Why do mathematicians sometimes assume famous conjectures in their research?
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I will use a specific example, but I mean in general. I went to a number theory conference and I saw one thing that surprised me: Nearly half the talks began with "Assuming the generalized Riemann Hypothesis..." Almost always, the crux of their argument depended on this conjecture.
Why would mathematicians perform research assuming a conjecture? By definition, it is not known to be true yet. In the off-chance that it turns out to be false, wouldn't all of the papers that assumed the conjecture be invalidated? I may be answering my own question, but I speculate that:
There is such strong evidence in support of the particular conjecture (Riemann Hypothesis in particular) and lack of evidence against it, that it is "safe" to assume it.
It's not so much about result obtained, but the methods and techniques used to prove it. Perhaps by assuming the conjecture, in the case of the Riemann Hypothesis, it leads to development of new techniques in analytic number theory.
conjectures motivation research
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
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show 7 more comments
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I will use a specific example, but I mean in general. I went to a number theory conference and I saw one thing that surprised me: Nearly half the talks began with "Assuming the generalized Riemann Hypothesis..." Almost always, the crux of their argument depended on this conjecture.
Why would mathematicians perform research assuming a conjecture? By definition, it is not known to be true yet. In the off-chance that it turns out to be false, wouldn't all of the papers that assumed the conjecture be invalidated? I may be answering my own question, but I speculate that:
There is such strong evidence in support of the particular conjecture (Riemann Hypothesis in particular) and lack of evidence against it, that it is "safe" to assume it.
It's not so much about result obtained, but the methods and techniques used to prove it. Perhaps by assuming the conjecture, in the case of the Riemann Hypothesis, it leads to development of new techniques in analytic number theory.
conjectures motivation research
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
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3. It can always be that the research leads to a disproof of the conjecture. :)
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– darij grinberg
Apr 28 '14 at 0:31
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I'm not an expert, but using the Riemann Hypothesis as an example, there's strong evidence in support of it, and why wait for it to be proved when you can already build new results upon it?
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– qwr
Apr 28 '14 at 0:33
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Part of the reason they're famous conjectures is because we believe they are likely to be true.
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– Hurkyl
Apr 28 '14 at 5:32
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There is also research that leads to practical applications and algorithms that just work well in practice, even when (partly) based on unproved hypothesis. Who would not use an O(n) algorithm to factor numbers whos runtime assesment depends on the riemann hypothesis, just because of that?
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– PlasmaHH
Apr 28 '14 at 10:29
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It's an anecdote, but I remember some theorem was proven this way: "Assume RH is true [...] then $P$. Assume RH is not true [...] then $P$. Therefore $P$".
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– Najib Idrissi
Apr 28 '14 at 12:28
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show 7 more comments
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I will use a specific example, but I mean in general. I went to a number theory conference and I saw one thing that surprised me: Nearly half the talks began with "Assuming the generalized Riemann Hypothesis..." Almost always, the crux of their argument depended on this conjecture.
Why would mathematicians perform research assuming a conjecture? By definition, it is not known to be true yet. In the off-chance that it turns out to be false, wouldn't all of the papers that assumed the conjecture be invalidated? I may be answering my own question, but I speculate that:
There is such strong evidence in support of the particular conjecture (Riemann Hypothesis in particular) and lack of evidence against it, that it is "safe" to assume it.
It's not so much about result obtained, but the methods and techniques used to prove it. Perhaps by assuming the conjecture, in the case of the Riemann Hypothesis, it leads to development of new techniques in analytic number theory.
conjectures motivation research
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
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I will use a specific example, but I mean in general. I went to a number theory conference and I saw one thing that surprised me: Nearly half the talks began with "Assuming the generalized Riemann Hypothesis..." Almost always, the crux of their argument depended on this conjecture.
Why would mathematicians perform research assuming a conjecture? By definition, it is not known to be true yet. In the off-chance that it turns out to be false, wouldn't all of the papers that assumed the conjecture be invalidated? I may be answering my own question, but I speculate that:
There is such strong evidence in support of the particular conjecture (Riemann Hypothesis in particular) and lack of evidence against it, that it is "safe" to assume it.
It's not so much about result obtained, but the methods and techniques used to prove it. Perhaps by assuming the conjecture, in the case of the Riemann Hypothesis, it leads to development of new techniques in analytic number theory.
conjectures motivation research
conjectures motivation research
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
edited Nov 6 '15 at 0:38
user147263
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
asked Apr 28 '14 at 0:29
Joseph DiNataleJoseph DiNatale
1,3731126
1,3731126
57
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3. It can always be that the research leads to a disproof of the conjecture. :)
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– darij grinberg
Apr 28 '14 at 0:31
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I'm not an expert, but using the Riemann Hypothesis as an example, there's strong evidence in support of it, and why wait for it to be proved when you can already build new results upon it?
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– qwr
Apr 28 '14 at 0:33
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Part of the reason they're famous conjectures is because we believe they are likely to be true.
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– Hurkyl
Apr 28 '14 at 5:32
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There is also research that leads to practical applications and algorithms that just work well in practice, even when (partly) based on unproved hypothesis. Who would not use an O(n) algorithm to factor numbers whos runtime assesment depends on the riemann hypothesis, just because of that?
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– PlasmaHH
Apr 28 '14 at 10:29
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It's an anecdote, but I remember some theorem was proven this way: "Assume RH is true [...] then $P$. Assume RH is not true [...] then $P$. Therefore $P$".
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– Najib Idrissi
Apr 28 '14 at 12:28
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show 7 more comments
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3. It can always be that the research leads to a disproof of the conjecture. :)
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– darij grinberg
Apr 28 '14 at 0:31
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I'm not an expert, but using the Riemann Hypothesis as an example, there's strong evidence in support of it, and why wait for it to be proved when you can already build new results upon it?
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– qwr
Apr 28 '14 at 0:33
3
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Part of the reason they're famous conjectures is because we believe they are likely to be true.
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– Hurkyl
Apr 28 '14 at 5:32
8
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There is also research that leads to practical applications and algorithms that just work well in practice, even when (partly) based on unproved hypothesis. Who would not use an O(n) algorithm to factor numbers whos runtime assesment depends on the riemann hypothesis, just because of that?
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– PlasmaHH
Apr 28 '14 at 10:29
26
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It's an anecdote, but I remember some theorem was proven this way: "Assume RH is true [...] then $P$. Assume RH is not true [...] then $P$. Therefore $P$".
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– Najib Idrissi
Apr 28 '14 at 12:28
57
57
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3. It can always be that the research leads to a disproof of the conjecture. :)
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– darij grinberg
Apr 28 '14 at 0:31
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3. It can always be that the research leads to a disproof of the conjecture. :)
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– darij grinberg
Apr 28 '14 at 0:31
10
10
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I'm not an expert, but using the Riemann Hypothesis as an example, there's strong evidence in support of it, and why wait for it to be proved when you can already build new results upon it?
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– qwr
Apr 28 '14 at 0:33
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I'm not an expert, but using the Riemann Hypothesis as an example, there's strong evidence in support of it, and why wait for it to be proved when you can already build new results upon it?
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– qwr
Apr 28 '14 at 0:33
3
3
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Part of the reason they're famous conjectures is because we believe they are likely to be true.
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– Hurkyl
Apr 28 '14 at 5:32
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Part of the reason they're famous conjectures is because we believe they are likely to be true.
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– Hurkyl
Apr 28 '14 at 5:32
8
8
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There is also research that leads to practical applications and algorithms that just work well in practice, even when (partly) based on unproved hypothesis. Who would not use an O(n) algorithm to factor numbers whos runtime assesment depends on the riemann hypothesis, just because of that?
$endgroup$
– PlasmaHH
Apr 28 '14 at 10:29
$begingroup$
There is also research that leads to practical applications and algorithms that just work well in practice, even when (partly) based on unproved hypothesis. Who would not use an O(n) algorithm to factor numbers whos runtime assesment depends on the riemann hypothesis, just because of that?
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– PlasmaHH
Apr 28 '14 at 10:29
26
26
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It's an anecdote, but I remember some theorem was proven this way: "Assume RH is true [...] then $P$. Assume RH is not true [...] then $P$. Therefore $P$".
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– Najib Idrissi
Apr 28 '14 at 12:28
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It's an anecdote, but I remember some theorem was proven this way: "Assume RH is true [...] then $P$. Assume RH is not true [...] then $P$. Therefore $P$".
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– Najib Idrissi
Apr 28 '14 at 12:28
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show 7 more comments
9 Answers
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For four main reasons:
If the famous conjecture is proven true, the demonstrated results are proven true too.
If the reasoning is correct but demonstrated results are proven false, the famous conjecture is proven false too.
Others may be able to prove further results based on the demonstrated results which may themselves be proven false, thus again proving the famous conjecture false.
Showing interesting consequences of the famous conjecture and surprising connections into other areas of research generates interest in proving that the famous conjecture is true.
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
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5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
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– Daniel Mahler
Apr 28 '14 at 5:11
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@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
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– Steve Jessop
Apr 28 '14 at 9:17
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Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
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– Alex Feinman
Apr 28 '14 at 14:28
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@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
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– starsplusplus
Apr 28 '14 at 15:13
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If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
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– Alex Feinman
Apr 28 '14 at 17:48
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The results would not be invalidated but would be rendered vacuous, i.e. true but no longer informative.
