At the Princeton graduate school, the physics department and the math department shared a common lounge, and every day at four o'clock we would have tea. It was a way of relaxing in the afternoon, in addition to imitating an English college. People would sit around playing Go, or discussing theorems. In those days topology was the big thing.
I still remember a guy sitting on the couch, thinking very hard, and another guy standing in front of him saying, "And therefore such-and-such is true.
"Why is that?" the guy on the couch asks.
"It's trivial! It's trivial!" the standing guy says, and he rapidly reels off a series of logical steps: "First you assume thus-and-so, then we have Kerchoff's this-and-that, then there's Waffenstoffer's Theorem, and we substitute this and construct that. Now you put the vector which goes around here and then thus-and-so . . ." The guy on the couch is struggling to understand all this stuff, which goes on at high speed for about fifteen minutes!
Finally the standing guy comes out the other end, and the guy on the couch says, "Yeah, yeah. It's trivial."
We physicists were laughing, trying to figure them out. We decided that "trivial" means "proved." So we joked with the mathematicians: "We have a new theorem -- that mathematicians can only prove trivial theorems, because every theorem that's proved is trivial."
The mathematicians didn't like that theorem, and I teased them about it. I said there are never any surprises -- that the mathematicians only prove things that are obvious.
From "Surely You're Joking, Mr. Feynman!": Adventures of a Curious Character
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In the first case, starting with p such that the highest power of 2 that divides p is an integer power of 2 (2^k for some integer k); then the highest power of 2 that divides p² is 2^2k; then the highest power of 2 that divides 2q² is also 2^2k; then the highest power of 2 that divides q is 2^(2k-1); therefore q must be a multiple of 2^(k-0.5); a noninteger power of 2.
This implies that there is a number 2^(0.5). It makes no claims as to whether or not this number is rational, or integer; it merely claims that such a number must exist. (Consider: if I had started instead with the equation x²-4=0, I would have ended up showing that a number of the form 4^(0.5) must exist - that number is rational, is indeed an integer).
Now, I think I can prove that an integer q which is a multiple of 2^(k-0.5) but which is not a multiple of 2^k, for integer k, does not exist; but I can only complete that proof by knowing in advance that 2^0.5 is irrational, so I can't use it to prove the irrationality of 2^0.5. I can easily prove that a rational number of the form 4^(k-0.5) for integer k does exist; indeed, an infinite number of such numbers exist (examples include 2, 8, 32).
No matter how forcefully that first passage conveys the irrationality of √2, it does not prove it.
The paragraph, of course, was talking about integer powers of 2 that divide p. As in, the largest number 2^k such that 2^k divides p and k is an integer.
The largest real power of 2 that divides p is, of course, p itself, as 2^log_2(p) = p.