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Science & Math

Euclid, Schmeuclid

euclid schmeuclid fallacy

Consider any two circles intersecting at points Q and R. Their diameters are QP and QS. PS intersects the circles at M and N.

Now, any angle inscribed in a semicircle is a right angle. This means that both QNP and QMS are right angles — and thus that there are two perpendiculars from Q to PS. Doesn’t it?

Mrs. Miniver’s Problem

mrs. miniver's problem

One of Jan Struther’s popular stories of the 1930s included the following passage:

She saw every relationship as a pair of intersecting circles. It would seem at first glance that the more they overlapped the better the relationship; but this is not so. Beyond a certain point the law of diminishing returns sets in, and there are not enough private resources left on either side to enrich the life that is shared. Probably perfection is reached when the area of the two outer crescents, added together, is exactly equal to that of the leaf-shaped piece in the middle. On paper there must be some neat mathematical formula for arriving at this; in life, none.

Interestingly, mathematicians who pursued the problem found that no precise solution is possible. With circles, as with relationships, we have to do the best we can.

Hard Bargaining

http://www.sxc.hu/photo/984541

In Robert Louis Stevenson’s story “The Bottle Imp,” the titular imp will grant its owner (almost) any wish, but if the owner dies with the bottle then he burns in hell. He may sell the bottle, but he must charge less than he paid for it, and the new buyer must understand these conditions.

Now, no one would buy such a bottle for 1 cent, as he could not then sell it again. (The imp can’t make you immortal, or support prices smaller than one cent, or alter the conditions.) And if 1 cent is too low a price, then so is 2 cents, for the same reason. And so on, apparently forever. It would be irrational to buy the bottle for any price.

But intuitively most people would consider $1,000 a reasonable price to pay for the use of a wish-granting genie. Who’s right?

See also Tug of War.

The Wealth Puzzle

A beggar asks a wealthy man for a penny. He says he wants to be a wealthy man himself someday.

“Well, consider,” says the wealthy man. “A single penny will not make you wealthy. And a second penny will not make you wealthy either.

“Indeed, there is no point at which an additional penny will make you a wealthy man. So I’m sorry, my friend, but your dream is impossible.”

Poetry in Motion

http://books.google.com/books?id=B7gEAAAAYAAJ&pg=PA404&dq="An+inextensible+heavy+chain"+maxwell&as_brr=1&ei=iQh6SoqCMZ6SygSHkbzKDA#v=onepage&q=&f=false

When he wasn’t taming electromagnetism, James Clerk Maxwell wrote verse. Here’s “A Problem in Dynamics,” composed in 1854:

An inextensible heavy chain
Lies on a smooth horizontal plane,
An impulsive force is applied at A,
Required the initial motion of K.

Let ds be the infinitesimal link,
Of which for the present we’ve only to think;
Let T be the tension, and T + dT
The same for the end that is nearest to B.
Let a be put, by a common convention,
For the angle at M ‘twixt OX and the tension;
Let Vt and Vn be ds‘s velocities,
Of which Vt along and Vn across it is;
Then Vn/Vt the tangent will equal,
Of the angle of starting worked out in the sequel.

In working the problem the first thing of course is
To equate the impressed and effectual forces.
K is tugged by two tensions, whose difference dT
Must equal the element’s mass into Vt.
Vn must be due to the force perpendicular
To ds‘s direction, which shows the particular
Advantage of using da to serve at your
Pleasure to estimate ds‘s curvature.
For Vn into mass of a unit of chain
Must equal the curvature into the strain.

Thus managing cause and effect to discriminate,
The student must fruitlessly try to eliminate,
And painfully learn, that in order to do it, he
Must find the Equation of Continuity.
The reason is this, that the tough little element,
Which the force of impulsion to beat to a jelly meant,
Was endowed with a property incomprehensible,
And was “given,” in the language of Shop, “inextensible.”
It therefore with such pertinacity odd defied
The force which the length of the chain should have modified,
That its stubborn example may possibly yet recall
These overgrown rhymes to their prosody metrical.
The condition is got by resolving again,
According to axes assumed in the plane.
If then you reduce to the tangent and normal,
You will find the equation more neat tho’ less formal.
The condition thus found after these preparations,
When duly combined with the former equations,
Will give you another, in which differentials
(When the chain forms a circle), become in essentials
No harder than those that we easily solve
In the time a T totum would take to revolve.

Now joyfully leaving ds to itself, a-
Ttend to the values of T and of a.
The chain undergoes a distorting convulsion,
Produced first at A by the force of impulsion.
In magnitude R, in direction tangential,
Equating this R to the form exponential,
Obtained for the tension when a is zero,
It will measure the tug, such a tug as the “hero
Plume-waving” experienced, tied to the chariot.
But when dragged by the heels his grim head could not carry aught,
So give a its due at the end of the chain,
And the tension ought there to be zero again.
From these two conditions we get three equations,
Which serve to determine the proper relations
Between the first impulse and each coefficient
In the form for the tension, and this is sufficient
To work out the problem, and then, if you choose,
You may turn it and twist it the Dons to amuse.