A result says If the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
If the Riemann hypothesis ultimately is seen to be false, then it is still true that if the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
"Are all cell phones in the classroom turned off?", asks the instructor. If it happens that there are no cell phones in the classroom, then the correct answer is "yes". That's "vacuous truth". This is one example showing how the concept of vacuous truth can be quite practical. The "yes" answer would no longer be informative if it were learned that no cell phones are in the classroom. And if no cell phones are in the classroom, it is quite probable that no one even knows that. You would only know that you don't have a cell phone; you wouldn't know that about all your classmates.
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
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I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
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– fgp
Apr 28 '14 at 0:54
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Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
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– Michael Joyce
Apr 28 '14 at 1:23
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@fgp : It seems to me that your comment says the same thing that my posted answer says.
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– Michael Hardy
Apr 28 '14 at 2:36
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@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
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– Steve Jessop
Apr 28 '14 at 9:20
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. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
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– Michael Hardy
Apr 30 '14 at 18:11
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One possible reason for assuming a conjecture and generating results is if you don't believe the conjecture and hope to eventually shoot it down.
A great historical example of this is the parallel postulate. This states:
Given a line and a point not on that line, there is exactly one line passing through this point which is also parallel to the given line.
It was widely believed that the part of Euclidean geometry presented by Euclid before this parallel postulate implied the postulate. In other words, many mathematicians believed that the parallel postulate was a redundant axiom. For over a thousand years, many mathematicians battled with this "conjecture".
Some started assuming that the parallel postulate was false and tried to get a contradiction. These researchers were led into a "weird" world where there are infinitely many parallel lines through the given point and triangles' angles add up to less than $180^circ$.
It wasn't until the work of János Bolyai and Nikolai Lobachevsky (Gauss anticipated both of them but didn't publish his findings) that it started to become clear that the parallel postulate did not depend on the other axioms and that it's negation could lead to other perfectly equally consistent kinds of geometry (i.e. hyperbolic and spherical geometry).
That's research. If you don't know something is true or not. Assume it or assume its negation and see where it takes you. Sometimes this answers your original question. Sometimes you get an answer you weren't expecting!
answered Apr 28 '14 at 1:40
Bill CookBill Cook
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I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
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– 6005
Apr 28 '14 at 16:14
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To provide a bit of a counterpoint to the other answers, here is an example of a similar situation, which might provide a better picture, since the conjecture in question has turned out to be false (so we are better able to get to a conclusion).
The conjecture in question is the Lusztig conjecture, which provides a formula for the character of a simple module for a reductive algebraic group in terms of the characters of certain "standard" modules (I will not go into details of the conjecture itself, other than to note that it is a characteristic $p$ analogue of the Kazhdan-Lusztig conjecture, which is known to be true).
This conjecture has been expected to be true for at least 20 years, so quite a few people have been working on consequences of the conjecture. A further reason a lot of work has been done assuming the conjecture is that it is known to hold when the characteristic is large enough relative to the Coxeter number of the group (due to work of Andersen, Jantzen and Soergel), so it seemed like just a matter of lowering this bound.
More precisely, the conjecture states that a certain formula should hold when $pgeq 2h-2$ (or possible $pgeq h$). This was then disproven in 2013 by Williamson, who in fact showed that it is not possible to provide any linear bound on $p$ in terms of the Coxeter number, such that the formula holds. Also, it seems that probably not even a polynomial bound will be possible, though as far as I know this still relies on some number theoretic conjectures (such as the existence of infinitely many primes among the Finonacci numbers).
About the work that assumed the conjecture: A large part of this work has not been at all wasted, and in fact much of it played into the disproof of the conjecture, which goes via a long chain of reasoning starting from the conjecture and using a lot of the work that had been done assuming it.
Also, while the results themselves might turn out not to be true, the methods used by them may well turn out to still be applicable, since it seems like the conjecture will still be "close" to being true (ie, it might only fail for a small number of the simple modules). So now a large amount of effort is going into understanding more precisely how the conjecture fails.
This is not to say that there are not plenty of results that must be viewed in a slightly different light now, since any assumption of the conjecture will severely limit the power of the results.
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
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I can give an individual perspective...first, Rh is a very, very special case. Hilbert thought it might still be unresolved in 1,000 years.
Now, I wrote a paper with Irving Kaplansky and Alexander Schiemann. We found all possible, umm, widgets, there being 913 of them. We had no proofs that some 22 of them really were genuine widgets. Maybe about 2012, a graduate student named Oliver sent me a preprint about that. It has perhaps appeared by now. Anyway, it is pdf number 9 at OLIVER. I guess it was originally a chapter in his dissertation. I liked it; it says that GRH implies the remaining 14 things really are widgets. To me, it says i was not stupid; it will take a massive result to disprove the widgetness of those things.
Meanwhile, I have written three or four papers with Alex. At dinner at a conference in early 2013, he said if I mentioned results implied by RH again he would lose all respect for me. So, I don't bring it up any more.
EEDDIITT: most of the information anyone could want on the widgets are at TERNARY. They are regular ternary quadratic forms. The oldest example is
$$ f(x,y,z) = x^2 + y^2 + z^2. $$
Gauss showed that we can solve $n=x^2 + y^2 + z^2 $ as long as $n neq 4^k (8w + 7).$ This is often called the Three Square Theorem, which I think is clever. there are a total of 102 such "regular" forms in the list i called Dickson_Diagonal; these are of the type $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 leq a leq b leq c$ and $gcd(a,b,c) = 1.$ Dickson also wrote in the numbers not represented. Very handy.
One of the remaining unproven is
$$ f(x,y,z) = 3 x^2 + 6 y^2 + 14 z^2 + 4 y z + 2 z x + 2 x y. $$ This form, with integer arguments, is never equal to $4m+1, 16m+4, 16m+10, 64m+40, 4^k(16m+2) $ but appears to represent everything else, checked very high by computer. Maybe, maybe not, but a GRH implies it.
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
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Out of curiosity, for the more technical, what are the widgets?
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– Steven Stadnicki
Apr 28 '14 at 1:01
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@StevenStadnicki, I will put in something short
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– Will Jagy
Apr 28 '14 at 1:02
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Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
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– corsiKa
Apr 30 '14 at 19:52
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@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
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– Will Jagy
Dec 25 '18 at 17:23
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In general, mathematicians are interested in proving conjectures, but some of them, like RH, have (ahem) proven resistant to this stratagem. Again in general, a problem whose truth cannot be established is uninteresting, but sometimes, it acquires connections that inspire curiosity. Then, even in the absence of a proof, people will put a lot of effort into building intuition about the importance of such a problem, with the goal of expanding mathematics until it can contain a proof, or expanding their conception of mathematics until it can contain an assessment of the truth.
For example, legend has it that the attempts to prove Fermat's last theorem (famously dismissed by Gauss as being just such a problem that's boring but hard) resulted in the development of algebraic number theory. This is the classic example of the first goal: building a theory that eventually leads to a proof.
RH is an example of the second goal (though there has been plenty of the first going on as well). It appears to be at the center of a lot of phenomena but, since no proof is forthcoming, one way to understand the problem is to understand its consequences. There are also generalizations, all equally (or even more so) conjectural, that live at the heart of big theories and conjectural mathematical programs. The point is that this conceptual activity is at least as important to mathematics as an intellectual pursuit as concrete progress, because without a narrative to give meaning to the theorems we prove, they are just boring.
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
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The difference between a famous conjecture and an axiom is not that large. An axiom is just an even older and more famous assumption than a famous conjecture.
In mathematics, you are just expected to be very clear what your assumptions are. Other mathematicians may be using different assumptions; it does not matter very much whether we say that they are working in a different theory, or that they have different famous conjectures in antecedents of their theorems.
Philosophically, each "axiom candidate" also needs some justification. There must be a reason why people tend to believe it, apart from a historical tradition. A large part of justification both for and against each axiom candidate is in the consequences that it would have if accepted. So somebody needs to research those consequences first and that's why one would explicitly assume Riemann Hypothesis before it's clear how it fits into established theories (or into the reality of the world in which we live).
An assumption can eventually be found to have a particular relation to a "base theory" that people do not even care to mention in their theorems.
- The assumption may be a theorem.
- The assumption may be a negation of a theorem.
- Independence of the assumption on the base theory may be provable, assuming the base theory is consistent. (Example: Axiom of Choice over ZF.)
- Independence of the assumption on the base theory may be impossible to prove, assuming the base theory is consistent. (Example: Axiom of Determinacy over ZF. The reason is that ZF+AD can prove the consistency of ZF, so the claim follows by the Second Goedel Incompleteness Theorem.)
We don't know (yet) whether Riemann Hypothesis falls into any of these categories; but that's not preventing anyone from assuming that it is likely true and publishing its interesting consequences.
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
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This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
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– user1729
Apr 29 '14 at 13:34
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(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
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– user1729
Apr 29 '14 at 13:49
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@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
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– Trevor Wilson
Apr 29 '14 at 18:58
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@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
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– Jirka Hanika
Apr 29 '14 at 22:11
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@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
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– Jirka Hanika
Apr 30 '14 at 14:39
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You can't expect a mathematician to find proof of a conjecture directly. Assuming that a conjecture is true is a creative approach to understanding the conjecture. There is no pretty systematic way to solve a problem.
answered Apr 29 '14 at 13:31
toplel32toplel32
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Stated assumptions allows one to focus other areas of research without getting bogged down in proving the assumption.
