Frank Morgan's Math Chat


June 21, 2001

A Mathematician at Heaven's Gate:

a play by Frank Morgan

"In my Father's house are many mansions. . . I go to prepare a place for you." - John 14:2

Scene I. An impeccable mathematician arrives at the Pearly Gates.


"Well, here I am."

"Oh yes, we'd been expecting you, but you're a little late."

"Oh, well see I had this idea for proving the Poincare conjecture before I died, and I had to check it out."


"It didn't work."


"Well, may I come in?"

"I have to apologize. We'd love to have you come in, but we're full."


"Every place taken."

"But isn't heaven infinite?"

"Oh yes, of course."

"Well then, how can you be full?"

"Well, we have infinitely many souls here."

"Oh, well, all you have to do is rearrange things a bit."

Blank stare.

"See, well, uh, could I talk to a mathematician?"

"Oh sorry, there aren't any here."

"Oh, I see. Well, souls do move around in heaven, don't they. They don't just stay in one place throughout eternity, do they?"

"No, of course not."

"Well then, I'm sure a place will open up. I'll just wait."

"Fine. Just go to the end of the line."

"Oh, how did I miss that. Boy, that's quite a line!"

"Yes, we've had many new arrivals."

"How long is it?"

"Oh, it's infinitely long."

"But you said it had an end."


"Oh right, that is possible. Well, we just need to rearrange the line and we can get everybody in eventually."

"You mean move some people who've been waiting longer behind others who arrived later?"


"That wouldn't be fair, would it?"



To himself: "Would it be possible to rearrange the line without moving anyone backwards? Of course the ones out past the first infinite, limit ordinal wouldn't care, since they'll never get in anyway. I wonder what the ordinal type of the line is. Wait a minute! The first uncountable ordinal and the answer to the Continuum Hypothesis/Conjecture is right out there in the line!

To the sentinel, breathlessly: "What's the first uncountable limit ordinal in the line. Is it the continuum?"

"Everything outside is countable [disdainfully], and not usually well-ordered."

He tries to peer inside: "Wow, it's spacious in there, probably infinite dimensional. What's that in that mirror? It's us outside! we're nothing."

"Just a tiny three-dimensional singularity."

He mutters to himself: "Probably a three-sphere." Immediately he gets all excited again. "Could it be an exotic three-sphere, a counterexample to the Poincare conjecture?"

To the sentinel: "I think maybe I'll look around out here a bit."

"As you like. Don't get lost. It's hard to find your way back here."

He mutters to himself: "Could that be a clue? Maybe it is exotic."

To be continued.

American Olympiad Scholar, eleventh grader Michael Hamburg, was selected for the most elegant solution on the USA Mathematical Olympiad. See the Clay Mathematics Institute webpage at

Old Challenge. If the party affiliation of each of the 100 US Senators were determined by flipping a fair coin (heads Republican, tails Democrat), what would be the chance of a 50-50 split? Is it surprising that a 50-50 split actually occurred this year?

Answer. The chance of a 50-50 split would be about 8%, so you might expect it to happen once in a while. Joe Conrad reports that the previous even split occurred in the 47th Congress (1881-1883), where each major party had 37 Senators and there were two others ( Before the 86th Congress (1959-1961) there were fewer than 100 seats, so the probability was higher. Indeed, as Raymond Hettinger points out, when there where only 13 new states (colonies), the chance would have been about 15.5%, so the Founding Fathers may have expected deadlock to occur more frequently. Consequently, the Vice-President's tie-breaking role might have had more significance. (The 1st Congress (1789-1791) actually had 18 pro-administration Senators and 8 anti-administration Senators.)

To compute the probability as 8%, Joshua Green figures that "the number of ways in which there can be 50 Democrats and 50 Republicans is the same as the number of ways in which we can choose 50 Senators (i.e., to make them Democrats and everyone else Republican). From the Binomial Theorem, we know that there are

100!/50!50! = 100891344545564193334812497256

ways to do this. As each Senator can be assigned one of 2 party affiliations, the total number of ways to assign affiliations to the entire Senate is

2 100 = 1267650600228229401496703205376.

Thus, the probability of having a 50-50 split is the quotient of these two big numbers, or about

.079589237387 = 7.9589237387 %."

New Challenge (Nick Chung). While my wife was away, I invited n married couples to a party last weekend. An odd number of the guests were friends (acquaintances) of mine. Indeed, everyone had an odd number of friends at the party. What is the smallest possible value of n?

MATH CHAT welcomes its new editor, Richard Hamilton, new Managing Editor of MAA Online. Hamilton succeeds Carol Baxter, who is now overseeing the Association Management Software conversion among other things. Math Chat extends to Baxter best wishes and thanks for her faithful support and friendship.

Copyright 2001, Frank Morgan.

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