CTK Insights

22 Jul

Beautiful and Practical

I found a remarkable talk given by Robert Lang pointed to at the Math Frolic! blog. Dr. Robert J. Lang is an American physicist who is also one of the foremost origami artists and theorists in the world. Among other achievementsm he is known for having proved the completeness of Huzita–Hatori axioms and developing paper folding algorithms. This is what his 16-minute 2008 TED Talk is about:

As often happens in math and art, Robert remarked, the pursuit of beautiful leads to practically useful discoveries. As an example, he helped design a folded of a solar battery that was to provide power to a deep space explaration craft. He concludes his talk by expressing a belief that sooner or later origami will be instrumental in saving human lives.

20 Jul

The Sum of Cubes Formula

The formula for the sum of the consecutive cubes of integers is one of the most elegant in elementary mathematics:

 1^3 + 2^3 + 3^3 + \ldots + n^3 = (1 + 2 + 3 + \ldots + n)^2.

Taking into account a better known formula for the sum of plain integers

 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2}

the formula for cubes can be rewritten succinctly as

 1^3 + 2^3 + 3^3 + \ldots + n^3 = [\frac{n(n+1)}{2}]^2.

The sum of the cubes is a square of a triangular number.

There are general approaches that lead to summation formulas for other powers of integers, or, as they are called, power series. Here I’ll use a straightforward induction argument.

For n = 1, the identity is trivial: 1 = [1(1 + 1)/2]^2. Assume it holds for n = k and let n = k +1.

 1^3 + 2^3 + 3^3 + \ldots + k^3 + (k+1)^3 = [\frac{k(k+1)}{2}]^2 + (k+1)^3.

Thus, this proof by induction is reduced to establishing the following identity

 [\frac{k(k+1)}{2}]^2 + (k+1)^3 = [\frac{(k+1)(k+2)}{2}]^2.

Indeed, with a little effort, it could be verified that both sides are equal to

 \frac{n^4 + 6n^3 + 13n^2 + 12n + 4}{4}.

Now, as an application of the summation formula, let’s solve the following problem (C. W. Trigg, Mathematical Quickies, Dover, 1985, #51): Solve the equation in positive integers

 \frac{1^{3} + 3^{3} + 5^{3} + \ldots + (2n-1)^{3}}{2^{3} + 4^{3} + 6^{3} + \ldots + (2n)^{3}} = \frac{199}{242}.

Solution →

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Adding 1 to both sides of the equation

 \frac{1^{3} + 3^{3} + 5^{3} + \ldots + (2n-1)^{3}}{2^{3} + 4^{3} + 6^{3} + \ldots + (2n)^{3}} = \frac{199}{242}

gives

 \frac{1^{3} + 2^{3} + 5^{3} + \ldots + (2n)^{3}}{8(1^{3} + 2^{3} + 3^{3} + \ldots + n^{3})} = \frac{441}{242},

or, taking into account the summation formula for the cubes,

 \frac{[2n(2n+1)/2]^2}{8\cdot [n(n+1)/2]^2} = \frac{21^2}{2\cdot 11^2}.

Taking the square root

 \frac{2n+1}{n+1} = \frac{21}{11},

with a unique and obvious solution n = 10.

19 Jul

Radical Simplification

In a related post I have shown that \sqrt[3]{2 \pm \sqrt{5}} = \frac{1 \pm \sqrt{5}}{2}. Without resorting to this proof, I am going to show that \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} = 1.

Let a = \sqrt[3]{2 + \sqrt{5}}, b = \sqrt[3]{2 - \sqrt{5}} and  x = a + b. Then

x^3 = (a + b)^3 = a^3 + b^3 + 3ab(a + b) = 4 -3x,

which gives us a third degree equation in x: x^3 +3x - 4 = 0. By direct verification x = 1 is one of the roots of that equation. Factoring gives x^3 +3x - 4 = (x - 1)(x^2 + x + 4). The second factor which is a quadratic polynomial is always positive because x^2 +x + 4 = (x + \frac{1}{2})^2 + 3\frac{3}{4} and, therefore has no real roots. However, x is clearly real and is then the only real root of the cubic equation x^3 +3x - 4 = 0, which is 1. Of necessity x = 1 and we are done.