By stating "Assuming the generalized Riemann Hypothesis.." at the beginning, they are not stating that the generalized Riemann Hypothesis is fact. They are merely informing the reader/viewer that this research is based on that idea and that the research might be invalidated if the assumption is shown to be false.
answered May 1 '14 at 15:36
GraxGrax
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9 Answers
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$begingroup$
For four main reasons:
If the famous conjecture is proven true, the demonstrated results are proven true too.
If the reasoning is correct but demonstrated results are proven false, the famous conjecture is proven false too.
Others may be able to prove further results based on the demonstrated results which may themselves be proven false, thus again proving the famous conjecture false.
Showing interesting consequences of the famous conjecture and surprising connections into other areas of research generates interest in proving that the famous conjecture is true.
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
$endgroup$
28
$begingroup$
5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
$endgroup$
– Daniel Mahler
Apr 28 '14 at 5:11
5
$begingroup$
@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:17
8
$begingroup$
Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
$endgroup$
– Alex Feinman
Apr 28 '14 at 14:28
9
$begingroup$
@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
$endgroup$
– starsplusplus
Apr 28 '14 at 15:13
3
$begingroup$
If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
$endgroup$
– Alex Feinman
Apr 28 '14 at 17:48
|
show 5 more comments
$begingroup$
For four main reasons:
If the famous conjecture is proven true, the demonstrated results are proven true too.
If the reasoning is correct but demonstrated results are proven false, the famous conjecture is proven false too.
Others may be able to prove further results based on the demonstrated results which may themselves be proven false, thus again proving the famous conjecture false.
Showing interesting consequences of the famous conjecture and surprising connections into other areas of research generates interest in proving that the famous conjecture is true.
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
$endgroup$
28
$begingroup$
5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
$endgroup$
– Daniel Mahler
Apr 28 '14 at 5:11
5
$begingroup$
@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:17
8
$begingroup$
Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
$endgroup$
– Alex Feinman
Apr 28 '14 at 14:28
9
$begingroup$
@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
$endgroup$
– starsplusplus
Apr 28 '14 at 15:13
3
$begingroup$
If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
$endgroup$
– Alex Feinman
Apr 28 '14 at 17:48
|
show 5 more comments
$begingroup$
For four main reasons:
If the famous conjecture is proven true, the demonstrated results are proven true too.
If the reasoning is correct but demonstrated results are proven false, the famous conjecture is proven false too.
Others may be able to prove further results based on the demonstrated results which may themselves be proven false, thus again proving the famous conjecture false.
Showing interesting consequences of the famous conjecture and surprising connections into other areas of research generates interest in proving that the famous conjecture is true.
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
$endgroup$
For four main reasons:
If the famous conjecture is proven true, the demonstrated results are proven true too.
If the reasoning is correct but demonstrated results are proven false, the famous conjecture is proven false too.
Others may be able to prove further results based on the demonstrated results which may themselves be proven false, thus again proving the famous conjecture false.
Showing interesting consequences of the famous conjecture and surprising connections into other areas of research generates interest in proving that the famous conjecture is true.
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
edited Jan 19 '16 at 8:05
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
answered Apr 28 '14 at 0:49
David SchwartzDavid Schwartz
1,57911012
1,57911012
28
$begingroup$
5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
$endgroup$
– Daniel Mahler
Apr 28 '14 at 5:11
5
$begingroup$
@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:17
8
$begingroup$
Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
$endgroup$
– Alex Feinman
Apr 28 '14 at 14:28
9
$begingroup$
@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
$endgroup$
– starsplusplus
Apr 28 '14 at 15:13
3
$begingroup$
If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
$endgroup$
– Alex Feinman
Apr 28 '14 at 17:48
|
show 5 more comments
28
$begingroup$
5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
$endgroup$
– Daniel Mahler
Apr 28 '14 at 5:11
5
$begingroup$
@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:17
8
$begingroup$
Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
$endgroup$
– Alex Feinman
Apr 28 '14 at 14:28
9
$begingroup$
@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
$endgroup$
– starsplusplus
Apr 28 '14 at 15:13
3
$begingroup$
If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
$endgroup$
– Alex Feinman
Apr 28 '14 at 17:48
28
28
$begingroup$
5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
$endgroup$
– Daniel Mahler
Apr 28 '14 at 5:11
$begingroup$
5. Conjectures become famous because mathematicians believe they are true. Mathematicians tend to be pretty smart. So the conditional proof provides circumstantial evidence for your conjecture until you find a real proof or a counter example.
$endgroup$
– Daniel Mahler
Apr 28 '14 at 5:11
5
5
$begingroup$
@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:17
$begingroup$
@DanielMahler: 5 is the least reputable though. Where that is the reason you're employing proof by academic reputation in place of real proof. I think it could not stand as the only reason for a particular paper, whereas any of David's four could stand as the only reason for a particular paper.
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:17
8
8
$begingroup$
Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
$endgroup$
– Alex Feinman
Apr 28 '14 at 14:28
$begingroup$
Assertion #2 above is dangerously incorrect...the results can be proven false for another reason.
$endgroup$
– Alex Feinman
Apr 28 '14 at 14:28
9
9
$begingroup$
@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
$endgroup$
– starsplusplus
Apr 28 '14 at 15:13
$begingroup$
@AlexFeinman Only if there is a mistake in the proof. If "if A then B" has been proven and you then find "not B", then it is certain that "not A". However - and I think this is what you're getting at - it is possible that there was a mistake in the second proof, meaning that "if A then B" was actually not true.
$endgroup$
– starsplusplus
Apr 28 '14 at 15:13
3
3
$begingroup$
If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
$endgroup$
– Alex Feinman
Apr 28 '14 at 17:48
$begingroup$
If you prove Z taking X and Y as your conjectures, it's hard to know whether X or Y was the false conjecture. That's what I meant.
$endgroup$
– Alex Feinman
Apr 28 '14 at 17:48
|
show 5 more comments
$begingroup$
The results would not be invalidated but would be rendered vacuous, i.e. true but no longer informative.
A result says If the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
If the Riemann hypothesis ultimately is seen to be false, then it is still true that if the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
"Are all cell phones in the classroom turned off?", asks the instructor. If it happens that there are no cell phones in the classroom, then the correct answer is "yes". That's "vacuous truth". This is one example showing how the concept of vacuous truth can be quite practical. The "yes" answer would no longer be informative if it were learned that no cell phones are in the classroom. And if no cell phones are in the classroom, it is quite probable that no one even knows that. You would only know that you don't have a cell phone; you wouldn't know that about all your classmates.
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
$endgroup$
1
$begingroup$
I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
$endgroup$
– fgp
Apr 28 '14 at 0:54
33
$begingroup$
Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
$endgroup$
– Michael Joyce
Apr 28 '14 at 1:23
2
$begingroup$
@fgp : It seems to me that your comment says the same thing that my posted answer says.
$endgroup$
– Michael Hardy
Apr 28 '14 at 2:36
19
$begingroup$
@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:20
3
$begingroup$
. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
$endgroup$
– Michael Hardy
Apr 30 '14 at 18:11
|
show 1 more comment
$begingroup$
The results would not be invalidated but would be rendered vacuous, i.e. true but no longer informative.
A result says If the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
If the Riemann hypothesis ultimately is seen to be false, then it is still true that if the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
"Are all cell phones in the classroom turned off?", asks the instructor. If it happens that there are no cell phones in the classroom, then the correct answer is "yes". That's "vacuous truth". This is one example showing how the concept of vacuous truth can be quite practical. The "yes" answer would no longer be informative if it were learned that no cell phones are in the classroom. And if no cell phones are in the classroom, it is quite probable that no one even knows that. You would only know that you don't have a cell phone; you wouldn't know that about all your classmates.
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
$endgroup$
1
$begingroup$
I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
$endgroup$
– fgp
Apr 28 '14 at 0:54
33
$begingroup$
Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
$endgroup$
– Michael Joyce
Apr 28 '14 at 1:23
2
$begingroup$
@fgp : It seems to me that your comment says the same thing that my posted answer says.
$endgroup$
– Michael Hardy
Apr 28 '14 at 2:36
19
$begingroup$
@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:20
3
$begingroup$
. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
$endgroup$
– Michael Hardy
Apr 30 '14 at 18:11
|
show 1 more comment
$begingroup$
The results would not be invalidated but would be rendered vacuous, i.e. true but no longer informative.
A result says If the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
If the Riemann hypothesis ultimately is seen to be false, then it is still true that if the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
"Are all cell phones in the classroom turned off?", asks the instructor. If it happens that there are no cell phones in the classroom, then the correct answer is "yes". That's "vacuous truth". This is one example showing how the concept of vacuous truth can be quite practical. The "yes" answer would no longer be informative if it were learned that no cell phones are in the classroom. And if no cell phones are in the classroom, it is quite probable that no one even knows that. You would only know that you don't have a cell phone; you wouldn't know that about all your classmates.
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
$endgroup$
The results would not be invalidated but would be rendered vacuous, i.e. true but no longer informative.
A result says If the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
If the Riemann hypothesis ultimately is seen to be false, then it is still true that if the Riemann hypothesis is true, then blah blah blah mumbo jumbo.
