A logic exercise by Lewis Carroll: What conclusion can be drawn from these premises?
From Lee Sallows:
(Thanks, Lee!)
The ancient Chinese philosopher Gongsun Long appeared to claim that a white horse is not a horse:
Is ‘a white horse is not horse’ assertible?
Advocate: It is.
Objector: How?
Advocate: ‘Horse’ is that by means of which one names the shape. ‘White’ is that by means of which one names the color. What names the color is not what names the shape. Hence, one may say ‘white horse is not horse.’
Objector: If there are white horses, one cannot say that there are no horses. If one cannot say that there are no horses, doesn’t that mean that there are horses? For there to be white horses is for there to be horses. How could it be that the white ones are not horses?
Advocate: If one wants horses, that extends to yellow or black horses. But if one wants white horses, that does not extend to yellow or black horses. Suppose that white horses were horses. Then what one wants [in the two cases] would be the same. If what one wants were the same, then ‘white’ would not differ from ‘horse.’ If what one wants does not differ, then how is it that yellow or black horses are acceptable in one case and unacceptable in the other case? It is clear that acceptable and unacceptable are mutually contrary. Hence, yellow and black horses are the same, one can respond that there are horses, but one cannot respond that there are white horses. Thus, it is evident that white horses are not horses.
Interpretations vary; one explanation is that the conundrum blurs the distinction between identity and class, exploiting an ambiguity in the Chinese language — certainly the expressions “white horse” and “horse” do not have identical meanings, but one can refer to a subset of the other.
Whether the philosopher was serious isn’t clear. His other paradoxes include “When no thing is not the pointed-out, to point out is not to point out” and “There is no 1 in 2.”
More trouble with horse color.
03/08/2024 UPDATE: A Swedish Facebook meme of 2012: Horses are a fruit that does not exist. (Thanks, Mikael.)
By Lee Sallows: If the letters BJFGSDNRMLATPHOCIYVEU are assigned to the integers -10 to 10, then:
J+A+N+U+A+R+Y = -9+0-4+10+0-3+7 = 1 F+E+B+R+U+A+R+Y = -8+9-10-3+10+0-3+7 = 2 M+A+R+C+H = -2+0-3+5+3 = 3 A+P+R+I+L = 0+2-3+6-1 = 4 M+A+Y = -2+0+7 = 5 J+U+N+E = -9+10-4+9 = 6 J+U+L+Y = -9+10-1+7 = 7 A+U+G+U+S+T = 0+10-7+10-6+1 = 8 S+E+P+T+E+M+B+E+R = -6+9+2+1+9-2-10+9-3 = 9 O+C+T+O+B+ER = 4+5+1+4-10+9-3 = 10 N+O+V+E+M+B+E+R = -4+4+8+9-2-10+9-3 = 11 D+E+C+E+M+B+E+R = -5+9+5+9-2-10+9-3 = 12
Similarly, if -7 to 7 are assigned SROEMUNFIDYHTAW, then SUNDAY to SATURDAY take on ordinal values. See Alignment.
(David Morice, “Kickshaws,” Word Ways 24:2 [May 1991], 105-116.)
Hungarian mathematician József Kürschák offered this “proof without words” that a regular dodecagon inscribed in a unit circle has area 3. If the circle is inscribed in a square, the resulting figure can be tiled by triangles of two families — 16 equilateral triangles whose sides are equal to those of the dodecagon and 32 isosceles triangles with angles 15°-15°-150° and longest side 1. The area of the large square is 4, and the triangles that make up the dodecagon can be rearranged to fill 3 of its quadrants (see the video below). So the area of the dodecagon is 3/4 of 4, or 3.
(To see that the square and the dodecagon can be tiled as claimed, see Alexander Bogomolny’s discussion here.)
(Gerald L. Alexanderson and Kenneth Seydel, “Kürschak’s Tile,” Mathematical Gazette 62:421 [October 1978], 192-196.)
[Wittgenstein] once greeted me with the question: ‘Why do people say that it was natural to think that the sun went round the earth rather than that the earth turned on its axis?’ I replied: ‘I suppose, because it looked as if the sun went round the earth.’ ‘Well,’ he asked, ‘what would it have looked like if it had looked as if the earth turned on its axis?’
— G.E.M. Anscombe, An Introduction to Wittgenstein’s Tractatus, 1959
Francis Galton quantified boredom. At a tedious meeting in 1885, he observed that the distance between neighboring heads in the listless crowd began to vary:
When the audience is intent each person forgets his muscular weariness and skin discomfort, and he holds himself rigidly in the best position for seeing and hearing. As this is practically identical for persons who sit side by side, their bodies are parallel, and again, as they sit at much the same distances apart, their heads are correspondingly equidistant. But when the audience is bored the several individuals cease to forget themselves and they begin to pay much attention to the discomforts attendant on sitting long in the same position. They sway from side to side, each in his own way, and the intervals between their faces, which lie at the free end of the radius formed by their bodies, with their seat as the centre of rotation varies greatly.
He wasn’t able to estimate this numerically, but he did find another measure: He counted about 50 fidgets per minute in each section of 50 people. “The audience was mostly elderly; the young would have been more mobile.” He urged “observant philosophers” at dull meetings to estimate “the frequency, amplitude, and duration of the fidgets of their fellow-sufferers” in hopes that “they may acquire the new art of giving numerical expression to the amount of boredom expressed by the audience generally during the reading of any particular memoir.”
(Francis Galton, “The Measure of Fidget,” Nature 32:817 [June 25, 1885], 174–175.)
In 1743, Swedish organ maker Daniel Stråhle published this method to calculate the sounding lengths of strings in a musical tuning with 12 pitches per octave that’s close to equal temperament. Draw segment QR 12 units long and establish it as the base of an isosceles triangle with sides of length 24. Find point P on OQ seven units above Q and draw a line through it from R to M such that PM = RP. Now MR is the string length of the lowest sounding pitch, MP is the pitch one octave higher, and the points labeled 2 through 12 give the endpoints for successive semitones within the octave.
Stråhle, who had no mathematical training, said he’d established the method after “some thought and a great number of attempts.” Exactly how he came up with it is not known.
English mapmaker John Ogilby completed a startling project in 1675: a road atlas of 17th-century Britain, offering strip maps of most of the major routes in England and Wales. He wrote to Charles II:
I have Attempted to Improve Our Commerce and Correspondency at Home, by Registring and Illustrating Your Majesty’s High-Ways, Directly and Transversely, as from Shoare to Shoare, so to the Prescrib’d Limts of the Circumambient Ocean, from the Great Emporium and Prime Center of the Kingdom, Your Royal Metropolis.
It used a consistent scale of one inch per mile, with each mile comprising 1760 yards, a standard that later mapmakers would follow. You can see the whole atlas here.
From reader Éric Angelini:
What distinguishes this sequence of integers?
10, 9, 18, 10, 17, 10, 9, 10, 16, 10, 9, 18 …
Adding 9 to each successive digit and inserting a comma after the result reproduces the original sequence:
1 + 9 = 10
0 + 9 = 9
9 + 9 = 18
1 + 9 = 10
8 + 9 = 17, etc.
This example (A369603 in OEIS) is lexicographically the earliest such sequence beginning with 10.
(Thanks, Éric.)