Reference

  1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #35
16 Jul

Counting on one hand and on two

A human hand carries five fingers; two hands have ten of them. Undoubtedly, this fact is responsible for the universal adoption of the decimal system.

Children learn to count by counting fingers, first to 5 on one hand and then to 10 on two hands. However, there is a simple way to count to 10 with just 5 fingers of one hand and to 100 using both hands.

This is how:

one two three four five
six seven eight nine

A folded fist may stand for 10 if you do not plan to use the second hand, or for 0 if you do:

 
ten/zero

The second hand is used in the same manner but for counting 10s:

 
  ten eleven
 
  twenty six fifty five

(There is an interesting discussion on the web concerning finger counting.)

16 Jul

Fractional representatives – logistic nightmare


George G. Szpiro, author of Numbers Rule: The Vexing Mathematics of Democracy, from Plato to the Present is a mathematician and journalist living in Switzerland.

Numbers Rule focuses on key figures in the development of democracy and on the mathematics of voting, elections, and apportionment that they developed. Szpiro pays particular attention to the paradoxes that arise, and discusses them through examples,

wrote Steven J. Brams, New York University – himself an expert on democracy related mathematics.

The quality of Szpiro’s writing can be tasted at his recent op-ed at the History News Network.

This particular article deals with the history of the problem of apportionment. In the article, the author highlights what became known as the Alabama and Population paradoxes. Szpiro lucidly explains how these and other paradoxes arise due to the rounding errors, inevitable evil of the process of dividing a limited number of seats among the representatives of such a big country as the USA. He makes it clear that mathematical tool set is not powerful enough to resolve the problems of democratic representation.

Towards the end of the article, Szpiro suggests – I believe rather light heartedly – to resolve the problem of paradoxes and the constitutionally mandated fairness of the representation by avoiding rounding (in all possible ways) and by simply assigning to the districts a fractional number of representatives depending on the size of the population. This is rather obviously an impractical solution. The House will have to grow (by 25 representatives in Szpiro’s estimate). Each new representative will be entitled on a full size staff. But budgetary considerations aside, presently the voting goes sometimes along the party lines and sometimes by creating coalitions. Some representatives have more power than others – due to their caring of important committees – and thus are better positioned to promote their (or their home states’) interests. So it is not exactly the case of 1 representative, 1 vote, anyway. Fractional representation will, in all likelihood, only deepen the divide. And of course it will create additional conflicts and power struggles. A state with 20.75 representatives will send 21 persons to serve in the House. Will there be anyone to carry the .75 power or will it be applied on a rotation basis?

There are just a couple of concerns that fractional representation will lead too. This is probably where mathematics should draw a line. All the power of mathematics notwithstanding, the representation will never be 100% fair. Politics may be a dirty job; the history teachers us that some dirt is necessary to make this job functional.

Szpiro’s article has been hightlighted at the Princeton University Press blog and elsewhere on the web.

16 Jul

Math teachers at play carnival #28, P.S.

Due to a freak accident, the 28th issue of the Math teachers at play carnival went out two days before a submission deadline. With sincerest apologies I’d like to mention the submissions that arrived after the accident.

John Golden, a.k.a. Math Hombre, celebrates a full circle day (6/28) with an article on Similarity and π.

John Cook, in his The Endeavor blog, writes about the Lincoln Index which is commonly applied to population estimates. As John shows, the index reliably estimates the number of errors that cropped up in a computer program – the place the bugs live in and thrive.

15 Jul

Areas in a Square

The NCTM holds to his promise to post interesting Problems to Ponder in the NCTM’s newsletter. Here’s one from the July 15 issue.

