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Feynman on Fermat's Last Theorem

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Re: Feynman on Fermat's Last Theorem

#91

Post by jsprogrammer »

Chinjut wrote: post #70
jsprogrammer wrote: post #66

Earlier quoted context omitted.

If you imagine it as a logical statement, represented physically in writing: P ~P Then, I think it is clear the statement is not true. NP is trivially definitionally not equivalent (equal) to P.

NP is most certainly not "trivially definitionally not equivalent (equal) to P". (Do you think everyone puzzling over the question of whether P = NP is a moron? You are aware there is a million dollar prize for a proof one way or the other, yes?). What definition are you using for NP?

P=deterministic algorithms in polynomial time

NP=non-deterministic algorithms in polynomial time

See, it is a definitionally trivial logical negation.

That's the joke!

Edit: Also, why cannot I report/flag your post? Your personal malignment is uncalled for.

Re: Feynman on Fermat's Last Theorem

#92

Post by jsprogrammer »

eru wrote: post #71
jsprogrammer wrote: post #66

Earlier quoted context omitted.

If you imagine it as a logical statement, represented physically in writing: P ~P Then, I think it is clear the statement is not true. NP is trivially definitionally not equivalent (equal) to P.

Do you know what P and NP mean? What's your ~ supposed to mean? (You seem to be implying logical negation. But P is not meant to be a logical statement, thus can't be negated.) Plus what Chinjut said.

~ is a standard character for logical negation.

See my replies elsewhere.

Re: Feynman on Fermat's Last Theorem

#93

Post by tvural »

Chinjut wrote: post #76
tvural wrote: post #75

Earlier quoted context omitted.

Well, Feynman is assuming that x^n + y^n is as likely as any other number to be an nth power. He shows that under this assumption, FLT is very likely true. So if FLT is false, it probably wouldn't be due to some coincidental counterexample. There would have to be a mathematical structure forcing x^n + y^n to be an nth power in some cases. In your example, the mathematical structure forcing x^n to be an nth power is o…

Fair enough on your first paragraph. (Though what distinguishes mathematical facts which are "coincidence" from mathematical facts forced true by mathematical structure?) Actually, I'd say it is odd to find this sort of analysis to give great confidence about the results, not just because it ignores the possibility of structure, but also because it ignores the possibility of coincidence! After all, there are some thi…

I don't consider the example you gave to be a coincidence, because if you know the definition of e then you know why one of those numbers could line up really, really well with e.

I'm basically saying that Feynman's argument rules out seemingly random counterexamples like 27^5 + 84^5 + 110^5 + 133^5 = 144^5, which disproved the sum of powers conjecture.

Re: Feynman on Fermat's Last Theorem

#94

Post by fareesh »

Chinjut wrote: post #89
fareesh wrote: post #78

My math is passably college level, so most of these proofs go over my head. I've always thought of FLT as stating "You cannot make n-3 cubes of side z by stacking/tiling n-3 cubes of side x and y", at which point it becomes something of a tessellation problem in 3 dimensions, which would seem to me to be the intuitive way to go about proving this.

Sorry, I'm having difficulty following you. What do you mean by an "n-3 cube"? Or are you actually talking about n - 3 many different cubes?

My bad - I meant a^(n-3) many different cubes of side a plus b^(n-3) many different cubes of side b needing to cover the total volume occupied by c^(n-3) many different cubes of side c. In other words it is a way to try and find a proof by thinking in terms of shapes and using some kind of tesselation method.You could do the same in squares too I suppose.

Re: Feynman on Fermat's Last Theorem

#95

Post by Chinjut »

fareesh wrote: post #94
Chinjut wrote: post #89

Earlier quoted context omitted.

Sorry, I'm having difficulty following you. What do you mean by an "n-3 cube"? Or are you actually talking about n - 3 many different cubes?

My bad - I meant a^(n-3) many different cubes of side a plus b^(n-3) many different cubes of side b needing to cover the total volume occupied by c^(n-3) many different cubes of side c. In other words it is a way to try and find a proof by thinking in terms of shapes and using some kind of tesselation method.You could do the same in squares too I suppose.

Or just in lines... a^n many lines of length 1 + b^n many lines of length 1 to cover the total length of c^n many lines of length 1.

Or just in n-dimensional cubes from the start: 1 n-dimensional cube of side-length a + 1 n-dimensional cube of side-length b to cover the volume of 1 n-dimensional cube of side-length c.

