Taking an unproven but “seemingly true” statement as an axiom
$begingroup$
I'm sorry for this uneducated question, but I've been thinking of this for a few hours and I couldn't find anything on the topic. Perhaps it is just a failure on my part and a limitation of my experience, but assuming not:
Could we take something like Goldbach's conjecture as an axiom, and go from there to prove new theorems? Does this ever happen, or do mathematicians build up structures from both assumptions (one structure assumes the conjecture is true, and one false, and then go on to build up theorems and additional structures from there).
I use Goldbach's conjecture as an example because it has a lot of "evidence" (not really but maybe empirically, even though math has nothing to do with empiricism, I'm just using Goldbach but we could use any conjecture) to err on the side of it being true, so perhaps assuming it's true, then going on to use it to prove other things, may result in some pretty results. If it turns out that the conjecture is false, we just throw out the results and the system. But assuming a system that includes "Goldbachs conjecture is true" as an axiom is consistent, then why wouldn't we be able to use the system? Do mathematicians ever do this, or is consider bad form (I can see why this would be completely banned).
Basically, could I say "Let us assume Goldbach's conjecture is true. Then..." and go one to build up books of math and results and theorems using this?
first-order-logic predicate-logic axioms goldbachs-conjecture
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
$endgroup$
add a comment |
$begingroup$
I'm sorry for this uneducated question, but I've been thinking of this for a few hours and I couldn't find anything on the topic. Perhaps it is just a failure on my part and a limitation of my experience, but assuming not:
Could we take something like Goldbach's conjecture as an axiom, and go from there to prove new theorems? Does this ever happen, or do mathematicians build up structures from both assumptions (one structure assumes the conjecture is true, and one false, and then go on to build up theorems and additional structures from there).
I use Goldbach's conjecture as an example because it has a lot of "evidence" (not really but maybe empirically, even though math has nothing to do with empiricism, I'm just using Goldbach but we could use any conjecture) to err on the side of it being true, so perhaps assuming it's true, then going on to use it to prove other things, may result in some pretty results. If it turns out that the conjecture is false, we just throw out the results and the system. But assuming a system that includes "Goldbachs conjecture is true" as an axiom is consistent, then why wouldn't we be able to use the system? Do mathematicians ever do this, or is consider bad form (I can see why this would be completely banned).
Basically, could I say "Let us assume Goldbach's conjecture is true. Then..." and go one to build up books of math and results and theorems using this?
first-order-logic predicate-logic axioms goldbachs-conjecture
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
$endgroup$
4
$begingroup$
Yes, people do this all the time, e.g. with the Riemann hypothesis as in Eevee Trainer's answer. You just state the result as "Goldbach's conjecture implies X" instead of as "X."
$endgroup$
– Qiaochu Yuan
Dec 13 '18 at 5:25
4
$begingroup$
This sort of thing is common in set theory and other areas of logic, where we have tools to prove that certain statements are unprovable, and it may make sense to study the mathematical universes where one of those statements holds, and also those where it fails.
$endgroup$
– Andrés E. Caicedo
Dec 13 '18 at 5:29
add a comment |
$begingroup$
I'm sorry for this uneducated question, but I've been thinking of this for a few hours and I couldn't find anything on the topic. Perhaps it is just a failure on my part and a limitation of my experience, but assuming not:
Could we take something like Goldbach's conjecture as an axiom, and go from there to prove new theorems? Does this ever happen, or do mathematicians build up structures from both assumptions (one structure assumes the conjecture is true, and one false, and then go on to build up theorems and additional structures from there).
I use Goldbach's conjecture as an example because it has a lot of "evidence" (not really but maybe empirically, even though math has nothing to do with empiricism, I'm just using Goldbach but we could use any conjecture) to err on the side of it being true, so perhaps assuming it's true, then going on to use it to prove other things, may result in some pretty results. If it turns out that the conjecture is false, we just throw out the results and the system. But assuming a system that includes "Goldbachs conjecture is true" as an axiom is consistent, then why wouldn't we be able to use the system? Do mathematicians ever do this, or is consider bad form (I can see why this would be completely banned).