"Are all cell phones in the classroom turned off?", asks the instructor. If it happens that there are no cell phones in the classroom, then the correct answer is "yes". That's "vacuous truth". This is one example showing how the concept of vacuous truth can be quite practical. The "yes" answer would no longer be informative if it were learned that no cell phones are in the classroom. And if no cell phones are in the classroom, it is quite probable that no one even knows that. You would only know that you don't have a cell phone; you wouldn't know that about all your classmates.
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
answered Apr 28 '14 at 0:39
Michael HardyMichael Hardy
1
1
1
$begingroup$
I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
$endgroup$
– fgp
Apr 28 '14 at 0:54
33
$begingroup$
Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
$endgroup$
– Michael Joyce
Apr 28 '14 at 1:23
2
$begingroup$
@fgp : It seems to me that your comment says the same thing that my posted answer says.
$endgroup$
– Michael Hardy
Apr 28 '14 at 2:36
19
$begingroup$
@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:20
3
$begingroup$
. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
$endgroup$
– Michael Hardy
Apr 30 '14 at 18:11
|
show 1 more comment
1
$begingroup$
I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
$endgroup$
– fgp
Apr 28 '14 at 0:54
33
$begingroup$
Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
$endgroup$
– Michael Joyce
Apr 28 '14 at 1:23
2
$begingroup$
@fgp : It seems to me that your comment says the same thing that my posted answer says.
$endgroup$
– Michael Hardy
Apr 28 '14 at 2:36
19
$begingroup$
@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:20
3
$begingroup$
. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
$endgroup$
– Michael Hardy
Apr 30 '14 at 18:11
1
1
$begingroup$
I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
$endgroup$
– fgp
Apr 28 '14 at 0:54
$begingroup$
I don't disagree, but one should note that all statements of the form "if the Riemann hypothesis is true then ..." become trivially true should the Riemann hypothesis be disproven. If $T vdash lnot P$, then $T vdash P rightarrow varphi$ for every possible $varphi$, simply because $bot rightarrow varphi$ is a tautology for every $varphi$.
$endgroup$
– fgp
Apr 28 '14 at 0:54
33
33
$begingroup$
Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
$endgroup$
– Michael Joyce
Apr 28 '14 at 1:23
$begingroup$
Unless you're in a logic class. Then , if there were no cell phones on, the students would answer the question about cell phones, "I don't know", "I don't know", ..., "I don't know", "Yes, all the cell phones are turned off." :)
$endgroup$
– Michael Joyce
Apr 28 '14 at 1:23
2
2
$begingroup$
@fgp : It seems to me that your comment says the same thing that my posted answer says.
$endgroup$
– Michael Hardy
Apr 28 '14 at 2:36
$begingroup$
@fgp : It seems to me that your comment says the same thing that my posted answer says.
$endgroup$
– Michael Hardy
Apr 28 '14 at 2:36
19
19
$begingroup$
@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:20
$begingroup$
@MichaelJoyce: you ignore the possibility that one of the students has answered "I don't know" because they haven't checked their own phone at all. So, suppose that (without looking at it) each student places their own phone on their head...
$endgroup$
– Steve Jessop
Apr 28 '14 at 9:20
3
3
$begingroup$
. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
$endgroup$
– Michael Hardy
Apr 30 '14 at 18:11
$begingroup$
. . . anyway the cell phone example is an immensely better example than stuff about unicorns and about all mountains made of pure gold and so forth. Those examples make the concept of vacuous truth appear disconnected from reality and not practical. They fail to illustrate its value. So much so that their use in pedagogy borders on malpractice.
$endgroup$
– Michael Hardy
Apr 30 '14 at 18:11
|
show 1 more comment
$begingroup$
One possible reason for assuming a conjecture and generating results is if you don't believe the conjecture and hope to eventually shoot it down.
A great historical example of this is the parallel postulate. This states:
Given a line and a point not on that line, there is exactly one line passing through this point which is also parallel to the given line.
It was widely believed that the part of Euclidean geometry presented by Euclid before this parallel postulate implied the postulate. In other words, many mathematicians believed that the parallel postulate was a redundant axiom. For over a thousand years, many mathematicians battled with this "conjecture".
Some started assuming that the parallel postulate was false and tried to get a contradiction. These researchers were led into a "weird" world where there are infinitely many parallel lines through the given point and triangles' angles add up to less than $180^circ$.
It wasn't until the work of János Bolyai and Nikolai Lobachevsky (Gauss anticipated both of them but didn't publish his findings) that it started to become clear that the parallel postulate did not depend on the other axioms and that it's negation could lead to other perfectly equally consistent kinds of geometry (i.e. hyperbolic and spherical geometry).
That's research. If you don't know something is true or not. Assume it or assume its negation and see where it takes you. Sometimes this answers your original question. Sometimes you get an answer you weren't expecting!
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
$endgroup$
3
$begingroup$
I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
$endgroup$
– 6005
Apr 28 '14 at 16:14
add a comment |
$begingroup$
One possible reason for assuming a conjecture and generating results is if you don't believe the conjecture and hope to eventually shoot it down.
A great historical example of this is the parallel postulate. This states:
Given a line and a point not on that line, there is exactly one line passing through this point which is also parallel to the given line.
It was widely believed that the part of Euclidean geometry presented by Euclid before this parallel postulate implied the postulate. In other words, many mathematicians believed that the parallel postulate was a redundant axiom. For over a thousand years, many mathematicians battled with this "conjecture".
Some started assuming that the parallel postulate was false and tried to get a contradiction. These researchers were led into a "weird" world where there are infinitely many parallel lines through the given point and triangles' angles add up to less than $180^circ$.
It wasn't until the work of János Bolyai and Nikolai Lobachevsky (Gauss anticipated both of them but didn't publish his findings) that it started to become clear that the parallel postulate did not depend on the other axioms and that it's negation could lead to other perfectly equally consistent kinds of geometry (i.e. hyperbolic and spherical geometry).
That's research. If you don't know something is true or not. Assume it or assume its negation and see where it takes you. Sometimes this answers your original question. Sometimes you get an answer you weren't expecting!
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
$endgroup$
3
$begingroup$
I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
$endgroup$
– 6005
Apr 28 '14 at 16:14
add a comment |
$begingroup$
One possible reason for assuming a conjecture and generating results is if you don't believe the conjecture and hope to eventually shoot it down.
A great historical example of this is the parallel postulate. This states:
Given a line and a point not on that line, there is exactly one line passing through this point which is also parallel to the given line.
It was widely believed that the part of Euclidean geometry presented by Euclid before this parallel postulate implied the postulate. In other words, many mathematicians believed that the parallel postulate was a redundant axiom. For over a thousand years, many mathematicians battled with this "conjecture".
Some started assuming that the parallel postulate was false and tried to get a contradiction. These researchers were led into a "weird" world where there are infinitely many parallel lines through the given point and triangles' angles add up to less than $180^circ$.
It wasn't until the work of János Bolyai and Nikolai Lobachevsky (Gauss anticipated both of them but didn't publish his findings) that it started to become clear that the parallel postulate did not depend on the other axioms and that it's negation could lead to other perfectly equally consistent kinds of geometry (i.e. hyperbolic and spherical geometry).
That's research. If you don't know something is true or not. Assume it or assume its negation and see where it takes you. Sometimes this answers your original question. Sometimes you get an answer you weren't expecting!
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
$endgroup$
One possible reason for assuming a conjecture and generating results is if you don't believe the conjecture and hope to eventually shoot it down.
A great historical example of this is the parallel postulate. This states:
Given a line and a point not on that line, there is exactly one line passing through this point which is also parallel to the given line.
It was widely believed that the part of Euclidean geometry presented by Euclid before this parallel postulate implied the postulate. In other words, many mathematicians believed that the parallel postulate was a redundant axiom. For over a thousand years, many mathematicians battled with this "conjecture".
Some started assuming that the parallel postulate was false and tried to get a contradiction. These researchers were led into a "weird" world where there are infinitely many parallel lines through the given point and triangles' angles add up to less than $180^circ$.
It wasn't until the work of János Bolyai and Nikolai Lobachevsky (Gauss anticipated both of them but didn't publish his findings) that it started to become clear that the parallel postulate did not depend on the other axioms and that it's negation could lead to other perfectly equally consistent kinds of geometry (i.e. hyperbolic and spherical geometry).
That's research. If you don't know something is true or not. Assume it or assume its negation and see where it takes you. Sometimes this answers your original question. Sometimes you get an answer you weren't expecting!
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
answered Apr 28 '14 at 1:40
Bill CookBill Cook
23.1k4869
23.1k4869
3
$begingroup$
I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
$endgroup$
– 6005
Apr 28 '14 at 16:14
add a comment |
3
$begingroup$
I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
$endgroup$
– 6005
Apr 28 '14 at 16:14
3
3
$begingroup$
I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
$endgroup$
– 6005
Apr 28 '14 at 16:14
$begingroup$
I particularly like the example of the parallel postulate because it shows another reason: what if it turns out the result you are assuming is in fact unprovable one way or the other? Then you might as well assume it as an axiom.
$endgroup$
– 6005
Apr 28 '14 at 16:14
add a comment |
$begingroup$
To provide a bit of a counterpoint to the other answers, here is an example of a similar situation, which might provide a better picture, since the conjecture in question has turned out to be false (so we are better able to get to a conclusion).
