Feynman on Fermat's Last Theorem (2016)

http://www.lbatalha.com/blog/feynman-on-fermats-last-theorem#email

by slbenfica • 9 years ago

97 28 9 years ago

lbatalha.com

chmaynard• 9 years ago

Apparently the author also hosts this interesting website: http://www.fermatslibrary.com

throwanem3• 9 years ago

the main job of theoretical physics is to prove yourself wrong as soon as possible.

The main job of practical engineering, too.

auggierose3• 9 years ago

Nah, I'd say the job of practical engineering is to provide reliable, correct solutions in a timely, in-budget manner.

If you are talking about theoretical engineering, I agree :D

throwanem3• 9 years ago

the job of practical engineering is to provide reliable, correct solutions in a timely, in-budget manner

Sure. And I'm sure you'll agree this job is easier to do in direct proportion to the celerity with which we eliminate impractical approaches.

faitswulff• 9 years ago

Feynman's approach seems very reminiscent of Jake Vanderplas's Statistics for Hackers talk[0] as opposed to the purely theoretical physicist approach that the author notes at the end.

[0] https://www.youtube.com/watch?v=Iq9DzN6mvYA

merricksb• 9 years ago

Previous HN discussion around the time of publication just over a year ago:

https://news.ycombinator.com/item?id=12018221

dang• 9 years ago

Also https://news.ycombinator.com/item?id=14355834.

Gregaros3• 9 years ago

Richard Feynman was probably one of the most talented physicists of the 20th century.

Starts off bold.

joombaga• 9 years ago

Do you disagree? Would it help for me to show you the probability is strictly non-zero?

Gregaros3• 9 years ago

Sarcasm. "Probably one of the most" is quite a few weasel words for somebody universally included in any list of significant physicists.

kgwgk• 9 years ago

One of how many? "One of the three most talented" would be a bold statement, "one of the 300 most talented" would be an understatement. (Edit: now it occurs to me that you might find that opening sentence an understatement?)

Gregaros3• 9 years ago

You missed my sarcasm. I did indeed think it was quite an understatement / poor writing to be so afraid to say "one of the best physicists of the 20th century".

jordigh3• 9 years ago

Sigh, this proof bugs me so much.

Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integrate this probability over the sum x^n + y^n, see that the probability of this being an nth power is also very low.

I guess this is close enough for government work, but it's so utterly fallacious. For example, the distance between n^2 and (n-1)^2 is 2n - 1. That "means" that the "probability" of N being a perfect square is about 1/(sqrt(2N - 1)). This probability also goes to zero in the limit as N goes to infinity. Not very quickly, but it does.

Does that mean that square numbers don't exist?

We have many examples of conjectures being disproved by very large counterexamples:

https://www.quora.com/What-is-an-example-of-a-conjecture-tha...

ykler• 9 years ago

You forgot to sum over N

jordigh3• 9 years ago

Yeah, okay, I messed up. Let's argue in a different way: there are approximately sqrt(N) squares between 1 and N, so let's call sqrt(N)/N = 1/sqrt(N) the "probability" that N itself is square. That also goes to zero as N goes to infinity. If we pick higher powers of N, it goes to zero much quicker.

todd8• 9 years ago

Interesting, and startling to me when I first learned of it, is the probabilistic method [1], now not infrequently used in combinatorics to prove the existence of a solution to some question without constructing it. One proves that the probability of existence is strictly greater than zero.

[1] https://www.amazon.com/Probabilistic-Method-Discrete-Mathema...

jordigh3• 9 years ago

Yeah, that's really different, though. It's essentially a way to simplify counting. But the "converse" of the probabilistic method doesn't work: zero probability doesn't mean inexistence in an infinite set.

pmiller2• 9 years ago

I have to agree that this argument is utterly unconvincing, even as a reason to believe FLT may be true intuitively. It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0. Indeed, until Wiles finally proved FLT, the relative lack of progress on Siegel's conjecture on the infinitude of regular primes would tend to lend support to the view that a counterexample might exist somewhere out there.

alimw• 9 years ago

I haven't read it closely, but it looks to me as if the calculation estimates the expected number of counterexamples rather than the "measure" of them (however you've chosen to define that).

thaumasiotes• 9 years ago

It simply dismisses with a wink and a nod the idea that there might be one single counterexample out there for one single exponent (or, a finite number of counterexamples for a finite number of exponents). In other words, it does nothing to argue against the set of counterexamples being of measure 0.

Those "other words" are substantially weaker than the statement before them. If x^n + y^n equaled z^n for every integer x, y, z, and n, the set of counterexamples to Fermat's last theorem would still be of measure 0. Nothing can argue against the set of counterexamples being of measure 0, because the entire problem space itself has measure 0, and the set of counterexamples is necessarily a subset of the problem space.

nshepperd2• 9 years ago

Compute an approximate distance between nth powers, interpret this as the probability of an integer being an nth power, integrate this probability over the sum x^n + y^n, see that the probability of this being an nth power is also very low.

You're mixing up "N is an nth power" with "there exists an N that is an nth power". The purpose over summing over all the integers is to obtain the latter from the former, and it's not necessarily very small even if the former is small.

kumarvvr• 9 years ago

Geniuses stand on the shoulders of giants and look further ahead, while I look at giants and assume a fetal position and cower in fear. Sigh !

jwilk• 9 years ago

Here's a more straight-forward (but equally hand-wavy) way to calculate the probability that N is a perfect power:

P(N) ≈ (number of perfect powers near N) / (size of the neighborhood)

≈ (ⁿ√(N + r) − ⁿ√N) / r, for some smallish r

≈ d/dN (ⁿ√N)

= ⁿ√N / nN

Feynman on Fermat's Last Theorem (2016)

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