Basically, could I say "Let us assume Goldbach's conjecture is true. Then..." and go one to build up books of math and results and theorems using this?
first-order-logic predicate-logic axioms goldbachs-conjecture
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
$endgroup$
I'm sorry for this uneducated question, but I've been thinking of this for a few hours and I couldn't find anything on the topic. Perhaps it is just a failure on my part and a limitation of my experience, but assuming not:
Could we take something like Goldbach's conjecture as an axiom, and go from there to prove new theorems? Does this ever happen, or do mathematicians build up structures from both assumptions (one structure assumes the conjecture is true, and one false, and then go on to build up theorems and additional structures from there).
I use Goldbach's conjecture as an example because it has a lot of "evidence" (not really but maybe empirically, even though math has nothing to do with empiricism, I'm just using Goldbach but we could use any conjecture) to err on the side of it being true, so perhaps assuming it's true, then going on to use it to prove other things, may result in some pretty results. If it turns out that the conjecture is false, we just throw out the results and the system. But assuming a system that includes "Goldbachs conjecture is true" as an axiom is consistent, then why wouldn't we be able to use the system? Do mathematicians ever do this, or is consider bad form (I can see why this would be completely banned).
Basically, could I say "Let us assume Goldbach's conjecture is true. Then..." and go one to build up books of math and results and theorems using this?
first-order-logic predicate-logic axioms goldbachs-conjecture
first-order-logic predicate-logic axioms goldbachs-conjecture
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
asked Dec 13 '18 at 4:54
regalrenderregalrender
212
212
4
$begingroup$
Yes, people do this all the time, e.g. with the Riemann hypothesis as in Eevee Trainer's answer. You just state the result as "Goldbach's conjecture implies X" instead of as "X."
$endgroup$
– Qiaochu Yuan
Dec 13 '18 at 5:25
4
$begingroup$
This sort of thing is common in set theory and other areas of logic, where we have tools to prove that certain statements are unprovable, and it may make sense to study the mathematical universes where one of those statements holds, and also those where it fails.
$endgroup$
– Andrés E. Caicedo
Dec 13 '18 at 5:29
add a comment |
4
$begingroup$
Yes, people do this all the time, e.g. with the Riemann hypothesis as in Eevee Trainer's answer. You just state the result as "Goldbach's conjecture implies X" instead of as "X."
$endgroup$
– Qiaochu Yuan
Dec 13 '18 at 5:25
4
$begingroup$
This sort of thing is common in set theory and other areas of logic, where we have tools to prove that certain statements are unprovable, and it may make sense to study the mathematical universes where one of those statements holds, and also those where it fails.
$endgroup$
– Andrés E. Caicedo
Dec 13 '18 at 5:29
4
4
$begingroup$
Yes, people do this all the time, e.g. with the Riemann hypothesis as in Eevee Trainer's answer. You just state the result as "Goldbach's conjecture implies X" instead of as "X."
$endgroup$
– Qiaochu Yuan
Dec 13 '18 at 5:25
$begingroup$
Yes, people do this all the time, e.g. with the Riemann hypothesis as in Eevee Trainer's answer. You just state the result as "Goldbach's conjecture implies X" instead of as "X."
$endgroup$
– Qiaochu Yuan
Dec 13 '18 at 5:25
4
4
$begingroup$
This sort of thing is common in set theory and other areas of logic, where we have tools to prove that certain statements are unprovable, and it may make sense to study the mathematical universes where one of those statements holds, and also those where it fails.
$endgroup$
– Andrés E. Caicedo
Dec 13 '18 at 5:29
$begingroup$
This sort of thing is common in set theory and other areas of logic, where we have tools to prove that certain statements are unprovable, and it may make sense to study the mathematical universes where one of those statements holds, and also those where it fails.
$endgroup$
– Andrés E. Caicedo
Dec 13 '18 at 5:29
add a comment |
1 Answer
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I have seen this occur with the Riemann hypothesis (albeit second-hand, I haven't looked into it, but it was mentioned on Numberphile).
It seems it's generally believed that the Riemann hypothesis is "probably true," even though the proof evades us (obviously). However, a number of proofs hinge on this - "Assuming the Riemann hypothesis is true, then (this result) is true," or something of that flavor.
That's part of why a proof of the Riemann hypothesis is actually so sought-for as to justify a million-dollar bounty - because proving the Riemann hypothesis would in turn prove a bunch of other theorems that hinge on its truth.
