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Paradox in the Teaching of Mathematics a Pythagorean Problem

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Paradox in the Teaching of Mathematics a Pythagorean Problem Paradox in the Teaching of Mathematics Larry G. Blaine Plymouth State University Plymouth, New Hampshire, USA [email protected] Abstract: This paper presents an argument for the use of paradoxes in the teaching of mathematics, as a means of raising interest and sharpening reasoning skills. Based on the author’s experience in and out of the classroom, four representative examples of contrasting types have been selected and analyzed. A distinction is made between problems whose solutions violate intuition but are not otherwise questionable (not true paradoxes, in the logical sense) and those that are deeply enigmatic. This will be a talk about two subjects, which we refer to temporarily as S1 and S2, and how they are regarded by students and by the public at large. S1 is useful sometimes, but it is dull and even painful to study. Many would avoid it altogether if given the choice. Not a few regard forced study of S1 as a form of torment. S2, on the other hand, is dazzling and delightful, sometimes magical. No one is ever forced to learn S2. I am speaking of course about two faces of the same subject- mathematics. We, as teachers and practitioners, are painfully aware of how familiar our students, friends, and acquaintances are with S1. Reflection and experience, though, reveal that most people are quite susceptible to the charms of S2, if introduced properly. I make no attempt here to be systematic. I mean merely to display a few paradoxical problems that I have found to be fascinating to curious but mathematically unsophisticated people, inside and outside the classroom. The level of technical skill required to understand them is for the most part low. Except for the first example, they have been written about at length. Nevertheless, my experience has been that most of them are unfamiliar to most teachers. They are offered in a spirit of seduction, for use in drawing students and others into a love for our noble art. Now a few words about terminology. A paradox in mathematics or logic may be regarded as a question or problem for which the application of correct reasoning leads to a self-evidently absurd conclusion, or to two or more contradictory conclusions. More whimsically, paradox has been defined as “truth standing on its head to attract attention.” For this note we adopt the following classification: (i) Puzzle Paradoxes are problems with solutions that are certainly correct, but that nevertheless violate our intuition. (ii) Pseudo-Paradoxes may seem enigmatic at first, but on examination turn out to be trivial or ambiguously posed. (They still may be very much worth thinking about.) (iii) True paradoxes are described above. Two examples of type (i) are given, and one each of (ii) and (iii). Note, though, that experts sometimes disagree violently as to which of (ii) or (iii) applies. That includes two of the problems below- caveat lector! A Pythagorean Problem This first problem gives an example of an easy computation that leads to a startling result. You are to build a bridge that goes horizontally from point A to point B. A and B are exactly one kilometer apart. This bridge is to be very simple, consisting of one perfectly straight, perfectly thin steel rod. Unfortunately, the rod you have is 10 cm. too long, and you cannot cut it. You decide to bend it slightly at its midpoint, so that the bridge consists of two linear segments. In crossing from A to B, one ascends to the middle, then descends to B. How high above the horizontal is the middle of the bridge? Having posed the question, the teacher should do two things immediately, before any calculations are started. First, emphasize how utterly tiny the extra 10 cm. is. “If this blackboard were 1 kilometer long, the extra 10 cm. would be the thickness of four pieces of paper” or something of the sort. Second, ask for an estimate. Almost all students expect the height to be quite small- typically about 5 or 6 cm. The Pythagorean Theorem tells a different story. Call the height h, and consider the right triangle formed by half the bridge, the horizontal line, and a vertical line at the middle. With meters as the unit, we have h2 + 5002 = 500.052, or h = 500.052 5002 . At this point, a calculator is convenient, but not essential. The familiar factorization of the difference of two squares gives h = .051000.05 = 50.0025 , and we conclude that the height exceeds 7 meters! There is nothing special about 10 cm; a similar calculation may be carried out for any excess length. If the rod is only one millimeter too long, the height is, unbelievably, about 70 cm. It is of course absurd to speak of an actual bridge here, but this type of problem is common in surveying, astronomy, etc. Before leaving our bridge, let us mention in passing a few related problems that are beyond the scope of this paper but may interest teachers of elementary analysis. It is a nice exercise to write a formula for h(x), height as a function of excess length x, and to examine the behavior of the derivative h’(x) as x → 0+. Even nicer is to approximate the height the bridge reaches, if it makes a circular arc from A to B. Approximate, because no exact solution is possible- Some rootfinding algorithm must be used at the end. A Card Problem This is a classic problem, or rather class of problems, with an interesting history. The oldest version known to me is described by J. E. Littlewood [4]. It is: East and West are partners at bridge. East’s hand contains no aces. It is known, however, that West has at least one ace. What is the probability p that West has more than one ace? This problem is slightly complicated, requiring knowledge of the game of bridge, and some skill in manipulating combinations. We will come back to it. First, we give a much simpler variant that illustrates the same principles. A deck consists of four cards- two aces and two others. We may suppose it is {A♠, A♣, 2♣, 2♠}. You are to be dealt two of these cards. What is the probability p that you will get both aces, assuming you get at least one ace? What is the probability q that you will get both aces, assuming you get A♠? It is obvious (is it not?) that the two probabilities are the same. After all, if we assume you get an ace, we may as well suppose it to be A♠ for purposes of calculation. So, all hands being equally likely, we just list them. There are six: {A♠, A♣}, {A♠, 2♣}, {A♠, 2♠}, {A♣, 2♣}, {A♣, 2♠}, and {2♣, 2♠}. Five contain at least one ace, one of those five contains both aces, and thus p = 1/5. Three contain A♠, one of those three contains both aces, and thus q = 1/3. With our intuition thus outraged, let us return to Littlewood’s version. He writes that it was proposed as an examination problem at Cambridge around 1911, but that two mathematicians who reviewed it came to different conclusions. One of them found the probability p for the problem as stated, but the other found the probability q that West has more than one ace, assuming he has A♠. Littlewood shows that p must be smaller than q, but does not give the actual numbers. Here they are. In the spirit of 1911, all calculations were done at first by hand, an hour’s plodding work. To find p, we must consider a deck of 39 cards, 4 of which are aces. (East already has 13 cards, none of which are aces.) Of C(39, 13) possible hands for West, C(35, 13) contain no ace at all, so C(39, 13) – C(35, 13) is the number of hands containing at least one ace. Of those, 4×C(35, 12) contain exactly one ace, and the rest contain two or more aces. Assuming all hands to be equally likely, calculation shows that p ≈ 49.8% . Finding q is easier. Consider a deck of 38 cards, 3 of which are aces, and see how hands of 12 of them may be distributed to West. (West already has A♠.) There are C(38, 12) such hands, C(35, 12) of which contain no ace. The probability of A♠ being West’s only ace is thus C(35, 12)/C(38, 12) = 650/2109, and it follows that q = 1459/2105 ≈ 69.2%. I confesses that, despite knowing in advance that p ≠ q, it was shocking at first to see how much they differ! A good discussion of problems of this general kind may be found in [3]. The Thompson Lamp Most, including this writer, who have thought about the Thompson Lamp have decided that it does not present a true paradox. We thus classify it as type (ii). This discussion will be brief. The interested reader may consult [3] and [5] for more details and other opinions. The lamp behaves like this. (It is unnecessary to stress that it is not a real object, and that what follows is a pure thought experiment.) It has only two states, on and off, and the states are controlled by a single switch. It is turned on. After half a minute it is turned off. After another quarter-minute it is turned on again, and an eighth of a minute after that, it is turned back off. This process continues in the obvious way, each time interval half as long as the previous one.
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