (4) may have given the same answer on the paper. But because I know these students and because I saw how they did with the rest of their exam, I honestly suspect (2) and (4) are two different groups. In other words, many of the strongest students in the class got my desired answer and many of the students who struggled elsewhere on the exam got this problem correct. On the other hand, some very strong students fell into category (3).

All-in-all, I wish I’d not asked this question. I ended up giving almost all of the students full credit on the problem, whether they got they got an answer that I had intended to be “the right answer” (i.e., students (2) or (4)) or the more technically correct answer that I did not have in mind (3). But I’m upset that I gave this problem that could be equally interpreted in two different ways.

]]>I spent a while thinking about an example I could show them where this is the case. I found this Math Overflow link on this same question. They gave a few examples. Here’s one of them—the multi-function product rule for derivatives. The base case requires using the definitions of continuity and differentiability and the theorem that every function differentiable at a point is continuous at the point. It is not a simple “1=1” base case. The inductive step is much easier, although it is still not totally trivial because it requires using both the base case and the inductive hypothesis.

**Theorem.** *Let be functions that are differentiable at some point . Then *

**Proof. **This is a proof by induction on Let be the statement that given differentiable functions it follows that

Base case: Let and be differentiable at Recall that because is differentiable at it is continuous at Then

The penultimate equality follows from the continuity of at and the final inequality follows from the differentiability of and at Thus holds.

Inductive step. Suppose holds for some integer Let be functions that are differentiable at some point . Then By the base case, this equals By the inductive assumption, this equals

Thus, holds.

It follows that holds for all #

If you have another example—especially one that would be suitable for into-to-proof students (who don’t have an extensive math background)—post them in the comments.

]]>In 1667, James Gregory did the same, but he used areas: He discovered the following beautiful double-recurrence relation that can be used to compute the areas of inscribed and circumscribed *n*-gons:

**Gregory’s Theorem. ***Let I _{k} and C_{k} denote the areas of regular k-gons inscribed in and circumscribed around a given circle. Then for all n, I_{2}_{n} is the geometric mean of I_{n} and C_{n}, and C_{2n} is the harmonic mean of I_{2n} and C_{n}; that is,*

and

We can use these formulas to approximate π. For instance, a square inscribed in a unit circle has area *I*4=2 and a square circumscribed about the unit circle has area *C*_{4}=4. Applying the recurrence relations, we obtain the following sequence of bounds:

n |
In |
Cn |

4 | 2 | 4 |

8 | 2.828427125 | 3.313708499 |

16 | 3.061467459 | 3.182597878 |

32 | 3.121445152 | 3.151724907 |

64 | 3.136548491 | 3.144118385 |

128 | 3.140331157 | 3.14222363 |

256 | 3.141277251 | 3.141750369 |

512 | 3.141513801 | 3.141632081 |

1024 | 3.14157294 | 3.14160251 |

2048 | 3.141587725 | 3.141595118 |

This summer I tweeted this theorem:

Let I_n and C_n be the areas of the inscribed and circumscribed regular n-gons for the unit circle. In 1667James Gr… twitter.com/i/web/status/1…

—

Dave Richeson (@divbyzero) June 20, 2018

My friend Tom Edgar—a mathematician at Pacific Lutheran University and a master at finding “proofs without words”—emailed me to see if I wanted to try finding a proof without words of Gregory’s theorem. This is what we came up with.

The two parts of Gregory’s theorem follow from the two parts of the following lemma. We give the proof… without words.

**Lemma. ** and

*Proof.*

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Möbius bands have the surprising properties that they are one-sided and have only one edge. This inspired me to write the words MÖBIUS and BAND as ambigrams—in particular, as ambigrams that appear the same after a 180° rotation.

Then I created a Möbius band with the words running along the band.

If you want to make your own, you can download this printable pdf.

However, after I made the Möbius band, it occurred to me that we don’t actually need a Möbius band to achieve this effect with the ambigram. In particular, if you tape the band together as a cylinder, it still looks the same when it is flipped over.

Now, if you were really good, you could create a version of MÖBIUS and BAND that have the 180° rotation and look the same in a mirror. Then you could write the words on a transparent Möbius band. . . .

]]>It got a lot of interest, so I thought I’d try making my own. Here’s the final product.

You can download a printable pdf if you want to make your own.

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