In the figure below, quadrilateral ABCD is a square, and E is the midpoint of the side AD. How do the areas of regions I, II, III, and IV compare? Another way to think about this is to consider the question, What are the ratios of the areas of the four subdivisions, I : II : III : IV?

square split into four parts

This problem also has many interesting extensions for your students. For example:

  • What happens if E, instead of being the midpoint of AD, is located at the 1/3 mark on segment AD?

  • Suppose that the original figure ABCD is actually a rectangle that is not a square. Does this change affect the ratios of the respective areas?

    And you can probably pose some additional extensions of your own.

The problem has relevance to a construction problem that once captured the NCTM headlines.

Solution →

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

square split into four parts

Let O be the point of intersection of BD and CE. Then OB = 2·DO so that the altitude in ΔDEO from O is 1/3 the side AB. Since its base DE is half the side its area is 1/3·1/2·1/2 = 1/12 of the are of the square.

In ΔCDO, the altitude from O is 1/3 of BC hence its area is 1·1/3·1/2 = 1/6 the area of the square. Similarly, the area of ΔOBC is 1·2/3·/1 = 1/3 the area of the square.

The portion occupied by the rectangle ABOE equals 1 – 1/12 – 1/6 – 1/3 = 5/12. It follows that the areas are in the following ratios

1/12 : 1/6 : 1/3 : 5/12 = 1 : 2 : 4 : 5.
15 Jul

Does It Matter Which Winner You Saw?

In an early June issue of the NCTM newletter, President Mike Shaughnessy offered a Problem to Ponder:

Scenario: Students at your school have just finished competing in the qualifying round of a nationally sponsored contest on mathematical reasoning and sense making. When the work was scored, it turned out that four students at your school all had perfect preliminary papers—two girls and two boys. The school decided to hold a random drawing among these four students to select two of them to send to the national finals. The drawing takes place in the school auditorium. You show up late to the drawing, just as one of the winners—a girl—is leaving the stage amid cheers.

1. Suppose that the girl that you saw leaving the stage is the first winner. What is the probability that the second winner will also be a girl?
2. Suppose that the girl that you saw leaving the stage was the second winner. What is the probability that the first winner was also a girl?

The mid-July newsletter offered a solution, actually two of them. The two solutions for the second problem intentionally produce different answers, providing thus the food for thought.

Solution 1

If the girl that you saw on stage was actually the second winner, then the chance that the first winner was also a girl is 1/2, because before the first girl was picked, there were two girls and two boys, so the chance of a choosing a girl was then 2 in 4, or 1/2.

Solution 2

It doesn’t make any difference whether you saw the first girl or the second girl; the fact that you saw a girl winning at all means that there is only a 1 in 3 chance that the other winner was a girl, so, the probability that the first winner was also a girl if the girl that you see on the stage is the second winner is also 1/3.

Truth be told, when posting a solution at my site I have overlooked the first argument that led to the probability of 1/2. The reason I think was because I saw my task was to solve the problem, while Mike Shaughnessy – mindful to get math teachers excited – had the task of posing the problem. Coming up with two solutions that led to different answers should have certainly affected the level of interest of his readers.

As a result, I, too, has amended that page with two Bayesian approaches. The first solution of course should have raised a red flag for an attentive reader. Indeed, the argument never exploits the fact that the second winner was a girl, although it would have been strange if that information were just a red herring. Indeed, it is easily seen that it is not. Consider a situation where there are just 1 boy and 1 girl. They are selected in turns. You peek in, see a girl, and are told that this is the second selection. What is then the probability that the first selection was a girl? Zero, of course. This so, even if the a priori probability of a girl being selected on the first draw is obviously 1/2.

Here is a short argument leading to the probability of 1/3. There are 6 combinations of 2 items out of four, making the probability that both winners were the girls 1/6. If G and g are the events that the first/second selection was a girl then what we are looking for is the conditional probability P(G|g). But P(G|g) = P(Gg)/P(g). The a priori probabily that the second selection is a girl is the same as for a boy: P(g) = P(b) = 1/2. It then follows that P(G|g) = 1/6 ÷ 1/2 = 1/3.

