Futility Closet

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

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

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?

Solution to Two by Two:

Essentially the technique converts the first factor into binary, multiplies each of its constituents by the second factor, and sums the results.

Imagine that each line is associated with a power of 2: the first line with 2⁰, the second with 2¹,and so on.

The business in the first column, halving the first factor successively and crossing out those lines with even numbers, effectively reduces the first factor to its binary constituents — here, the lines that remain are those associated with 2⁰, 2⁵, and 2⁶, and, sure enough, 2⁰ + 2⁵ + 2⁶ = 97.

Now we need to multiply each of those constituents by the second factor, 23. In other words, we want to find:

(23 × 2⁰) + (23 × 2⁵) + (23 × 2⁶)

That’s what’s accomplished in the second column. Doubling the second factor successively, as we’ve done there, is equivalent to multiplying it by 2⁰ in the first line, by 2¹ in the second, and so on. That is, 23 = 23 × 2⁰, 46 = 23 × 2¹, etc. And the lines that haven’t been crossed out are precisely the ones we want:

23 (= 23 × 2⁰)
736 (= 23 × 2⁵)
1472 (= 23 × 2⁶)

So if we add those values, we’ll get the product of the original two numbers, which is what we sought: 23 + 736 + 1472 = 2231.

Here’s essentially what we’ve done, from the top:

97 × 23
= (2⁰ + 2⁵ + 2⁶) × 23
= (23 × 2⁰) + (23 × 2⁵) + (23 × 2⁶)
= 23 + 736 + 1472
= 2231

It works with any pair of numbers.

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:

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

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

(Answer)

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).

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?

Colin Rose makes mathematical works of art:

Many more at his website.

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.)

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?

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