So I guess among mathematicians this would be "kosher," but generally only with results that are regarded as "probably true," even if the proof evades us.
I did actually answer another question the other day here on MSE that concerned whether the infinite product
$$prod_{k=1}^infty 10^k = 0$$
Obviously untrue in the "usual topology of $mathbb{R}$", but as one commenter noted at the time it was not impossible that in other topologies of the reals that the result could hold true. So in a way, it seems like assuming an "obviously false" result to be true instead could result in, if nothing else, some interesting discussions.
This is all on the tenet that I'm just an amateur - I'm an undergrad, not a professional mathematician, so I'm mostly taking what I know and have seen and extrapolating from there. (For all I know, it's in bad taste to assume the Riemann hypothesis true, for example, I wouldn't know.) So take this last paragraph as a big ol' "grain of salt" warning.
A minor footnote, but is saying that you would take it as an "axiom" valid? If the other axioms disprove it somehow, that presents a problem, doesn't it? Though I suppose you could just replace the axioms that disprove it with others that wouldn't... I'm not sure, this is already a bit above my head so I'll shut up.
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
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add a comment |
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$begingroup$
I have seen this occur with the Riemann hypothesis (albeit second-hand, I haven't looked into it, but it was mentioned on Numberphile).
It seems it's generally believed that the Riemann hypothesis is "probably true," even though the proof evades us (obviously). However, a number of proofs hinge on this - "Assuming the Riemann hypothesis is true, then (this result) is true," or something of that flavor.
That's part of why a proof of the Riemann hypothesis is actually so sought-for as to justify a million-dollar bounty - because proving the Riemann hypothesis would in turn prove a bunch of other theorems that hinge on its truth.
So I guess among mathematicians this would be "kosher," but generally only with results that are regarded as "probably true," even if the proof evades us.
I did actually answer another question the other day here on MSE that concerned whether the infinite product
$$prod_{k=1}^infty 10^k = 0$$
Obviously untrue in the "usual topology of $mathbb{R}$", but as one commenter noted at the time it was not impossible that in other topologies of the reals that the result could hold true. So in a way, it seems like assuming an "obviously false" result to be true instead could result in, if nothing else, some interesting discussions.
This is all on the tenet that I'm just an amateur - I'm an undergrad, not a professional mathematician, so I'm mostly taking what I know and have seen and extrapolating from there. (For all I know, it's in bad taste to assume the Riemann hypothesis true, for example, I wouldn't know.) So take this last paragraph as a big ol' "grain of salt" warning.
A minor footnote, but is saying that you would take it as an "axiom" valid? If the other axioms disprove it somehow, that presents a problem, doesn't it? Though I suppose you could just replace the axioms that disprove it with others that wouldn't... I'm not sure, this is already a bit above my head so I'll shut up.
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
add a comment |
$begingroup$
I have seen this occur with the Riemann hypothesis (albeit second-hand, I haven't looked into it, but it was mentioned on Numberphile).
It seems it's generally believed that the Riemann hypothesis is "probably true," even though the proof evades us (obviously). However, a number of proofs hinge on this - "Assuming the Riemann hypothesis is true, then (this result) is true," or something of that flavor.
That's part of why a proof of the Riemann hypothesis is actually so sought-for as to justify a million-dollar bounty - because proving the Riemann hypothesis would in turn prove a bunch of other theorems that hinge on its truth.
So I guess among mathematicians this would be "kosher," but generally only with results that are regarded as "probably true," even if the proof evades us.
I did actually answer another question the other day here on MSE that concerned whether the infinite product
$$prod_{k=1}^infty 10^k = 0$$
Obviously untrue in the "usual topology of $mathbb{R}$", but as one commenter noted at the time it was not impossible that in other topologies of the reals that the result could hold true. So in a way, it seems like assuming an "obviously false" result to be true instead could result in, if nothing else, some interesting discussions.
This is all on the tenet that I'm just an amateur - I'm an undergrad, not a professional mathematician, so I'm mostly taking what I know and have seen and extrapolating from there. (For all I know, it's in bad taste to assume the Riemann hypothesis true, for example, I wouldn't know.) So take this last paragraph as a big ol' "grain of salt" warning.