14 Jul

Estimating Pi

The fact that π exists is due to the similarity of all circles. The ratio of the circumference to the diameter would not be constant otherwise.

The simple idea of similarity is commonly being reported as overly hard on children. I am confident that the problem must lie with the manner of presentation and not with the similarity per se. This is because children grow up in the world of models: toy cars, stuffed animals, house drawings, etc. To detect pattern – an activity in which the human brain has no peer – is to see similarities. This is to say that the notion of similarity should not be too difficult for the kids, i.e., if it is introduced properly. The difficulty usually lies in the algebraic expression of similarity not in the geometric concept.

I was very pleasantly surprised by a submission to the latest Math teachers at play carnival that featured a gentle introduction into the concept of similarity. Once explained and internalized, the concept was applied to the determination of π.

Here I would like to add my 2¢. It is a common practice to measure and tabulate the diameter and the circumference of several round objects and take the ratio. “Lo and behold” goes the argument, “the results are so close that not only may we estimate π but also convince ourselves that this fellow really exists.”

Such experimentation goes a long way as a verification of an abstract concept of the similarity of all circles. However, we can easily do much better in estimating the value of π.

Let D and C be a measurement of the radius and of the circumference of a circular object with D0 and C0 being the real (but unknown) values. We assume that D is close to D0 and C is close to C0 and hope further that the ratio C/D is close to C0D0.

The accuracy of the measurement, i.e., the estimate for |C0C| depends only on the measuring device. For example, if a ruler is used then the maximum error between the “real” and measured values may be assumed to be one half of the smallest division marked on the ruler. This is true for “big” and “small” lengths. If that max error is denoted Δ then |C0C| < Δ, for any circumference C.

How do we measure circumference? Simple: take a thread and wrap it once around the object, unwrap it, keep it stretched and measure the length of the thread. Obviously, if we wrap the thread several times over then the measurements will have to be divided by the number of turns the thread was wrapped around the object. If n is the number of turn then the the measured length is close to nC. How close? The estimate is still within the same precision marks as it was for a single turn, i.e., Δ. For an estimate C0 we then have |nC0nC| < Δ. As a consequence |C0C| < Δ/n.

Let’s see how it works. For a round object I took a carton tube, a leftover from a roll of paper towels:

I wrapped a thread around the tube several times and made the marks:

The diameter was measured to be D = 13/8. (All measurements are in inches.) Then I tabulated the results for 1, 2, 3, and 4 wraps of the thread:

  1 2 3 4
nC   43/8 21/2 31/2 41/2
nC/nD   3.3077 3.2308 3.1795 3.1538

(The numbers in the bottom row have been rounded to 4 digits.)

I did not have a long enough ruler to measure 5 or more turns but the trend could be easily observed with just four measurements: the ratios go closer to π as the number of turns of the thread around the object grows.

13 Jul

Math teachers at play carnival

This is a Math teachers at play carnival, issue #

a single image stereogram of number 28

which I am going to reveal shortly. See if you can make it from what is known as a single image stereogram. Try focusing your eyes behind the screen.

What is the number of this issue?

There are three books in my library that are exclusively devoted to listing all kinds of information about all kinds of numbers – as is well known no number is uninteresting. The granddaddy of them all (at least in the recent times) is The Penguin Dictionary of Curious and Interesting Numbers by David Wells. There is The Kingdom of Infinite Number by Bryan Bunch and the latest Lure of the Integers by Joe Roberts. There are marvelous websites, the wikipedia and Number gossip that place at one’s fingertips a huge number of properties of a multitude of numbers.

Well, in the announcement of the current issue I offered a simple problem whose solution is this issue’s number. Subtracting this number of years from 2010 gives a number whose last two digits are this number written with the digits swapped, i.e., in the reversed order. The carnival is in its prime, but the number, being triangular, is not. Now, what’s number is the next issue? There are two triangular numbers that are good candidates for the main condition: 28 and 55, for 2010 – 28 = 1982 and 2010 – 55 = 1955. With two identical digits, 55 does not match well the “reversed order” clause,leaving 28 as the only possible solution.