The conjecture in question is the Lusztig conjecture, which provides a formula for the character of a simple module for a reductive algebraic group in terms of the characters of certain "standard" modules (I will not go into details of the conjecture itself, other than to note that it is a characteristic $p$ analogue of the Kazhdan-Lusztig conjecture, which is known to be true).
This conjecture has been expected to be true for at least 20 years, so quite a few people have been working on consequences of the conjecture. A further reason a lot of work has been done assuming the conjecture is that it is known to hold when the characteristic is large enough relative to the Coxeter number of the group (due to work of Andersen, Jantzen and Soergel), so it seemed like just a matter of lowering this bound.
More precisely, the conjecture states that a certain formula should hold when $pgeq 2h-2$ (or possible $pgeq h$). This was then disproven in 2013 by Williamson, who in fact showed that it is not possible to provide any linear bound on $p$ in terms of the Coxeter number, such that the formula holds. Also, it seems that probably not even a polynomial bound will be possible, though as far as I know this still relies on some number theoretic conjectures (such as the existence of infinitely many primes among the Finonacci numbers).
About the work that assumed the conjecture: A large part of this work has not been at all wasted, and in fact much of it played into the disproof of the conjecture, which goes via a long chain of reasoning starting from the conjecture and using a lot of the work that had been done assuming it.
Also, while the results themselves might turn out not to be true, the methods used by them may well turn out to still be applicable, since it seems like the conjecture will still be "close" to being true (ie, it might only fail for a small number of the simple modules). So now a large amount of effort is going into understanding more precisely how the conjecture fails.
This is not to say that there are not plenty of results that must be viewed in a slightly different light now, since any assumption of the conjecture will severely limit the power of the results.
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
$endgroup$
add a comment |
$begingroup$
To provide a bit of a counterpoint to the other answers, here is an example of a similar situation, which might provide a better picture, since the conjecture in question has turned out to be false (so we are better able to get to a conclusion).
The conjecture in question is the Lusztig conjecture, which provides a formula for the character of a simple module for a reductive algebraic group in terms of the characters of certain "standard" modules (I will not go into details of the conjecture itself, other than to note that it is a characteristic $p$ analogue of the Kazhdan-Lusztig conjecture, which is known to be true).
This conjecture has been expected to be true for at least 20 years, so quite a few people have been working on consequences of the conjecture. A further reason a lot of work has been done assuming the conjecture is that it is known to hold when the characteristic is large enough relative to the Coxeter number of the group (due to work of Andersen, Jantzen and Soergel), so it seemed like just a matter of lowering this bound.
More precisely, the conjecture states that a certain formula should hold when $pgeq 2h-2$ (or possible $pgeq h$). This was then disproven in 2013 by Williamson, who in fact showed that it is not possible to provide any linear bound on $p$ in terms of the Coxeter number, such that the formula holds. Also, it seems that probably not even a polynomial bound will be possible, though as far as I know this still relies on some number theoretic conjectures (such as the existence of infinitely many primes among the Finonacci numbers).
About the work that assumed the conjecture: A large part of this work has not been at all wasted, and in fact much of it played into the disproof of the conjecture, which goes via a long chain of reasoning starting from the conjecture and using a lot of the work that had been done assuming it.
Also, while the results themselves might turn out not to be true, the methods used by them may well turn out to still be applicable, since it seems like the conjecture will still be "close" to being true (ie, it might only fail for a small number of the simple modules). So now a large amount of effort is going into understanding more precisely how the conjecture fails.
This is not to say that there are not plenty of results that must be viewed in a slightly different light now, since any assumption of the conjecture will severely limit the power of the results.
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
$endgroup$
add a comment |
$begingroup$
To provide a bit of a counterpoint to the other answers, here is an example of a similar situation, which might provide a better picture, since the conjecture in question has turned out to be false (so we are better able to get to a conclusion).
The conjecture in question is the Lusztig conjecture, which provides a formula for the character of a simple module for a reductive algebraic group in terms of the characters of certain "standard" modules (I will not go into details of the conjecture itself, other than to note that it is a characteristic $p$ analogue of the Kazhdan-Lusztig conjecture, which is known to be true).
This conjecture has been expected to be true for at least 20 years, so quite a few people have been working on consequences of the conjecture. A further reason a lot of work has been done assuming the conjecture is that it is known to hold when the characteristic is large enough relative to the Coxeter number of the group (due to work of Andersen, Jantzen and Soergel), so it seemed like just a matter of lowering this bound.
More precisely, the conjecture states that a certain formula should hold when $pgeq 2h-2$ (or possible $pgeq h$). This was then disproven in 2013 by Williamson, who in fact showed that it is not possible to provide any linear bound on $p$ in terms of the Coxeter number, such that the formula holds. Also, it seems that probably not even a polynomial bound will be possible, though as far as I know this still relies on some number theoretic conjectures (such as the existence of infinitely many primes among the Finonacci numbers).
About the work that assumed the conjecture: A large part of this work has not been at all wasted, and in fact much of it played into the disproof of the conjecture, which goes via a long chain of reasoning starting from the conjecture and using a lot of the work that had been done assuming it.
Also, while the results themselves might turn out not to be true, the methods used by them may well turn out to still be applicable, since it seems like the conjecture will still be "close" to being true (ie, it might only fail for a small number of the simple modules). So now a large amount of effort is going into understanding more precisely how the conjecture fails.
This is not to say that there are not plenty of results that must be viewed in a slightly different light now, since any assumption of the conjecture will severely limit the power of the results.
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
$endgroup$
To provide a bit of a counterpoint to the other answers, here is an example of a similar situation, which might provide a better picture, since the conjecture in question has turned out to be false (so we are better able to get to a conclusion).
The conjecture in question is the Lusztig conjecture, which provides a formula for the character of a simple module for a reductive algebraic group in terms of the characters of certain "standard" modules (I will not go into details of the conjecture itself, other than to note that it is a characteristic $p$ analogue of the Kazhdan-Lusztig conjecture, which is known to be true).
This conjecture has been expected to be true for at least 20 years, so quite a few people have been working on consequences of the conjecture. A further reason a lot of work has been done assuming the conjecture is that it is known to hold when the characteristic is large enough relative to the Coxeter number of the group (due to work of Andersen, Jantzen and Soergel), so it seemed like just a matter of lowering this bound.
More precisely, the conjecture states that a certain formula should hold when $pgeq 2h-2$ (or possible $pgeq h$). This was then disproven in 2013 by Williamson, who in fact showed that it is not possible to provide any linear bound on $p$ in terms of the Coxeter number, such that the formula holds. Also, it seems that probably not even a polynomial bound will be possible, though as far as I know this still relies on some number theoretic conjectures (such as the existence of infinitely many primes among the Finonacci numbers).
About the work that assumed the conjecture: A large part of this work has not been at all wasted, and in fact much of it played into the disproof of the conjecture, which goes via a long chain of reasoning starting from the conjecture and using a lot of the work that had been done assuming it.
Also, while the results themselves might turn out not to be true, the methods used by them may well turn out to still be applicable, since it seems like the conjecture will still be "close" to being true (ie, it might only fail for a small number of the simple modules). So now a large amount of effort is going into understanding more precisely how the conjecture fails.
This is not to say that there are not plenty of results that must be viewed in a slightly different light now, since any assumption of the conjecture will severely limit the power of the results.
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
answered Apr 28 '14 at 9:44
Tobias KildetoftTobias Kildetoft
16.8k14274
16.8k14274
add a comment |
add a comment |
$begingroup$
I can give an individual perspective...first, Rh is a very, very special case. Hilbert thought it might still be unresolved in 1,000 years.
Now, I wrote a paper with Irving Kaplansky and Alexander Schiemann. We found all possible, umm, widgets, there being 913 of them. We had no proofs that some 22 of them really were genuine widgets. Maybe about 2012, a graduate student named Oliver sent me a preprint about that. It has perhaps appeared by now. Anyway, it is pdf number 9 at OLIVER. I guess it was originally a chapter in his dissertation. I liked it; it says that GRH implies the remaining 14 things really are widgets. To me, it says i was not stupid; it will take a massive result to disprove the widgetness of those things.
Meanwhile, I have written three or four papers with Alex. At dinner at a conference in early 2013, he said if I mentioned results implied by RH again he would lose all respect for me. So, I don't bring it up any more.
EEDDIITT: most of the information anyone could want on the widgets are at TERNARY. They are regular ternary quadratic forms. The oldest example is
$$ f(x,y,z) = x^2 + y^2 + z^2. $$
Gauss showed that we can solve $n=x^2 + y^2 + z^2 $ as long as $n neq 4^k (8w + 7).$ This is often called the Three Square Theorem, which I think is clever. there are a total of 102 such "regular" forms in the list i called Dickson_Diagonal; these are of the type $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 leq a leq b leq c$ and $gcd(a,b,c) = 1.$ Dickson also wrote in the numbers not represented. Very handy.
One of the remaining unproven is
$$ f(x,y,z) = 3 x^2 + 6 y^2 + 14 z^2 + 4 y z + 2 z x + 2 x y. $$ This form, with integer arguments, is never equal to $4m+1, 16m+4, 16m+10, 64m+40, 4^k(16m+2) $ but appears to represent everything else, checked very high by computer. Maybe, maybe not, but a GRH implies it.
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
$endgroup$
2
$begingroup$
Out of curiosity, for the more technical, what are the widgets?
