Author!

How do I know that I’m not just a fictional character in some imagined story? What could I learn about myself that would prove that I’m real? “I am human, male, brunette, etc., but none of that helps,” writes UCLA philosopher Terence Parsons. “I see people, talk to them, etc., but so did Sherlock Holmes.”

Descartes would say that the very fact that I’m thinking about this shows that I exist: cogito ergo sum. But a fictional character could make the same argument. “Hamlet did think a great many things,” writes Jaakko Hintikka. “Does it follow that he existed?” Robert Nozick adds, “Could not any proof be written into a work of fiction and be presented by one of the characters, perhaps one named ‘Descartes’?”

Tweedledee tells Alice that she’s only a figment of the Red King’s dream. “If that there King was to wake,” adds Tweedledum, “you’d go out — bang! — just like a candle!”

Alice says, “Hush! You’ll be waking him, I’m afraid, if you make so much noise.”

“Well, it’s no use YOUR talking about waking him,” replies Tweedledum, “when you’re only one of the things in his dream. You know very well you’re not real.”

“It seems to me that this is a philosophical problem that deserves to be treated seriously on a par with issues like the reality of the external world and the existence of other minds,” Parsons writes. “I don’t know how to solve it.”

(Terence Parsons, Nonexistent Objects, 1980; Charles Crittenden, Unreality, 1991; Robert Nozick, “Fiction,” Ploughshares 6:3 [1980], 74-78; Jaakko Hintikka, “Cogito, Ergo Sum: Inference or Performance?”, Philosophical Review 71:1 [January 1962], 3-32.)

| Literature · Science & Math

Scattered Objects

http://commons.wikimedia.org/wiki/File:Auklet_flock_Shumagins_1986.jpg

If a flock of birds disperses gradually, at what point does it cease to be a flock?

“There is at the moment a pipe on my desk,” wrote MIT philosopher Richard Cartwright in 1987. “Its stem has been removed, but it remains a pipe for all that; otherwise no pipe could survive a thorough cleaning.”

But he also owned a two-volume set of John McTaggart’s The Nature of Existence, one volume of which was in Cambridge and the other in Boston. Do those two volumes still make one thing? If so, is there a “thing” composed of the Eiffel Tower and the Old North Church? Why not?

(From Cartwright’s Philosophical Essays.)

| Science & Math

A Golden March

fibonacci circle

Draw a circle whose circumference is the golden mean. Choose a point and label it 1, then move clockwise around the circle in steps of arc length 1, labeling the points 2, 3, and so on. At each step, the difference between each pair of adjacent numbers on the circle is a Fibonacci number.

| Science & Math

Misc

  • What time is it at the North Pole?
  • The shortest three-syllable word in English is W.
  • After the revolution, the French frigate Carmagnole used a guillotine as its figurehead.
  • 823502 + 381252 = 8235038125
  • PRICES: CRIPES!
  • “Conceal a flaw, and the world will imagine the worst.” — Martial

When Montenegro declared independence from Yugoslavia, its top-level domain changed from .yu to .me.

| History · Language · Quotations · Science & Math · Technology

The Richardson Effect

http://commons.wikimedia.org/wiki/File:Britain-fractal-coastline-200km.png
Image: Wikimedia Commons

How long is a coastline? If we measure with a long yardstick, we get one answer, but as we shorten the scale the total length goes up. For certain mathematical shapes, indeed, it goes up without limit.

English mathematician Lewis Fry Richardson discovered this perplexing result in the early 20th century while examining the relationship between the lengths of national boundaries and the likelihood of war. If the Spanish claim that the length of their border with Portugal is 987 km, and the Portuguese say it’s 1,214 km, who’s right? The ambiguity arises because a wiggly boundary occupies a fractional dimension — it’s something between a line and a surface.

“At one extreme, D = 1.00 for a frontier that looks straight on the map,” Richardson wrote. “For the other extreme, the west coast of Britain was selected because it looks like one of the most irregular in the world; it was found to give D = 1.25.”