28

is a distinguished number rich in properties and associations. 28 is happy, perfect, hexagonal, triangular, and what not … Here are a few less known (i.e., to me until recently) facts.

Bryan Bunch wrote of the discovery of the main asteroid belt. In 1776 the astronomer J. E. Bode published the discovery made four years earlier by J. D. Titius, that the distance of each of the planets from the sun is determined by a simple sequence which is now known as Bode’s Law. Write 0, 3 and then double the last number on each step: 0, 3, 6, 12, … Then add 4 to every term: 4, 7, 10, 16, 28, 48, … If the distance to from Earth to the Sub is taken 10 units, then Mercury is at 4, Venus at 7, and Mars at 16 units from the Sun. Jupiter, the next planet, is at distance of 52 (think of it as close to 48), skipping the expected 28.

On the New Year’s Day, 1801, the Italian astronomer G. Piazzi found a previously unknown, but very small, “planet”. The great K. F. Gauss devoted many hours to calculating its orbit and found that it corresponded to the number 28 in the Bode’s sequence. The “planet” was named Ceres and in time was shown to be the largest (950 km or 590 mi in diameter) among the myriad of asteroids in the main asteroid belt.

Joe Roberts gives several uncommon properties of 28. The least positive integer having h divisors is denoted A(h). Letting τ(n) be the number of divisors of n we say that n is minimal when A(τ(n)) = n; i.e., if n is the least positive integer having the number of divisors it has (p. 86).

There exist exactly 14 values of N for which the least common multiple of all integers from 1 through N is minimal. 28 is the largest integer with this property.

0.28 is commonly cited as an element of the 14-terms long solution to Steinhaus’ problem (Find an N-term sequence with terms in the closed interval [0, 1], such that the first two are in different halves of the interval, the first three are in different thirds, …, and all of them are in different Nths of the interval.)

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Sad news

One of the topmost Russian mathematician Vladimir Arnold, after whom one of the asteroids – Vladarnolda – in the main asteroid belt has been named, passed away on June 3, 2010. Arnold was an outspoken critic of the trend in math education known in the US as the New Math, while curiously, his teacher, Andrey Kolmogorov (indisputably one of the greatest mathematicians of the 20th century) was the initiator of that trend in the USSR. Arnold considered mathematics in part as a natural science. In his view, mathematics is being developed by trial and error, modeling and experimentation, with proofs coming last as a method of hypothesis verification. Both aspects of mathematics ought to be reflected in mathematics education.

In June I learned of passing in January 31 of Steve Fisk, professor of mathematics at Bowdoin College. Fisk became a math legend having dreamed up an almost trivial proof of Chvatal’s Art Gallery theorem while on a bus trip somewhere in Afganistan. The incident puts another big question mark to the notion that mathematics is the product of the left side of the brain.

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Interesting and relevant news

The controversial Russian mathematician Grigory Perelman has been awarded the Millennium Prize by the Clay Mathematics Institute in Cambridge, MA. The award honored his solving of the Poincaré conjecture, which characterized the three-dimensional (which is a four-dimensional object) sphere among other three-dimensional manifolds. As was not unexpected, Perelman turned down the $1 million prize; four years previously he declined to accept the Fields Medal. In a telephone interview, Perelman, 43, told Interfax: “I don’t like their decisions, I consider them unjust.” He judged his contribution not greater than that of Columbia University mathematician Richard Hamilton.

Jim Carlson, President of the Clay Institute, said institute officials will meet this fall to decide what to do with the prize money. “We have some ideas in mind,” he said. “We want to consider that carefully and make the best use possible of the money for the benefit of mathematics.”

Will they seek Perelman’s advice? I doubt it.

