Cut The Knot!by Alex Bogomolny |
Erro Ergo Disco
December 2003
... learn from other people's mistakes, life isn't long enough to make them all yourself. Harry S. Truman The way in which man approaches his failure determines what he will become. Karl Jaspers, |
I've been looking for a piece of wisdom echoing the Russian "one learns from mistakes." A search on the web for "mistakes and learn" brought up Harry Truman's quote. A search for "err and learn" produced a German proverb, "He who likes cherries soon learns to climb." As a substitute, I tried to create an authentically looking maxim. "Erro ergo disco", which in my view means "I err, therefore, I learn", was my first Latin sentence ever. The second was "Non errat non discat", which I believe stands for "He who does not err does not learn". The idea was to convey a positive message that it was OK to err. Mistakes give one a chance at active learning. This is well known. Finding mistakes is a useful and entertaining educational activity. Mistakes have been written about.
Rouse Ball's classic Mathematical Recreations and Essays contains sections on arithmetic and geometric fallacies and one on geometric paradoxes. In Rouse Ball's terminology, a fallacy is a deliberately erroneous argument that leads to an impossible conclusion. A paradox is a seemingly fallacious result whose oddity is due more to misperception than to a misleading argument (see, for example, Fibonacci Bamboozlement.)
The doyen of the current literature that deals with erroneous arguments in a systematic way is Lapses in Mathematical Reasoning whose first Russian edition appeared in 1938, the horrible time of purges in Russian history, when mandatory quotes by the fathers of the revolution served to demonstrate the correctness of the authors' message. Thus in Chapter 1 one is treated to V. I. Lenin's insight to the effect that "there exist collections of such mathematical sophisms, and they bring profit to schoolchildren". The word "sophism", that has the same meaning as Rouse Ball's "fallacy", is used interchangeably with word "sophistry" which, according to Lenin, is "... the grasping of the outer coincidence of cases outside the relation of events ..." The book offers fallacies from Arithmetic, Algebra, Geometry, Trigonometry and a few due to a mistreatment of approximate calculations.
Many of the fallacies collected by E. Barbeau in his Mathematical Fallacies, Flaws, and Flimflam are honest student errors plucked from homeworks and solutions to exam problems. B. Cipra's book is a light hearted attempt to help Calculus students avoid making errors and ease their detection based on little more than common sense. For further reading, Cipra recommends the long lost Pseudaria by Euclid that, according to Proclus, was apparently written with a similar goal in mind. Readily available books by Dewdney and Paulos treat real life examples of innocent and deliberate misuse of (mostly) mathematics of proportion, probability and statistics.
Probability and statistics are two branches of mathematics that, in the modern world of media and politics, are destined to play an increasingly important role in everyday life. More than other branches of mathematics, probability in its more elementary aspects appeals to common intuition and everyday experience. It is often possible to talk of probability without invoking the formal rules of mathematics, which leads to some pitfalls in reasoning. As has been keenly observed in the Principle and Standards for School Mathematics (NCTM, 2000, p. 254), "Misconceptions about probability have been held by many students but also by many adults."
Let's have a closer look at a problem of family statistics [Falk, p. 45]:
(1) | "Do men have more sisters than women?" |
An answer may be surmised from a few examples. Take a small family with two siblings: a boy and a girl. The boy has a sister, while the girl does not. In a family with a son and two daughters, the boy has two sisters, while the girls have only one each. A rule seems to emerge: a girl is excluded from sister counting, while boys count all the female siblings there are. From which the conclusion that men should have more sisters than women seems to follow naturally. However, this conclusion is wrong. Men have as many sisters as do women. An applet below offers a simulation that might help clarify the situation. The argument that follows seems to me sufficiently convincing to seal the result.
So why men have as many sisters as do women? Fix the number of children in a family and consider all possible variants. For example, a family with two children may have two boys, a boy and a girl, or two girls. The heterogeneous variant should be counted twice, because statistically it is twice as likely as either of the homogeneous combinations. (Think of a two coin experimentation.) The number of males' sisters is 0 for the first variant, is 1 for the two mixed variants, and 0 again for the family with two daughters. On the whole, 2 children families contribute 2 males' sisters. 2 children families also contribute 2 females' sisters that come from the family with two daughters. The average contribution of a 2 child family is 1/2 in both cases. For a family with 3 children we have the following table:
Combination | # of Combinations | Males' sisters | Females' sisters | |||
---|---|---|---|---|---|---|
boy, boy, boy | 1 | 0 | 0 | |||
boy, boy, girl | 3 | 2×1 | 0 | |||
boy, girl, girl | 3 | 2 | 2×1 | |||
girl, girl, girl | 1 | 0 | 3×2 |
On the whole, such families contribute 12 males' sisters (second and third rows) and 12 females' sisters (third and fourth rows). The average contribution per family is the same 12/8 to both counts. In general, too, families with n children contribute the same number, n(n-1)2^{n-2}, of males' and females' sisters, the average being n(n-1)/4 per family.