This is a mathematical notion, but it’s also a practical problem. On the fjord-addled panhandle of Alaska, the boundary with British Columbia was originally defined as “formed by a line parallel to the winding of the coast.” Who gets to define that? On the map below, the United States claimed the blue border, Canada wanted the red one, and British Columbia claimed the green. The yellow border was arbitrated in 1903.

alaska panhandle dispute

| History · Science & Math

Two Lists

Write out the positive powers of 10 in both base 2 and base 5:

powers of 10

Now for any integer n > 1, we’ll find exactly one number of length n somewhere on the two lists. They contain one 3-digit number, one 4-digit number, and so on forever — if n = 100 we find a 100-digit number in the 30th position on the base 2 list.

(This result first appeared in the 1994 Asian Pacific Mathematics Olympiad. I found it in Ravi Vakil’s A Mathematical Mosaic.)

Two further curious lists: If we write out the triangular numbers, those in positions 3, 33, etc. show a pattern:

T(3) = 6
T(33) = 561
T(333) = 55611
T(3,333) = 5556111
T(33,333) = 555561111
T(333,333) = 55555611111

Similarly:

T(6) = 21
T(66) = 2211
T(666) = 222111
T(6,666) = 22221111
T(66,666) = 2222211111
T(666,666) = 222222111111

(Thanks, Larry.)

| Science & Math

Some “Odd” Theorems

http://commons.wikimedia.org/wiki/File:One-seventh_area_triangle.svg

Draw any triangle and divide each leg into three equal segments. Connect each vertex to one of the trisection points on the opposite leg, as shown, and the triangle formed in the center will have 1/7 the area of the original triangle.

2/5 semicircle theorem

A square inscribed in a semicircle has 2/5 the area of a square inscribed in a circle of the same radius.

1/5 square theorem

Draw a square and connect each vertex to the midpoint of an opposite side, as shown. The square formed in the center will have 1/5 the area of the original square.

A “proof without words”:

1/5 square theorem - proof

Trisect each side of a triangle and join each vertex to the opposite trisection points. Then write a hexagram in the hexagon in the center. The area of the hexagram is 7/100 the area of the original triangle.

| Science & Math

Naturally

Steven Bartlett and Peter Suber’s Self-Reference: Reflections on Reflexivity contains a bibliography of works on reflexivity.

It includes an entry for Steven Bartlett and Peter Suber’s Self-Reference: Reflections on Reflexivity.

| Science & Math

Apportionment Paradoxes

http://commons.wikimedia.org/wiki/File:United_States_Capitol_-_west_front_edit.jpg

Until 1911, the U.S. House of Representatives grew along with the country. Accordingly, when the 1880 census showed an increase in population, C.W. Seaton, chief clerk of the census office, worked out apportionments for all House sizes between 275 and 350, in order to see which states would get the new seats.

He was in for a surprise. The method was straightforward: Take the total U.S. population and divide it by the proposed number of seats in the House, rounding all fractions down. This would dispose of most of the seats; any leftover seats would be awarded to the states whose fractional remainders had been highest. But Seaton discovered an oddity:

alabama paradox

If the House had 299 seats, Alabama would get 8 representatives (because its remainder, .646, was higher than that of Texas or Illinois). But if the House had 300 seats it would get only 7 (the extra representative would now go to Illinois, whose remainder had surpassed Alabama’s). The problem is that the “fair share” of a large state increases more quickly than that of a small state.

Seaton called this the Alabama paradox. A related problem is the population paradox: If the method above had been used in 1901 to reallocate 386 seats in the House, Virginia would have lost a seat to Maine even though the ratio of their populations had increased from 2.67 to 2.68:

population paradox

Here, even though the size of the House has not changed, a fast-growing state receives fewer representatives than a slow-growing one.

In 1982 mathematicians Michel Balinski and Peyton Young showed that if each party gets one of the two numbers closest to its fair share of seats, then any system of apportionment will run into one of these paradoxes. The solution, it seems clear, is to start cutting legislators into pieces.

(These data are from Hannu Nurmi’s Voting Paradoxes and How to Deal With Them, 1999. Balinski and Young’s book is Fair Representation: Meeting the Ideal of One Man, One Vote.)

| Science & Math · Society