$endgroup$
– Steven Stadnicki
Apr 28 '14 at 1:01
$begingroup$
@StevenStadnicki, I will put in something short
$endgroup$
– Will Jagy
Apr 28 '14 at 1:02
1
$begingroup$
Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
$endgroup$
– corsiKa
Apr 30 '14 at 19:52
1
$begingroup$
@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
$endgroup$
– Will Jagy
Dec 25 '18 at 17:23
add a comment |
$begingroup$
I can give an individual perspective...first, Rh is a very, very special case. Hilbert thought it might still be unresolved in 1,000 years.
Now, I wrote a paper with Irving Kaplansky and Alexander Schiemann. We found all possible, umm, widgets, there being 913 of them. We had no proofs that some 22 of them really were genuine widgets. Maybe about 2012, a graduate student named Oliver sent me a preprint about that. It has perhaps appeared by now. Anyway, it is pdf number 9 at OLIVER. I guess it was originally a chapter in his dissertation. I liked it; it says that GRH implies the remaining 14 things really are widgets. To me, it says i was not stupid; it will take a massive result to disprove the widgetness of those things.
Meanwhile, I have written three or four papers with Alex. At dinner at a conference in early 2013, he said if I mentioned results implied by RH again he would lose all respect for me. So, I don't bring it up any more.
EEDDIITT: most of the information anyone could want on the widgets are at TERNARY. They are regular ternary quadratic forms. The oldest example is
$$ f(x,y,z) = x^2 + y^2 + z^2. $$
Gauss showed that we can solve $n=x^2 + y^2 + z^2 $ as long as $n neq 4^k (8w + 7).$ This is often called the Three Square Theorem, which I think is clever. there are a total of 102 such "regular" forms in the list i called Dickson_Diagonal; these are of the type $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 leq a leq b leq c$ and $gcd(a,b,c) = 1.$ Dickson also wrote in the numbers not represented. Very handy.
One of the remaining unproven is
$$ f(x,y,z) = 3 x^2 + 6 y^2 + 14 z^2 + 4 y z + 2 z x + 2 x y. $$ This form, with integer arguments, is never equal to $4m+1, 16m+4, 16m+10, 64m+40, 4^k(16m+2) $ but appears to represent everything else, checked very high by computer. Maybe, maybe not, but a GRH implies it.
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
$endgroup$
2
$begingroup$
Out of curiosity, for the more technical, what are the widgets?
$endgroup$
– Steven Stadnicki
Apr 28 '14 at 1:01
$begingroup$
@StevenStadnicki, I will put in something short
$endgroup$
– Will Jagy
Apr 28 '14 at 1:02
1
$begingroup$
Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
$endgroup$
– corsiKa
Apr 30 '14 at 19:52
1
$begingroup$
@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
$endgroup$
– Will Jagy
Dec 25 '18 at 17:23
add a comment |
$begingroup$
I can give an individual perspective...first, Rh is a very, very special case. Hilbert thought it might still be unresolved in 1,000 years.
Now, I wrote a paper with Irving Kaplansky and Alexander Schiemann. We found all possible, umm, widgets, there being 913 of them. We had no proofs that some 22 of them really were genuine widgets. Maybe about 2012, a graduate student named Oliver sent me a preprint about that. It has perhaps appeared by now. Anyway, it is pdf number 9 at OLIVER. I guess it was originally a chapter in his dissertation. I liked it; it says that GRH implies the remaining 14 things really are widgets. To me, it says i was not stupid; it will take a massive result to disprove the widgetness of those things.
Meanwhile, I have written three or four papers with Alex. At dinner at a conference in early 2013, he said if I mentioned results implied by RH again he would lose all respect for me. So, I don't bring it up any more.
EEDDIITT: most of the information anyone could want on the widgets are at TERNARY. They are regular ternary quadratic forms. The oldest example is
$$ f(x,y,z) = x^2 + y^2 + z^2. $$
Gauss showed that we can solve $n=x^2 + y^2 + z^2 $ as long as $n neq 4^k (8w + 7).$ This is often called the Three Square Theorem, which I think is clever. there are a total of 102 such "regular" forms in the list i called Dickson_Diagonal; these are of the type $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 leq a leq b leq c$ and $gcd(a,b,c) = 1.$ Dickson also wrote in the numbers not represented. Very handy.
One of the remaining unproven is
$$ f(x,y,z) = 3 x^2 + 6 y^2 + 14 z^2 + 4 y z + 2 z x + 2 x y. $$ This form, with integer arguments, is never equal to $4m+1, 16m+4, 16m+10, 64m+40, 4^k(16m+2) $ but appears to represent everything else, checked very high by computer. Maybe, maybe not, but a GRH implies it.
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
$endgroup$
I can give an individual perspective...first, Rh is a very, very special case. Hilbert thought it might still be unresolved in 1,000 years.
Now, I wrote a paper with Irving Kaplansky and Alexander Schiemann. We found all possible, umm, widgets, there being 913 of them. We had no proofs that some 22 of them really were genuine widgets. Maybe about 2012, a graduate student named Oliver sent me a preprint about that. It has perhaps appeared by now. Anyway, it is pdf number 9 at OLIVER. I guess it was originally a chapter in his dissertation. I liked it; it says that GRH implies the remaining 14 things really are widgets. To me, it says i was not stupid; it will take a massive result to disprove the widgetness of those things.
Meanwhile, I have written three or four papers with Alex. At dinner at a conference in early 2013, he said if I mentioned results implied by RH again he would lose all respect for me. So, I don't bring it up any more.
EEDDIITT: most of the information anyone could want on the widgets are at TERNARY. They are regular ternary quadratic forms. The oldest example is
$$ f(x,y,z) = x^2 + y^2 + z^2. $$
Gauss showed that we can solve $n=x^2 + y^2 + z^2 $ as long as $n neq 4^k (8w + 7).$ This is often called the Three Square Theorem, which I think is clever. there are a total of 102 such "regular" forms in the list i called Dickson_Diagonal; these are of the type $f(x,y,z) = a x^2 + b y^2 + c z^2$ with $1 leq a leq b leq c$ and $gcd(a,b,c) = 1.$ Dickson also wrote in the numbers not represented. Very handy.
One of the remaining unproven is
$$ f(x,y,z) = 3 x^2 + 6 y^2 + 14 z^2 + 4 y z + 2 z x + 2 x y. $$ This form, with integer arguments, is never equal to $4m+1, 16m+4, 16m+10, 64m+40, 4^k(16m+2) $ but appears to represent everything else, checked very high by computer. Maybe, maybe not, but a GRH implies it.
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
edited Dec 25 '18 at 17:20
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
answered Apr 28 '14 at 0:49
Will JagyWill Jagy
103k5102200
103k5102200
2
$begingroup$
Out of curiosity, for the more technical, what are the widgets?
$endgroup$
– Steven Stadnicki
Apr 28 '14 at 1:01
$begingroup$
@StevenStadnicki, I will put in something short
$endgroup$
– Will Jagy
Apr 28 '14 at 1:02
1
$begingroup$
Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
$endgroup$
– corsiKa
Apr 30 '14 at 19:52
1
$begingroup$
@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
$endgroup$
– Will Jagy
Dec 25 '18 at 17:23
add a comment |
2
$begingroup$
Out of curiosity, for the more technical, what are the widgets?
$endgroup$
– Steven Stadnicki
Apr 28 '14 at 1:01
$begingroup$
@StevenStadnicki, I will put in something short
$endgroup$
– Will Jagy
Apr 28 '14 at 1:02
1
$begingroup$
Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
$endgroup$
– corsiKa
Apr 30 '14 at 19:52
1
$begingroup$
@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
$endgroup$
– Will Jagy
Dec 25 '18 at 17:23
2
2
$begingroup$
Out of curiosity, for the more technical, what are the widgets?
$endgroup$
– Steven Stadnicki
Apr 28 '14 at 1:01
$begingroup$
Out of curiosity, for the more technical, what are the widgets?
$endgroup$
– Steven Stadnicki
Apr 28 '14 at 1:01
$begingroup$
@StevenStadnicki, I will put in something short
$endgroup$
– Will Jagy
Apr 28 '14 at 1:02
$begingroup$
@StevenStadnicki, I will put in something short
$endgroup$
– Will Jagy
Apr 28 '14 at 1:02
1
1
$begingroup$
Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
$endgroup$
– corsiKa
Apr 30 '14 at 19:52
$begingroup$
Maybe I need to be more math literate... But I like the word "widgetness" it made me chuckle.
$endgroup$
– corsiKa
Apr 30 '14 at 19:52
1
1
$begingroup$
@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
$endgroup$
– Will Jagy
Dec 25 '18 at 17:23
$begingroup$
@CarstenS I updated the link to my ternary page. Lemke-Oliver's article has surely appeared by now. If I find a link to a preprint later I will adjust that one also.
$endgroup$
– Will Jagy
Dec 25 '18 at 17:23
add a comment |
$begingroup$
In general, mathematicians are interested in proving conjectures, but some of them, like RH, have (ahem) proven resistant to this stratagem. Again in general, a problem whose truth cannot be established is uninteresting, but sometimes, it acquires connections that inspire curiosity. Then, even in the absence of a proof, people will put a lot of effort into building intuition about the importance of such a problem, with the goal of expanding mathematics until it can contain a proof, or expanding their conception of mathematics until it can contain an assessment of the truth.