The Beaker Paradox

See this beaker? It contains 1 to 2 liters of water and 1 liter of wine. That means that the ratio of water to wine (call it r) is between 1 and 2. Thus there’s a 50 percent chance that r is between 1 and 3/2. Right?

But now consider the ratio of wine to water, or 1/r. That’s between 1/2 and 1, so there’s a 50 percent chance that 1/r is between 3/4 and 1.

Taking the reciprocal, that means there’s a 50 percent chance that r is between 1 and 4/3, which contradicts our earlier result. Where is the error?

The Bedford Level Experiment

http://books.google.com/books?id=oTUDAAAAQAAJ&pg=PA58&dq="earth+not+a+globe"&as_brr=1&ei=3DduSoqLIJDWygTo3PzqDg"

In 1838, Samuel Rowbotham waded into a drainage canal in Norfolk and sighted along its length with a telescope. Six miles away, an assistant held a flag three feet above the water. If the earth were round, its curvature should hide the flag from him. But he decided he could see it clearly. “It follows,” he wrote, “that the surface of standing water is not convex, and therefore that the Earth IS NOT A GLOBE!”

Rowbotham’s triumphant result stood until 1870, when naturalist, surveyor, and obvious crackpot Alfred Russel Wallace attempted to disprove the result. His endeavor ended only in a heated argument — and eventually a libel suit against the “planists.” (Round-earthers are clearly desperate men.)

In fairness, we must note that not all observations have agreed with Rowbotham’s. In 1896 a newspaper editor conducted a similar experiment in Illinois and discovered that the earth is concave. Clearly more work is needed.

High Roller

http://www.sxc.hu/photo/642737

Here’s a tip for your next craps game. You can find the odds of rolling any number with two dice by subtracting the number from 7, ignoring the sign, and subtracting the result from 6. The remainder is the number of chances out of 36 that the number will appear.

For example, there are (6 – (7 – 5)) = 4 chances in 36, or 1 chance in 9, that you’ll roll a 5.

Pythagoras Disproved

pythagoras disproved - 1

We’re told that, in any right triangle, a2 + b2 = c2. But consider:

pythagoras disproved - 2

In the figure above, the total length of the red line is 2(a/2) + 2(b/2), or a + b. And again:

pythagoras disproved - 3

Here the red line’s length is 4(a/4) + 4(b/4), which is still a + b.

With each iteration, the red line more closely approximates c, but its length remains a + b. At the limit, then, it seems, a + b = c. Was Pythagoras mistaken?

Two by Two

Here’s a curious way to multiply two numbers. Suppose we want to multiply 97 by 23. Write each at the head of a column. Now halve the first number successively, discarding remainders, until you reach 1, and double the second number correspondingly in its own column:

two by two - first image

Cross out each row that has an even number in the left column, and add the numbers that remain in the second column:

two by two - second image

That gives the right answer (97 × 23 = 2231). Why does it work?

Click for Answer

So Much for Entropy

This is rather amazing. Arrange a deck of cards in this order, top to bottom:

A♣, 8♥, 5♠, 4♦, J♣, 2♥, 9♠, 3♦, 7♣, Q♥, K♠, 6♦, 10♣,
A♥, 8♠, 5♦, 4♣, J♥, 2♠, 9♦, 3♣, 7♥, Q♠, K♦, 6♣, 10♥,
A♠, 8♦, 5♣, 4♥, J♠, 2♦, 9♣, 3♥, 7♠, Q♦, K♣, 6♥, 10♠,
A♦, 8♣, 5♥, 4♠, J♦, 2♣, 9♥, 3♠, 7♦, Q♣, K♥, 6♠, 10♦

Now:

  1. Cut the deck and complete the cut. Do this as many times as you like.
  2. Deal cards face down one at a time, stopping whenever you have a substantial pile.
  3. Riffle-shuffle the two packs back together again.

Despite all this, you’ll find that the resulting deck is made up of 13 successive quartets of four suits–and four consecutive straights, ace through king.

The reasons for this are fairly complex, so I’ll just call it magic. You’ll find a full analysis in Julian Havil’s Impossible? Surprising Solutions to Counterintuitive Conundrums (2008).

The Prisoners’ Paradox

Three condemned prisoners share a cell. A guard arrives and tells them that one has been pardoned.

“Which is it?” they ask.

“I can’t tell you that,” says the guard. “I can’t tell a prisoner his own fate.”

Prisoner A takes the guard aside. “Look,” he says. “Of the three of us, only one has been pardoned. That means that one of my cellmates is still sure to die. Give me his name. That way you’re not telling me my own fate, and you’re not identifying the pardoned man.”

The guard thinks about this and says, “Prisoner B is sure to die.”

Prisoner A rejoices that his own chance of survival has improved from 1/3 to 1/2. But how is this possible? The guard has given him no new information. Has he?

Narcissistic Numbers

Colin Rose makes mathematical works of art:

http://www.numq.com/pwn/
http://www.numq.com/pwn/
http://www.numq.com/pwn/
http://www.numq.com/pwn/

Many more at his website.

