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Feynman on Fermat's Last Theorem

lbatalha.com

31–40 of 113 posts

Re: Feynman on Fermat's Last Theorem

#31
Very cute argument, and very much in his style--he was famous, as the author notes, for heuristic arguments that weren't very formalizable but had a lot of beauty.

One story I heard is that a computer scientist tried to explain the P=NP? problem to him; Feymnan couldn't understand why this was a problem. It was obviously true that P != NP, what even needed proving?

Re: Feynman on Fermat's Last Theorem

#32
post #19

Earlier quoted context omitted.

Several of these don't work, as it's assumed you tackle smaller values in another way, before your -> infinity method kicks in, so most of these would be easily knocked off.

not sure I follow "the integers don't exist, because the percent of real numbers that are integers is effectively 0" is what the parent post is implying, which "works", in the sense that using this proof methodology "works"

But it's not the proof "methodology" given by Feynman. Fenyman's result also gives a moderate-sized probability for small N, then eliminates those through known mathematics.

Re: Feynman on Fermat's Last Theorem

#34
post #26
post #21

Earlier quoted context omitted.

The meaning here is: pick a positive integer N, what is the chance (aka probability) of it being an n-th power of another positive integer. And you are correct, this probability depends on N.

I don't get it. The probability must depend on how likely I am to select any specific integer, i.e. probability mass function of N. The probability cannot depend on the value of the random variable itself but could involve any parameters that define its distribution. I would consider something like the following a valid question: "Let N be a random integer between 0 and M-1 with uniform distribution. What is the prob…

Intuitively, you can think of N as a rough measure of the "ballpark" around where you are in the number line. As you move up to higher numbers, perfect powers get more and more scarce, so the probability of finding one by picking a number at random becomes lower.

Re: Feynman on Fermat's Last Theorem

#35
post #26
post #21

Earlier quoted context omitted.

The meaning here is: pick a positive integer N, what is the chance (aka probability) of it being an n-th power of another positive integer. And you are correct, this probability depends on N.

I don't get it. The probability must depend on how likely I am to select any specific integer, i.e. probability mass function of N. The probability cannot depend on the value of the random variable itself but could involve any parameters that define its distribution. I would consider something like the following a valid question: "Let N be a random integer between 0 and M-1 with uniform distribution. What is the prob…

One meaning is that summing the probabilities for N=1,2,3, ... and checking how many numbers actually meet the criteria yield the same result in the limit. More precisely, their ratio goes to 1 as N goes to infinity.

Edit: define f(N) as the number of numbers below N that match. Consider the function f(N)/N. We call a function of N the probability of N matching if it asymptomatically approaches f(N)/N.

Re: Feynman on Fermat's Last Theorem

#36
post #23

i have to say this never-ending fixation / worship of Feynman is a bit creepy

There's hardly any denying that Feynman was quite a character; he (sometimes spectacularly) failed to live up to our expectations as to how such a brilliant intellect ought to behave.

That is probably a significant reason for his enduring appeal; a streetwise, bongo-playing physicist with a talent for quips.

Re: Feynman on Fermat's Last Theorem

#37
post #18

Earlier quoted context omitted.

1 could be a prime, too, as for example Legendre, Lesbegue, Cayley (in the Encyclopædia Britannica), Kronecker, Hardy and Sagan stated at least once (see the URL I linked to earlier) One isn't typically called a prime for the same reason as mathematicians typically say 0^0 equals 1; it makes many theorems and proofs look better.

Hm. I'm a mathematician. I always thought it wasn't called prime because it breaks uniqueness for the fundamental theorem of arithmetic.

It makes it look pret29th!

Re: Feynman on Fermat's Last Theorem

#38

It's a nice exercise However Number Theory is a different beast altogether It has as much to do with "regular" math as English and Latin have in common, even though they are written with the same alphabet.

That's not exactly true. Density arguments similar to this are fairly common in number theory (even if the arguments aren't rigorous). Granted, one has to be careful with arguments like this, but they can often be useful.

Re: Feynman on Fermat's Last Theorem

#39
post #7
post #2

I wonder how many false conjectures could pass muster using this sort of probabilistic argument.

I guess every false conjecture can be made to pass it. The trick is to make the set of items searched in large enough. For example, to show that no elephants exist, start with the (infinite) set of all possible chromosome sets. The proportion of them that produces an elephant is zero. QED. Examples from mathematics: The number 42 does not exist (logic: pick an integer. The probability that it equals 42 is zero. QED)…

Whoa, hang on. Statistical arguments for unproven conjectures are bad, but this counterargument is as bad or worse, especially when you start talking about infinity. Just to address your first example:

> The number 42 does not exist (logic: pick an integer. The probability that it equals 42 is zero. QED)

I object! What is your probability distribution function over the integers? Your phrasing sort of implies a uniform distribution, but there is no such thing as a uniform distribution on an infinite set, and as soon as you pick a plausible pdf the argument stops working.

Re: Feynman on Fermat's Last Theorem

#40
post #26
post #21

Earlier quoted context omitted.

The meaning here is: pick a positive integer N, what is the chance (aka probability) of it being an n-th power of another positive integer. And you are correct, this probability depends on N.

I don't get it. The probability must depend on how likely I am to select any specific integer, i.e. probability mass function of N. The probability cannot depend on the value of the random variable itself but could involve any parameters that define its distribution. I would consider something like the following a valid question: "Let N be a random integer between 0 and M-1 with uniform distribution. What is the prob…

I think you should just think of it as the density and that's it.
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