Guillermo Bautista came up with the idea of a new Mathematics and Multimedia Blog Carnival. The first issue saw the light of day on July 12, 2010. Meanwhile, the Carnival of Mathematics saw its 67th issue.

Next Math teachers at play carnival will be hosted at The Number Warrior blog.

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From the trenches

Tom DeRosa of the I Want to Teach Forever blog shared his experience of teaching operations with fractions using the play cards and a Simple Graphic Organizer to “make fraction a little less painful to student.”

Bill James of the Discrete Ideas blog has put together a table of the most common fractions and showed how “Learning the first 12 fractions can make it super-easy to do division in your head and produce answers down to the 10ths or even 1000ths quickly and easily.”

I once observed that there is a way of introducing addition of fractions without a formal definition. Nothing more difficult than dividing a few apples between a group of boys. I also found a couple of fractions related bloopers that illustrate the persisting difficulty the populace is having with fractions.

Caroline Mukisa who is said to be on a mission to help parents to support their children Math learning and is a big fan of sneaking math into children’s (and parent’s) diets, authored a guest post at Sol Lederman’s Wild About Math blog. Caroline shows how the many fascinating facts gathered in a Guinness World Records book naturally lead to simple math inquiries. Caroline’s post reminded me of Eric Charlesworth’s book 225 Fantastic Facts Math Word Problems (Scholastic, 2001). He, too, tries to enliven the study of mathematics by exploiting various world records, incredible animal facts, historical anomalies, and such.

Denise from the Let’s play math blog offers an advice on how to start a math teacher blog. With her blog high ranking, who could know better. A must read for all aspiring bloggers.

Cindy from the love2learn2day finds absorbing activities for little kids even in the insect world. Seeing and identifying patterns at an early age contributes greatly to the mathematical development of children.

Guillermo P. Bautista Jr. submitted one of his great tutorials. This one is a very lucid Introduction to Permutations.

Tracy Beach made a well argued pitch for the DreamBox, a project of virtual manipulatives to help teach early numeracy.

Maria Droujkova and Linda Fahlberg-Stojanovska have co-hosted an Elluminate Workshop on using and teaching with GeoGebra. Materials for this workshop are available at GeoGebra’s wiki site.

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Math curiosities

Patrick Vennebush, the Online Projects Manager for NCTM, hosted a July 7 webinar, where he discussed Calculation Nation, an online world of math strategy games, which is part of the NCTM Illuminations project. In his personal blog, Math Jokes 4 Mathy Folks, Patrick ponders over an argument he came across recently on the web. It was said that “the Senate gives just 18% of the U.S. population the power to stop a bill from passing Congress. That is, if 50 Senators vote ‘no’ to a bill, then it fails, and the 25 least populous states represent just 18% of the population.” This piece of statistics served a basis for a claim that the US Senate was no longer necessary. Along the way, it was implied that the Senate might have been necessary when it was first created, to give a voice to smaller states. Patrick checked the statistics of the first Senate that had only 24 senators from 12 states. At the time, according to his calculations, a bill might have been stopped by the senators representing only 17% of the population. As he put it, “Please understand, I’m not arguing that the Senate should be retained or abolished. But by the numbers, it appears that the Senate might have been even less necessary in 1789 than it is today.”

Shecky Riemann from Math-Frolic shares his life-long fascination with the Galton Box (the device which is also known as a “quincunx.”) His post, A Museum Piece describes the appeal of the contraption held to his even in childhood: “… like a magic, unseen hand guiding the fate of those individual spheres — even though each one took a rather random, unpredictable journey, the end result was highly predictable and little-changing. Even as a youngster I sensed there was something profound in that.” The post contains a list of links to an explanation of the mathematics behind the workings of the device, a video of a demonstration and an interactive simulation.

James Pollack in his new blog takes a fresh look at the Birthday problem.

Mimi Yang does an investigation into the problem of the best viewing angle in an iMax theater. Regiomontanus would have loved her submission, complete with diagrams and graphs.