For example, legend has it that the attempts to prove Fermat's last theorem (famously dismissed by Gauss as being just such a problem that's boring but hard) resulted in the development of algebraic number theory. This is the classic example of the first goal: building a theory that eventually leads to a proof.
RH is an example of the second goal (though there has been plenty of the first going on as well). It appears to be at the center of a lot of phenomena but, since no proof is forthcoming, one way to understand the problem is to understand its consequences. There are also generalizations, all equally (or even more so) conjectural, that live at the heart of big theories and conjectural mathematical programs. The point is that this conceptual activity is at least as important to mathematics as an intellectual pursuit as concrete progress, because without a narrative to give meaning to the theorems we prove, they are just boring.
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
$endgroup$
add a comment |
$begingroup$
In general, mathematicians are interested in proving conjectures, but some of them, like RH, have (ahem) proven resistant to this stratagem. Again in general, a problem whose truth cannot be established is uninteresting, but sometimes, it acquires connections that inspire curiosity. Then, even in the absence of a proof, people will put a lot of effort into building intuition about the importance of such a problem, with the goal of expanding mathematics until it can contain a proof, or expanding their conception of mathematics until it can contain an assessment of the truth.
For example, legend has it that the attempts to prove Fermat's last theorem (famously dismissed by Gauss as being just such a problem that's boring but hard) resulted in the development of algebraic number theory. This is the classic example of the first goal: building a theory that eventually leads to a proof.
RH is an example of the second goal (though there has been plenty of the first going on as well). It appears to be at the center of a lot of phenomena but, since no proof is forthcoming, one way to understand the problem is to understand its consequences. There are also generalizations, all equally (or even more so) conjectural, that live at the heart of big theories and conjectural mathematical programs. The point is that this conceptual activity is at least as important to mathematics as an intellectual pursuit as concrete progress, because without a narrative to give meaning to the theorems we prove, they are just boring.
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
$endgroup$
add a comment |
$begingroup$
In general, mathematicians are interested in proving conjectures, but some of them, like RH, have (ahem) proven resistant to this stratagem. Again in general, a problem whose truth cannot be established is uninteresting, but sometimes, it acquires connections that inspire curiosity. Then, even in the absence of a proof, people will put a lot of effort into building intuition about the importance of such a problem, with the goal of expanding mathematics until it can contain a proof, or expanding their conception of mathematics until it can contain an assessment of the truth.
For example, legend has it that the attempts to prove Fermat's last theorem (famously dismissed by Gauss as being just such a problem that's boring but hard) resulted in the development of algebraic number theory. This is the classic example of the first goal: building a theory that eventually leads to a proof.
RH is an example of the second goal (though there has been plenty of the first going on as well). It appears to be at the center of a lot of phenomena but, since no proof is forthcoming, one way to understand the problem is to understand its consequences. There are also generalizations, all equally (or even more so) conjectural, that live at the heart of big theories and conjectural mathematical programs. The point is that this conceptual activity is at least as important to mathematics as an intellectual pursuit as concrete progress, because without a narrative to give meaning to the theorems we prove, they are just boring.
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
$endgroup$
In general, mathematicians are interested in proving conjectures, but some of them, like RH, have (ahem) proven resistant to this stratagem. Again in general, a problem whose truth cannot be established is uninteresting, but sometimes, it acquires connections that inspire curiosity. Then, even in the absence of a proof, people will put a lot of effort into building intuition about the importance of such a problem, with the goal of expanding mathematics until it can contain a proof, or expanding their conception of mathematics until it can contain an assessment of the truth.
For example, legend has it that the attempts to prove Fermat's last theorem (famously dismissed by Gauss as being just such a problem that's boring but hard) resulted in the development of algebraic number theory. This is the classic example of the first goal: building a theory that eventually leads to a proof.
RH is an example of the second goal (though there has been plenty of the first going on as well). It appears to be at the center of a lot of phenomena but, since no proof is forthcoming, one way to understand the problem is to understand its consequences. There are also generalizations, all equally (or even more so) conjectural, that live at the heart of big theories and conjectural mathematical programs. The point is that this conceptual activity is at least as important to mathematics as an intellectual pursuit as concrete progress, because without a narrative to give meaning to the theorems we prove, they are just boring.
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
answered Apr 28 '14 at 4:48
Ryan ReichRyan Reich
5,4011627
5,4011627
add a comment |
add a comment |
$begingroup$
The difference between a famous conjecture and an axiom is not that large. An axiom is just an even older and more famous assumption than a famous conjecture.
In mathematics, you are just expected to be very clear what your assumptions are. Other mathematicians may be using different assumptions; it does not matter very much whether we say that they are working in a different theory, or that they have different famous conjectures in antecedents of their theorems.
Philosophically, each "axiom candidate" also needs some justification. There must be a reason why people tend to believe it, apart from a historical tradition. A large part of justification both for and against each axiom candidate is in the consequences that it would have if accepted. So somebody needs to research those consequences first and that's why one would explicitly assume Riemann Hypothesis before it's clear how it fits into established theories (or into the reality of the world in which we live).
An assumption can eventually be found to have a particular relation to a "base theory" that people do not even care to mention in their theorems.
- The assumption may be a theorem.
- The assumption may be a negation of a theorem.
- Independence of the assumption on the base theory may be provable, assuming the base theory is consistent. (Example: Axiom of Choice over ZF.)
- Independence of the assumption on the base theory may be impossible to prove, assuming the base theory is consistent. (Example: Axiom of Determinacy over ZF. The reason is that ZF+AD can prove the consistency of ZF, so the claim follows by the Second Goedel Incompleteness Theorem.)
We don't know (yet) whether Riemann Hypothesis falls into any of these categories; but that's not preventing anyone from assuming that it is likely true and publishing its interesting consequences.
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
$endgroup$
2
$begingroup$
This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
$endgroup$
– user1729
Apr 29 '14 at 13:34
$begingroup$
(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
$endgroup$
– user1729
Apr 29 '14 at 13:49
1
$begingroup$
@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
$endgroup$
– Trevor Wilson
Apr 29 '14 at 18:58
1
$begingroup$
@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
$endgroup$
– Jirka Hanika
Apr 29 '14 at 22:11
2
$begingroup$
@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
$endgroup$
– Jirka Hanika
Apr 30 '14 at 14:39
|
show 6 more comments
$begingroup$
The difference between a famous conjecture and an axiom is not that large. An axiom is just an even older and more famous assumption than a famous conjecture.
In mathematics, you are just expected to be very clear what your assumptions are. Other mathematicians may be using different assumptions; it does not matter very much whether we say that they are working in a different theory, or that they have different famous conjectures in antecedents of their theorems.
Philosophically, each "axiom candidate" also needs some justification. There must be a reason why people tend to believe it, apart from a historical tradition. A large part of justification both for and against each axiom candidate is in the consequences that it would have if accepted. So somebody needs to research those consequences first and that's why one would explicitly assume Riemann Hypothesis before it's clear how it fits into established theories (or into the reality of the world in which we live).
An assumption can eventually be found to have a particular relation to a "base theory" that people do not even care to mention in their theorems.
- The assumption may be a theorem.
- The assumption may be a negation of a theorem.
- Independence of the assumption on the base theory may be provable, assuming the base theory is consistent. (Example: Axiom of Choice over ZF.)
- Independence of the assumption on the base theory may be impossible to prove, assuming the base theory is consistent. (Example: Axiom of Determinacy over ZF. The reason is that ZF+AD can prove the consistency of ZF, so the claim follows by the Second Goedel Incompleteness Theorem.)
We don't know (yet) whether Riemann Hypothesis falls into any of these categories; but that's not preventing anyone from assuming that it is likely true and publishing its interesting consequences.
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
$endgroup$
2
$begingroup$
This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
$endgroup$
– user1729
Apr 29 '14 at 13:34
$begingroup$
(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
$endgroup$
– user1729
Apr 29 '14 at 13:49
1
$begingroup$
@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
$endgroup$
– Trevor Wilson
Apr 29 '14 at 18:58
1
$begingroup$
@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
$endgroup$
– Jirka Hanika
Apr 29 '14 at 22:11
2
$begingroup$
@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
$endgroup$
– Jirka Hanika
Apr 30 '14 at 14:39
|
show 6 more comments
$begingroup$
The difference between a famous conjecture and an axiom is not that large. An axiom is just an even older and more famous assumption than a famous conjecture.
In mathematics, you are just expected to be very clear what your assumptions are. Other mathematicians may be using different assumptions; it does not matter very much whether we say that they are working in a different theory, or that they have different famous conjectures in antecedents of their theorems.
Philosophically, each "axiom candidate" also needs some justification. There must be a reason why people tend to believe it, apart from a historical tradition. A large part of justification both for and against each axiom candidate is in the consequences that it would have if accepted. So somebody needs to research those consequences first and that's why one would explicitly assume Riemann Hypothesis before it's clear how it fits into established theories (or into the reality of the world in which we live).
An assumption can eventually be found to have a particular relation to a "base theory" that people do not even care to mention in their theorems.
- The assumption may be a theorem.
- The assumption may be a negation of a theorem.
- Independence of the assumption on the base theory may be provable, assuming the base theory is consistent. (Example: Axiom of Choice over ZF.)