Indeed, in a family with s girls and (n-s) boys, the boys have a total of s(n-s) sisters, while the girls have altogether s(s-1) sisters. Assume that boys and girls come into the world with equal probabilities of 1/2 and that the birth events are independent. Then there are 2^{n} ways a family with n children might have come about. Of this, C(n,s) - the binomial coefficient "n choose s" - is the number of n children families with s daughters. Therefore, the average number of sisters boys from n children families have is given by
B_{n} = 2^{-n} ∑_{s}C(n,s)·s(n-s).
Similarly, the average number of girls' sisters in such families is
G_{n} = 2^{-n} ∑_{s}C(n,s)·s(s-1).
Both sums are easily computed with generating functions.
Let B_{n}(x,y) = (x+y)^{n}/2^{n} and G_{n}(x) = (1+x)^{n}/2^{n}. Then
B_{n} = (d^{2}/dx dy)B_{n}(1,1), while
G_{n} = (d^{2}/dx^{2})G_{n}(1).
Which immediately implies B_{n} = G_{n} = n(n-1)/4.
Ruma Falk in whose book I found the above example offers several explanations of why men have the same number of sisters as women. I confess that some of them went counter to my intuition. The reason might be in that she tried to convince the reader through a verbal discussion rather than just prove the result. For example, (1) is equivalent to another question,
(2) | "Do men have more sisters than brothers?" |
And this one is reduced to
(3) | "Does a random child have as many sisters as brothers?" |
Removal of "a random child" leaves a family short of one child, for which (3) becomes
(4) | "Does an average family have as many daughters as sons?" |
The answer to the latter is a sound "Yes, of course." All it takes now is to step backwards. The first move from (4) to (3) is easy. So the answer to (3) is "Yes." The move from (3) to (2) is based on the idea of statistical independence. Since child births and their sexes are independent of each other even in a single family, it should not matter what sexes have been queried about. This was hard for me to accept. Why? I do not know. One possible reason may be in that the answer to (2) can be "No" without logically contradicting (3). If men had more sisters than brothers then naturally women would have more brothers than sisters, and all in the same proportion, so that it would be possible to move from (2) to (3) without implying their equivalence.
The step from (1) to (2) is also logical. Since, on average, the number of men's brothers is the same as the number of women's sisters, (1) and (2) are equivalent. When I say "logical" I mean "based on reason and that part of theory of probability that I accept viscerally". This does not cover the equivalence of (2) and (3) which seems to require a leap of faith.
Now, the extent of the discussion of that problem by the author is reassuring. The difficulty is rather common. (Here's one reader's story.) She almost says this much. People are liable to have mental blocks. The following anecdote shows that even superior minds have them occasionally.
Andrew Vazsonyi begins his memoir with a story of his first encounter with the infamous Monty Hall problem. The solution he reluctantly arrived at went against his gut feeling. Curiously, his friend Paul Erdös had even more difficult time accepting the solution. He just grew upset on hearing Vazsonyi's explanations. A few days later, Erdös met R. Graham and they arrived at a solution that made everything perfectly clear to Erdös. He telephoned Vazsonyi to share the news and the solution. As the latter observed, "It made no sense at all. Apparently, Graham and Erdös had developed a private language that I wasn't privy to."
The story is amazing not only because of Erdös' stature in the world of mathematics in general, but, in particular, because the probabilistic method Erdös invented in collaboration with Alfred Rényi may be considered his most lasting legacy [Aigner, p. 187]. The method applies the theory of probability to obtain number theoretic results.
The idea, however, might not be that new. What would he say if that mysterious proof that did not fit in the margins of Fermat's book, was a probabilistic one? As the author of a recent online publication suggests, the possibility is very real indeed. After all, Fermat is considered one of the originators of the probability theory. (I am grateful to Maxim R. for bringing this publication to my attention.)
Since the proof was never published, we may only speculate. The author of the publication in fact does.
Assume x^{n} + y^{n} = z^{n}, for n > 2, and x, y, z positive integers. Write this as
(FLT) | (x/z)^{n} + (y/z)^{n} = 1. |
Think now of two events A and B with probabilities
P(A + B) = 1. |
Let A^{c} denote the event complimentary to A, and similarly B^{c}.
P(A^{c}·B^{c}) = 0.
The rest is easy.
0 | = P(A^{c}·B^{c}) |
= P(A^{c})·P(B^{c}) | |
= (1 - (x/z)^{n})·(1 - (y/z)^{n}). |
Therefore either x = z or y = z, so that (FLT) has only trivial solutions. A contradicion. That's all there is to it.
Curiously, the author does not see any fault with the above, except for the fact that the derivation also works for
A question to ponder.
References
- M. Aigner, G. M. Ziegler, Proofs from THE BOOK, Springer, 1999
- E. J. Barbeau, Mathematical Fallacies, Flaws, and Flimflam, MAA, 2000
- V. M. Bradis et al, Lapses in Mathematical Reasoning, Dover, 1999
- B. Cipra, Misteaks ... and how to find them before the teacher does ..., A K Peters, 2000
- A. K. Dewdney, 200% of Nothing, John Wiley & Sons, 1993
- R. Falk, Understanding Probability and Statistics, A K Peters, 1993
- J. A. Paulos, A Mathematician Reads the Newspaper, Anchor Books, 1996
- W. W. Rouse Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, Dover, 1987
- A. Vazsonyi, Which Door has the Cadillac, Writers Club Press, 2002
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