Real mathematics occurs on most elementary levels. In a well known story, J. Piaget tells of a child who grew excited after discovering that the result of counting is independent of the order in which items have been counted. Another example that never fails to surprise both children and grownups is the question of whether it is possible to draw the same curve on two very different surfaces, say, a sphere and a cube. (For a cube, stipulate that a vertex must be inside the curve.) I would like to solicit such examples of “minimal mathematics”. I have put together a post Most elementary aha! moments to which comments can be added.

Sam Alexander from xamuel.com submitted an article on a variant of König’s Lemma that deals with trees in a graph theoretical sense. A tree is a connected graph without loops, meaning there is always one and only one way to get from any one node to any other along the edges of the graph. One of the nodes is usually chosen to be a root. The length of the path from a node to the root defines the level of that node. König’s Lemma says that an infinite tree all of whose levels are finite must contain an infinite branch. Xamuel applies the lemma to family trees under an apparent assumption that the big crunch (see, for example, S. Hawking’s A Brief History of Time) is never coming, and to games on an infinite board.

Vi Hart co-authored a paper with Erik and Martin Demaine, Computational Balloon Twisting: The Theory of Balloon Polyhedra. One motivation for balloon twisting is education. Balloon twisting is fun: the activity can both entertain and engage children of all ages. Thus balloon twisting can be a vehicle for teaching mathematical concepts inherent in balloons.

Ballons

Vi Hart offers step by step instructions for creating ballon polyhedrons. She also wrote a music piece for a music box on a Möbius strip.

Mobius strip music

This kind of musical composition is known as a canon. In a canon, copies of a single theme are played by the various participating voices. The copies can be shifted relative to the main theme, played backwards, or be inverted. The difficulty is of course to have the various voices coalesce in a harmonious whole. Douglas Hofstadter’s Gödel, Escher, Bach gives a literary expression to this idea while exploiting the manner in which it permeates the mathematics of K. Gödel, the art of M. C. Escher and the music of J. S. Bach.

I am grateful to Victor Gutenmacher for pointing me to the following youTube video

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Humor

A joke by V. Arnold:

Lemma: when communists put you to death, you lose on average half of your expected lifetime.

On a lighter note, my wife came across a couple of jokes in a Beginner’s Spanish Reader book. Both are probably folk tales.

A farmer went to market and bought 4 donkeys. He rode one and drove the rest home. Upon reaching his house he decided to check on the acquisition. He counted: 1, 2, 3 in bewilderment. Agitated, he called his wife. “Look here! I know I bought 4 donkeys but there are only three of them now: 1, 2, 3.” The wife looked and replied, “How strange! You see only three and I see 5 donkeys: 1, 2, 3, 4, 5.”

A student of mathematics was visiting his parents on a Christmas break. At breakfast, his mother put on the table a plate with two hard-boiled eggs. The son decided to play a joke on his father. After all, wasn’t he a very advanced math student?! Having removed and hidden one egg, he asked the father how many eggs were there. Naturally, the father counted 1. The student put the hidden egg back on the plate and asked the same question. “2″, replied the father. “So, in all, there are 3 eggs, right?”, said the student. “We counted one egg before, and two eggs later, and together this makes three eggs.” At this point, his mother intervened. “You are absolutely right, there are three eggs indeed. But let’s eat already. I’ll take the egg nearest to me and the brown one will go to your father. You’ll have the third one.”

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P.S.

With sincerest and humble apologies to all the carnival participants – reader and authors – I am adding submissions that were received before the deadline but after the carnival issue went out due to an unfortunate accident. What I was thinking when hitting the Publish button two days earlier? I am certain I was not thinking about pressing that button.

John Golden, a.k.a. Math Hombre, celebrates a full circle day (6/28) with an article on Similarity and π.

John Cook, in his The Endeavor blog, writes about the Lincoln Index which is commonly applied to population estimates. As John shows, the index reliably estimates the number of errors that cropped up in a computer program – the place the bugs live in and thrive.

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