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Likely because MathML isn't widely supported, and has even been removed from Chrome.
MathJax works an absolute treat.
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MathJax works an absolute treat.
Except on Android where the math takes up more horizontal space than the layout engine thinks it should, causing overlap with text after the math.
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I think jordigh is saying that the method is not statistically sound, i.e. that it will not (necessarily) give accurate probability estimates. They're not criticizing the method simply for being statistical.
jordigh hasn't given support for that claim.
So I find his small probability, however tiny, that out of all possible integer tuples none of them are a counterexample to be utterly unconvincing and ultimately misguided.
To be clear: my complaint is that there is no way to turn this kind of argument into an actual proof. We could salvage other physicist arguments, but not this one. Probability zero on an infinite set cannot mean inexistence. And the rest of what he's doing, trying to determine that counterexamples must be rare, is "well duh, we knew that, because we've been looking for them."
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
Can you please read the site guidelines and follow them? https://news.ycombinator.com/newsguidelines.html
_DON'T DOWN VOTE JUST BECAUSE YOU CAN'T DO MATH_ The proof Fermat hinted to was about the difference between squares. All whole numbers taken to a power greater than two (n^3) can be represented as the difference between two whole squares (x^2 - y^2). These differences can then be shown as the sum of consecutive odd numbers: 2^3 = 3^2 - 1^2 = (1+3+5) - (1) = 8, 3^3 = 6^2 - 3^2 = (1+3+5+7+9+11) - (1+3+5) = 27, 4^3 = 1…
~~Lemma 1~~:
z^n can be written as a difference of squares:
z^n = x^2 - y^2 = (x+y)(x-y), leading to the system of equations
x + y = z^(n-1)
x - y = z
which may be solved for x and y:
x = (z^(n-1) + z)/2
y = x - z
QED.
~~Lemma 2~~:
Any number squared can be written as a sum of sequantial odd numbers starting at 1.
(n)^2 = (n-1)^2 + (2(n-1)+1) = \sum_{k=0}^{n-1}(2k + 1)
QED.
~~~Fermat's Last Theorem~~~:
z^n = x^n + y^n has no solutions for n>2, and positive z, x, y.
Without loss of generality, assume a > b, then rewrite z^n as a sum of sequential odd numbers starting at b:
z^n = a^2 - b^2 = \sum_{k=b}^{a-1}(2k+1)
x^n and y^n can similarly be written as a sum of sequenatal odd numbers:
x^n = c^2 - d^2 = \sum_{k=d}^{c-1}(2k+1)
y^n = e^2 - f^2 = \sum{k=f}^{e-1}(2k+1)
By substitution to the Theorem's equation:
\sum_{k=b}^{a-1}(2k+1) = \sum_{k=d}^{c-1}(2k+1) + \sum_{f}^{e-1}(2k+1)
Which is true if and only if there are no gaps in the bounds of summation on the right side, so d = b, c = f, and e = a. But then
a^2 - b^2 = (f^2 + (-a^2))^n + (b^2 + (-f^2))^n
and this is certainly not true by the binomial theorem. We've reached a contradiction so QED.
^^ This is where I'm stuck. I don't actually know why that would not be true by the binomial theorem? It seems like simple expansion should do it.
> Feynman concluded: “for my money Fermat’s theorem is true”. > "the main job of theoretical physics is to prove yourself wrong as soon as possible." Great example of the main difference between mathematicians and theoretical physicists . This reminds me of another magician, Enrico Fermi, who was also an extremely good mathematician but didn't pursue rigor or precision for the sake of it: 20% was good enough precisio…
> > Feynman concluded: “for my money Fermat’s theorem is true”. > > "the main job of theoretical physics is to prove yourself wrong as soon as possible." > Great example of the main difference between mathematicians and theoretical physicists. Actually, I'm not sure I agree: even before Wiles's proof, almost every mathematician would have been willing to wager, at least conversationally, on the truth of FLT; and math…
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> > Feynman concluded: “for my money Fermat’s theorem is true”. > > "the main job of theoretical physics is to prove yourself wrong as soon as possible." > Great example of the main difference between mathematicians and theoretical physicists. Actually, I'm not sure I agree: even before Wiles's proof, almost every mathematician would have been willing to wager, at least conversationally, on the truth of FLT; and math…
The difference lies in the fact that absolute rigor to assess truths is not as fundamental in theoretical physics as it is in mathematics. Uncertainty is accepted. Physics puts a premium on empirical results and intuition over the more formal treatments common in mathematics (many important results/tools are not mathematically well-defined e.g. Feynman path-integral in d > 1).
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The difference lies in the fact that absolute rigor to assess truths is not as fundamental in theoretical physics as it is in mathematics. Uncertainty is accepted. Physics puts a premium on empirical results and intuition over the more formal treatments common in mathematics (many important results/tools are not mathematically well-defined e.g. Feynman path-integral in d > 1).
Agreed! I didn't mean to claim that there isn't a difference, for there is a wide one; only that the two particular quotes chosen seemed (unlike most other things Feynman said!) not to illustrate them.
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It's completely arrogant to assume that because it hasn't been solved by "better" people that I couldn't solve it.
It looks like you edited this comment, but I'm serious. I'm trying to understand your proof, but I'm having trouble seeing what the steps are for higher powers than 3 or 4. I already know the proofs for then cases n=3 and n=4, but I can't see how what you say works in the case, say, n=5, or n=13. Seriously, can you walk us through the steps of why x^5+y^5=z^5 has no (non-trivial) solutions? And to be fair, there are…
p=2, n=5, x=1/2(p+1)p^((n-1)/2), y=1/2(p-1)p^((n-1)/2)
Try changing the values and playing with it until you understand that there will always be x^2 - y^2 for all p^n, n>2.
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It's completely arrogant to assume that because it hasn't been solved by "better" people that I couldn't solve it.
>about: Fuck you, hater. Oh, you're that guy.