- Independence of the assumption on the base theory may be impossible to prove, assuming the base theory is consistent. (Example: Axiom of Determinacy over ZF. The reason is that ZF+AD can prove the consistency of ZF, so the claim follows by the Second Goedel Incompleteness Theorem.)
We don't know (yet) whether Riemann Hypothesis falls into any of these categories; but that's not preventing anyone from assuming that it is likely true and publishing its interesting consequences.
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
$endgroup$
The difference between a famous conjecture and an axiom is not that large. An axiom is just an even older and more famous assumption than a famous conjecture.
In mathematics, you are just expected to be very clear what your assumptions are. Other mathematicians may be using different assumptions; it does not matter very much whether we say that they are working in a different theory, or that they have different famous conjectures in antecedents of their theorems.
Philosophically, each "axiom candidate" also needs some justification. There must be a reason why people tend to believe it, apart from a historical tradition. A large part of justification both for and against each axiom candidate is in the consequences that it would have if accepted. So somebody needs to research those consequences first and that's why one would explicitly assume Riemann Hypothesis before it's clear how it fits into established theories (or into the reality of the world in which we live).
An assumption can eventually be found to have a particular relation to a "base theory" that people do not even care to mention in their theorems.
- The assumption may be a theorem.
- The assumption may be a negation of a theorem.
- Independence of the assumption on the base theory may be provable, assuming the base theory is consistent. (Example: Axiom of Choice over ZF.)
- Independence of the assumption on the base theory may be impossible to prove, assuming the base theory is consistent. (Example: Axiom of Determinacy over ZF. The reason is that ZF+AD can prove the consistency of ZF, so the claim follows by the Second Goedel Incompleteness Theorem.)
We don't know (yet) whether Riemann Hypothesis falls into any of these categories; but that's not preventing anyone from assuming that it is likely true and publishing its interesting consequences.
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
edited May 6 '14 at 14:53
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
answered Apr 29 '14 at 12:53
Jirka HanikaJirka Hanika
1315
1315
2
$begingroup$
This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
$endgroup$
– user1729
Apr 29 '14 at 13:34
$begingroup$
(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
$endgroup$
– user1729
Apr 29 '14 at 13:49
1
$begingroup$
@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
$endgroup$
– Trevor Wilson
Apr 29 '14 at 18:58
1
$begingroup$
@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
$endgroup$
– Jirka Hanika
Apr 29 '14 at 22:11
2
$begingroup$
@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
$endgroup$
– Jirka Hanika
Apr 30 '14 at 14:39
|
show 6 more comments
2
$begingroup$
This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
$endgroup$
– user1729
Apr 29 '14 at 13:34
$begingroup$
(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
$endgroup$
– user1729
Apr 29 '14 at 13:49
1
$begingroup$
@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
$endgroup$
– Trevor Wilson
Apr 29 '14 at 18:58
1
$begingroup$
@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
$endgroup$
– Jirka Hanika
Apr 29 '14 at 22:11
2
$begingroup$
@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
$endgroup$
– Jirka Hanika
Apr 30 '14 at 14:39
2
2
$begingroup$
This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
$endgroup$
– user1729
Apr 29 '14 at 13:34
$begingroup$
This isn't quite right. An axiom should be independent of the other axioms. For example, in set theory the axiom of choice is independent of the axioms of ZF, as is the continuum hypothesis. So we can take them as axioms. Assuming that the Riemann Hypothesis is true is different - we do not know if it is independent of our axioms or not! Something cannot be both true and not true. This is inconsistent.
$endgroup$
– user1729
Apr 29 '14 at 13:34
$begingroup$
(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
$endgroup$
– user1729
Apr 29 '14 at 13:49
$begingroup$
(Interestingly, if you did prove that the RH was independent of our axioms then it would be true. If it was false then there would be a counter-example, and so would be dependent which is a constriction. I suppose that this assumes the law of the excluded middle, but that is a sensible thing to assume...)
$endgroup$
– user1729
Apr 29 '14 at 13:49
1
1
$begingroup$
@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
$endgroup$
– Trevor Wilson
Apr 29 '14 at 18:58
$begingroup$
@user1729 I don't think that your requirement that the axioms of a system be independent is universally accepted. The axioms of ZF, as usually stated, are not independent of each other: the separation axioms follow from the replacement axioms (in the context of the other axioms,) and moreover some of the replacement axioms follow from other replacement axioms, and some of the separation axioms follow from other separation axioms.
$endgroup$
– Trevor Wilson
Apr 29 '14 at 18:58
1
1
$begingroup$
@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
$endgroup$
– Jirka Hanika
Apr 29 '14 at 22:11
$begingroup$
@user1729 - Independence of axioms is just a technicality. Cantor's letters to Dedekind care to list power set and pairing as basic set theoretic axioms, even though he assumed unrestricted comprehension as well at that time. But a power set axiom is just a trivial application of unrestricted comprehension. He must have noticed and he certainly didn't care in the least. He just wanted the basic assumptions to be non-controversial.
$endgroup$
– Jirka Hanika
Apr 29 '14 at 22:11
2
2
$begingroup$
@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
$endgroup$
– Jirka Hanika
Apr 30 '14 at 14:39
$begingroup$
@user1729 - Relative consistency is not a holy cow either. ZF cannot prove the relative consistency of ZF + AD (the Axiom of Determinacy). Still, AD is customarily referred to as an axiom, while RH as a hypothesis. Work done with assuming either has basically the same status; a considerable amount of mathematics shares one or the other, and is valuable in identifying what AD or RH really means.
$endgroup$
– Jirka Hanika
Apr 30 '14 at 14:39
|
show 6 more comments
$begingroup$
You can't expect a mathematician to find proof of a conjecture directly. Assuming that a conjecture is true is a creative approach to understanding the conjecture. There is no pretty systematic way to solve a problem.
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
$endgroup$
add a comment |
$begingroup$
You can't expect a mathematician to find proof of a conjecture directly. Assuming that a conjecture is true is a creative approach to understanding the conjecture. There is no pretty systematic way to solve a problem.
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
$endgroup$
add a comment |
$begingroup$
You can't expect a mathematician to find proof of a conjecture directly. Assuming that a conjecture is true is a creative approach to understanding the conjecture. There is no pretty systematic way to solve a problem.
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
$endgroup$
You can't expect a mathematician to find proof of a conjecture directly. Assuming that a conjecture is true is a creative approach to understanding the conjecture. There is no pretty systematic way to solve a problem.
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
answered Apr 29 '14 at 13:31
toplel32toplel32
1111
1111
add a comment |
add a comment |
$begingroup$
Stated assumptions allows one to focus other areas of research without getting bogged down in proving the assumption.
By stating "Assuming the generalized Riemann Hypothesis.." at the beginning, they are not stating that the generalized Riemann Hypothesis is fact. They are merely informing the reader/viewer that this research is based on that idea and that the research might be invalidated if the assumption is shown to be false.
answered May 1 '14 at 15:36
GraxGrax
1112
$endgroup$
add a comment |
$begingroup$
Stated assumptions allows one to focus other areas of research without getting bogged down in proving the assumption.
By stating "Assuming the generalized Riemann Hypothesis.." at the beginning, they are not stating that the generalized Riemann Hypothesis is fact. They are merely informing the reader/viewer that this research is based on that idea and that the research might be invalidated if the assumption is shown to be false.
answered May 1 '14 at 15:36
GraxGrax
1112
$endgroup$
add a comment |
$begingroup$
Stated assumptions allows one to focus other areas of research without getting bogged down in proving the assumption.
By stating "Assuming the generalized Riemann Hypothesis.." at the beginning, they are not stating that the generalized Riemann Hypothesis is fact. They are merely informing the reader/viewer that this research is based on that idea and that the research might be invalidated if the assumption is shown to be false.
answered May 1 '14 at 15:36
GraxGrax
1112
$endgroup$
Stated assumptions allows one to focus other areas of research without getting bogged down in proving the assumption.
By stating "Assuming the generalized Riemann Hypothesis.." at the beginning, they are not stating that the generalized Riemann Hypothesis is fact. They are merely informing the reader/viewer that this research is based on that idea and that the research might be invalidated if the assumption is shown to be false.
answered May 1 '14 at 15:36
GraxGrax
1112
answered May 1 '14 at 15:36
GraxGrax
1112
answered May 1 '14 at 15:36
GraxGrax
1112
answered May 1 '14 at 15:36
GraxGrax
1112
1112
add a comment |
add a comment |
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57
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3. It can always be that the research leads to a disproof of the conjecture. :)
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– darij grinberg
Apr 28 '14 at 0:31
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I'm not an expert, but using the Riemann Hypothesis as an example, there's strong evidence in support of it, and why wait for it to be proved when you can already build new results upon it?
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– qwr
Apr 28 '14 at 0:33
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Part of the reason they're famous conjectures is because we believe they are likely to be true.
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– Hurkyl
Apr 28 '14 at 5:32
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There is also research that leads to practical applications and algorithms that just work well in practice, even when (partly) based on unproved hypothesis. Who would not use an O(n) algorithm to factor numbers whos runtime assesment depends on the riemann hypothesis, just because of that?
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– PlasmaHH
Apr 28 '14 at 10:29
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It's an anecdote, but I remember some theorem was proven this way: "Assume RH is true [...] then $P$. Assume RH is not true [...] then $P$. Therefore $P$".
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– Najib Idrissi
Apr 28 '14 at 12:28