Of course, both of these don't amount to much different than just saying a^n + b^n = c^n directly. :)

I don't suspect that decomposing it into a^(n - k) a^k + b^(n - k) b^k = c^(n - k) c^k for arbitrary k is particularly helpful, and, for reasons as illustrated above, I suspect the aid to intuition from thinking in terms of volume and tessellations geometrically isn't very great, but, I'm not an expert on Fermat's Last Theorem. (For all I know, something like this does get used in the proof...)

Re: Feynman on Fermat's Last Theorem

#96

Post by Chinjut »

jsprogrammer wrote: post #91
Chinjut wrote: post #70

Earlier quoted context omitted.

NP is most certainly not "trivially definitionally not equivalent (equal) to P". (Do you think everyone puzzling over the question of whether P = NP is a moron? You are aware there is a million dollar prize for a proof one way or the other, yes?). What definition are you using for NP?

P=deterministic algorithms in polynomial time NP=non-deterministic algorithms in polynomial time See, it is a definitionally trivial logical negation. That's the joke! Edit: Also, why cannot I report/flag your post? Your personal malignment is uncalled for.

It's true I did not understand you were joking, but I didn't personally malign you either. If you are referring to the word "moron", I didn't call you a moron; I was shocked that you thought P vs. NP was so trivial, and asked whether you therefore thought everyone puzzling over it was a moron.

Re: Feynman on Fermat's Last Theorem

#97

Post by Chinjut »

jsprogrammer wrote: post #90
k__ wrote: post #85

Earlier quoted context omitted.

I think this was a joke

Yes, we are talking about an anecdote from the book, "Surely You're Joking, Mr. Feynman!": Adventures of a Curious Character ? Why would Feynman not have published a full(er) proof?

I presume you mean the idea of physical examples in general comes up in "Surely You're Joking"; I can't find any discussion of P vs. NP (or Fermat's Last Theorem, for that matter) within it.

As for why Feynman didn't publish a full(er) proof: a full(er) proof of what? He didn't have anything near a proof of P != NP or of Fermat's Last Theorem... There was nothing of the sort to publish.

Re: Feynman on Fermat's Last Theorem

#98

Post by jsprogrammer »

Chinjut wrote: post #96
jsprogrammer wrote: post #91

Earlier quoted context omitted.

P=deterministic algorithms in polynomial time NP=non-deterministic algorithms in polynomial time See, it is a definitionally trivial logical negation. That's the joke! Edit: Also, why cannot I report/flag your post? Your personal malignment is uncalled for.

It's true I did not understand you were joking, but I didn't personally malign you either. If you are referring to the word "moron", I didn't call you a moron; I was shocked that you thought P vs. NP was so trivial, and asked whether you therefore thought everyone puzzling over it was a moron.

I didn't say you maligned me. You implied others were morons (or, at least that I would think that).

You should be directing your disbelief at Feynman's remains. He is the one that made the claim. I just explained the joke.

Anyway, it is still true that the physical model of P != NP is definitionally trivial.

Re: Feynman on Fermat's Last Theorem

#99

Post by jordigh »

tgb wrote: post #86
jordigh wrote: post #83

Earlier quoted context omitted.

Those are two separate events. I'm talking about the event n = 42, which makes sense for all [n]. You're talking about the event n >= 42, which is a very different event.

No, I'm not talking about integrating the probability that n >= 42. What's the chance that a random number in [n] is 42? It's 1/n, if n is at least 42. Sum that over all x in [n] and you get 1, i.e. the probability that there is a number in [n] that is 42.

You're still talking about different events. The event that there is 42 in [n] is different from the even that x = 42.

Re: Feynman on Fermat's Last Theorem

#100

Post by Chinjut »

tvural wrote: post #93
Chinjut wrote: post #76

Earlier quoted context omitted.

Fair enough on your first paragraph. (Though what distinguishes mathematical facts which are "coincidence" from mathematical facts forced true by mathematical structure?) Actually, I'd say it is odd to find this sort of analysis to give great confidence about the results, not just because it ignores the possibility of structure, but also because it ignores the possibility of coincidence! After all, there are some thi…

I don't consider the example you gave to be a coincidence, because if you know the definition of e then you know why one of those numbers could line up really, really well with e. I'm basically saying that Feynman's argument rules out seemingly random counterexamples like 27^5 + 84^5 + 110^5 + 133^5 = 144^5, which disproved the sum of powers conjecture.

It seems to Feynman's argument works via the opposite of ruling out coincidences of that sort. It shows "If nothing surprising happens, then we expect very nearly zero counterexamples to Fermat's Last Theorem"... But the "surprising" in "If nothing surprising happens" encompasses both surprising structure (things which are true for some deep, clean reason) AND surprising numeric coincidences (things which are true for no good reason, but just happen, surprisingly, to line up in a nice, unexpected way).
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