A minor footnote, but is saying that you would take it as an "axiom" valid? If the other axioms disprove it somehow, that presents a problem, doesn't it? Though I suppose you could just replace the axioms that disprove it with others that wouldn't... I'm not sure, this is already a bit above my head so I'll shut up.
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
add a comment |
$begingroup$
I have seen this occur with the Riemann hypothesis (albeit second-hand, I haven't looked into it, but it was mentioned on Numberphile).
It seems it's generally believed that the Riemann hypothesis is "probably true," even though the proof evades us (obviously). However, a number of proofs hinge on this - "Assuming the Riemann hypothesis is true, then (this result) is true," or something of that flavor.
That's part of why a proof of the Riemann hypothesis is actually so sought-for as to justify a million-dollar bounty - because proving the Riemann hypothesis would in turn prove a bunch of other theorems that hinge on its truth.
So I guess among mathematicians this would be "kosher," but generally only with results that are regarded as "probably true," even if the proof evades us.
I did actually answer another question the other day here on MSE that concerned whether the infinite product
$$prod_{k=1}^infty 10^k = 0$$
Obviously untrue in the "usual topology of $mathbb{R}$", but as one commenter noted at the time it was not impossible that in other topologies of the reals that the result could hold true. So in a way, it seems like assuming an "obviously false" result to be true instead could result in, if nothing else, some interesting discussions.
This is all on the tenet that I'm just an amateur - I'm an undergrad, not a professional mathematician, so I'm mostly taking what I know and have seen and extrapolating from there. (For all I know, it's in bad taste to assume the Riemann hypothesis true, for example, I wouldn't know.) So take this last paragraph as a big ol' "grain of salt" warning.
A minor footnote, but is saying that you would take it as an "axiom" valid? If the other axioms disprove it somehow, that presents a problem, doesn't it? Though I suppose you could just replace the axioms that disprove it with others that wouldn't... I'm not sure, this is already a bit above my head so I'll shut up.
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
I have seen this occur with the Riemann hypothesis (albeit second-hand, I haven't looked into it, but it was mentioned on Numberphile).
It seems it's generally believed that the Riemann hypothesis is "probably true," even though the proof evades us (obviously). However, a number of proofs hinge on this - "Assuming the Riemann hypothesis is true, then (this result) is true," or something of that flavor.
That's part of why a proof of the Riemann hypothesis is actually so sought-for as to justify a million-dollar bounty - because proving the Riemann hypothesis would in turn prove a bunch of other theorems that hinge on its truth.
So I guess among mathematicians this would be "kosher," but generally only with results that are regarded as "probably true," even if the proof evades us.
I did actually answer another question the other day here on MSE that concerned whether the infinite product
$$prod_{k=1}^infty 10^k = 0$$
Obviously untrue in the "usual topology of $mathbb{R}$", but as one commenter noted at the time it was not impossible that in other topologies of the reals that the result could hold true. So in a way, it seems like assuming an "obviously false" result to be true instead could result in, if nothing else, some interesting discussions.
This is all on the tenet that I'm just an amateur - I'm an undergrad, not a professional mathematician, so I'm mostly taking what I know and have seen and extrapolating from there. (For all I know, it's in bad taste to assume the Riemann hypothesis true, for example, I wouldn't know.) So take this last paragraph as a big ol' "grain of salt" warning.
A minor footnote, but is saying that you would take it as an "axiom" valid? If the other axioms disprove it somehow, that presents a problem, doesn't it? Though I suppose you could just replace the axioms that disprove it with others that wouldn't... I'm not sure, this is already a bit above my head so I'll shut up.
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
answered Dec 13 '18 at 5:06
Eevee TrainerEevee Trainer
5,9781936
5,9781936
add a comment |
add a comment |
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$begingroup$
Yes, people do this all the time, e.g. with the Riemann hypothesis as in Eevee Trainer's answer. You just state the result as "Goldbach's conjecture implies X" instead of as "X."
$endgroup$
– Qiaochu Yuan
Dec 13 '18 at 5:25
4
$begingroup$
This sort of thing is common in set theory and other areas of logic, where we have tools to prove that certain statements are unprovable, and it may make sense to study the mathematical universes where one of those statements holds, and also those where it fails.
$endgroup$
– Andrés E. Caicedo
Dec 13 '18 at 5:29