Overtime

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

Allen, Brown, and Carr run a shop. At least one of them must always be present to mind the shop, and when Allen leaves he always takes Brown with him.

This means that, if Carr were out, these statements would be true simultaneously:

  1. If Allen is out then Brown is in (to mind the shop).
  2. If Allen is out then Brown is out (because Allen always takes Brown with him).

This is a contradiction; they cannot be true together. Therefore, logically, Carr must always be in the shop.

(From Lewis Carroll.)

It Tolls for Thee

Suspend a metal coat hanger on a length of string, wrap the ends of the string around your index fingers, and insert your fingers in your ears. Now swing the hanger gently against the edge of table. What do you hear?

No Spin Zone

If the Earth did move at a tremendous speed, how could we keep a grip on it with our feet? We could walk only very, very slowly; and should find it slipping rapidly under our footsteps. Then, which way is it turning? If we walked in the direction of its tremendous speed, it would push us on terribly rapidly. But if we tried to walk against its revolving–? Either way we should be terribly giddy, and our digestive processes impossible.

– Margaret Missen, The Sun Goes Round the Earth, quoted in Patrick Moore, Can You Speak Venusian?, 1972

Unite and Conquer

fraction products

Frequent Flyer

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

In 1957 ornithologists marked a Manx shearwater, a migrating seabird, on Bardsey Island off Wales.

In April 2002 they discovered the same bird was still alive and still gamely flying to South America each winter.

In the intervening 45 years, they calculated, it had covered 5 million miles.

See also Longest Migration.

You’re Welcome

You’d pay $1,000 to witness my mastery of the black arts, wouldn’t you? Of course you would.

  1. Buy a brand-new deck of cards.
  2. Discard the jokers, cut the deck 13 times, and deal it into 13 piles.
  3. Now stand back … Ph’nglui mglw’nafh C’thulhu R’lyeh wgah’nagl fhtagn!
  4. Look at the cards. Presto! They have magically grouped themselves by value — all the aces are in one pile, kings in another, etc.

You owe me $1,000.

The Long and the Short of It

In July 1838, Charles Darwin was considering whether to propose to his cousin, Emma Wedgwood. Ever the rationalist, drew up a balance sheet:

darwin marriage balance sheet

At the bottom he wrote “Marry – Marry – Marry Q.E.D.” They were wed in January.

Finders’ Fees

http://commons.wikimedia.org/wiki/File:Knuth-check2.png

Donald Knuth is so revered among computer scientists that they won’t cash his checks.

Knuth offers a standard reward of $2.56 (one “hexadecimal dollar”) to the first finder of each error in his published books. Since 1981 he has written more than $20,000 in checks, but most of the recipients have simply framed them as points of pride.

“There’s one man who lives near Frankfurt who would probably have more than $1,000 if he cashed all the checks I’ve sent him,” Knuth said in an October 2001 lecture. “Even if everybody cashed their checks, it would still be more than worth it to me to know that my books are getting better.”

Epimenides Soused

Somebody had told me of a dealer in gin who, having had his attention roused to the enormous waste of liquor caused by the unsteady hands of drunkards, invented a counter which, through a simple set of contrivances, gathered into a common reservoir all the spillings that previously had run to waste. … It struck me, therefore, on reviewing this case, that the more the people drank, the more they would titubate, by which word it was that I expressed the reeling and stumbling of intoxication. … [T]he more they titubated, the more they would spill; and the more they spilt, the more, it is clear, they did not drink. … Yet, again, if they drank nothing worth speaking of, how could they titubate? Clearly they could not; and, not titubating, they could have had no reason for spilling, in which case they must have drunk the whole–that is, they must have drunk to the whole excess imputed, which doing, they were dead drunk, and must have titubated to extremity, which doing, they must have spilt nearly the whole. … ‘And so round again,’ as my lord the bishop pleasantly expresses it, in secula seculorum.

– Thomas de Quincey, Essays on Philosophical Writers, 1856

Math Notes

math notes

Hands Up?

Beginning poker players are often shown a table like this:

"poker frequencies - no wild cards

It’s straightforward enough, assigning a hierarchy to the hands based on the likelihood of their appearance. But a strange thing happens when wild cards are introduced. Suppose we add one wild joker:

poker frequencies - one wild joker

Now three of a kind is more likely (and thus less valuable) than two pair. Well, can we just reverse their places in the table? No, we can’t, because the wild card permits some players to reinterpret their hands. If you’re holding 6♠ 6♥ 7♣ 10♦ plus the joker, and we change the table, you’ll simply decide you’re holding two pair rather than three of a kind. So will everyone in your position. In fact, if we recalculate the odds with this expectation, we find that two pair has again become the more likely hand (13:1 vs. 34:1).

This can go on all day. Whenever a hand is declared “rare” it becomes popular — and thus not rare. The bottom line is that when wild cards are allowed, it becomes impossible to rank hands based on frequency.

From Julian Havil, Impossible?, 2008.