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August 5 Information
Are there theorems of number theory that have so far needed Calculus for being proven?
Well, I suspect Prime number theorem is not purely number-theoretical, because its very formulation involves terms (like the natural logarithm as you've already pointed out) which are not recognized by Peano System. Anyways, thank you for two theorems, which are purely number-theoretical:
Chen's theorem, and
Friedlander–Iwaniec theorem. Please notice that all of the other theorems are not purely number-theoretical.
HOTmag (
talk)
15:23, 5 August 2016 (UTC)reply
The whole subject of
analytic number theory is the application of
mathematical analysis to number theory. My impression is that it's the bulk of number theory these days. Probably most recent results in number theory are answers to your question.
Oh, thank you for reminding me this important theorem. Now I wonder if it's provable in Peano system, because if it's not, then it can be used as a much simpler example (than Goodstein's Theorem) for theorems - which are unprovable in Peano system - even though they are formulated in Peano language.
HOTmag (
talk)
22:26, 6 August 2016 (UTC)reply
I think that's totally open as of now. Unless I haven't heard about it (which is totally possible; I'm not all that looped in these days), no one has reported either a proof in
Peano arithmetic or a reason there shouldn't be one. A very plausible state of affairs is that the analytic proofs can be converted by sufficiently careful technical attention into a PA proof, but that it would be so tedious that no one would ever actually do it. I don't know whether anyone has made a convincing argument that it ought (or ought not) to be possible. --
Trovatore (
talk)
22:35, 6 August 2016 (UTC)reply
Thx. Now I also wonder, if this theorem can really be formulated in Peano language. How would you formulate the term ∃(a,b,c)(a^b=c) in Peano language?
HOTmag (
talk)
22:45, 6 August 2016 (UTC)reply
The information you've supplied is really a bit sketchy, but I'm sure you've made an effort to give me all you've had, so I thank you anyway.
HOTmag (
talk)
23:24, 6 August 2016 (UTC)reply
Well, no, I could have thought about it more and supplied more details, but that would have taken more time. If you're interested in that sort of detail, it's better to work it out for yourself anyway. --
Trovatore (
talk)
00:26, 7 August 2016 (UTC)reply
As for the Prime Number Theorem, it turned out to be "purely number-theoretical". See our article, the section on
Elementary Proofs and the refs. It's provable in a somewhat weaker system than Peano & later verified by automatic methods even. The Erdos-Selberg "elementary" (in a vaguer mathematical sense meaning no complex analysis) proof was a key step.
John Z (
talk)
02:14, 13 August 2016 (UTC)reply
calculating the average odds?
Hello,
this is not homework, just a problem that has been bothering me.
Say the chance of a hurricane (in the next year) is 10%, the chance of an earthquake is 20% and the chance of a blizzard is 30%.
I want to know the odds that only one of these things is going to happen.
I know how to do that in a very "sisyphean" way:
[(1-0.1)*(1-0.2)*0.3]+[(1-0.1)*0.2*(1-0.3)]+[0.1*(1-0.2)*(1-0.3)]=0.398 so that's 39.8%.
The problem is this way is not practical if you have 20 different factors instead of 3.
Do you know a faster way to solve this sort of problem? maybe some equation that takes the sum of the odds, or the average odds, or the product of the odds, or the number of different factors?
It's just a rewriting of your own calculation (I swapped the order). P × 0.1/(1-0.1) is your 0.1*(1-0.2)*(1-0.3), while P × 0.2/(1-0.2) is your (1-0.1)*0.2*(1-0.3), and P × 0.3/(1-0.3) is your (1-0.1)*(1-0.2)*0.3.
PrimeHunter (
talk)
01:07, 6 August 2016 (UTC)reply
To clarify even further: Suppose you have 5 events, and let be the probability of each of them not happening. The original method involves calculating each of which scales quadratically with the number of events. But all these terms are similar - they can each be found from by eliminating one of the variables. So you calculate once and then instead of calculating , you calculate . This scales linearly. --
Meni Rosenfeld (
talk)
10:27, 7 August 2016 (UTC)reply
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the
current reference desk pages.
August 5 Information
Are there theorems of number theory that have so far needed Calculus for being proven?
Well, I suspect Prime number theorem is not purely number-theoretical, because its very formulation involves terms (like the natural logarithm as you've already pointed out) which are not recognized by Peano System. Anyways, thank you for two theorems, which are purely number-theoretical:
Chen's theorem, and
Friedlander–Iwaniec theorem. Please notice that all of the other theorems are not purely number-theoretical.
HOTmag (
talk)
15:23, 5 August 2016 (UTC)reply
The whole subject of
analytic number theory is the application of
mathematical analysis to number theory. My impression is that it's the bulk of number theory these days. Probably most recent results in number theory are answers to your question.
Oh, thank you for reminding me this important theorem. Now I wonder if it's provable in Peano system, because if it's not, then it can be used as a much simpler example (than Goodstein's Theorem) for theorems - which are unprovable in Peano system - even though they are formulated in Peano language.
HOTmag (
talk)
22:26, 6 August 2016 (UTC)reply
I think that's totally open as of now. Unless I haven't heard about it (which is totally possible; I'm not all that looped in these days), no one has reported either a proof in
Peano arithmetic or a reason there shouldn't be one. A very plausible state of affairs is that the analytic proofs can be converted by sufficiently careful technical attention into a PA proof, but that it would be so tedious that no one would ever actually do it. I don't know whether anyone has made a convincing argument that it ought (or ought not) to be possible. --
Trovatore (
talk)
22:35, 6 August 2016 (UTC)reply
Thx. Now I also wonder, if this theorem can really be formulated in Peano language. How would you formulate the term ∃(a,b,c)(a^b=c) in Peano language?
HOTmag (
talk)
22:45, 6 August 2016 (UTC)reply
The information you've supplied is really a bit sketchy, but I'm sure you've made an effort to give me all you've had, so I thank you anyway.
HOTmag (
talk)
23:24, 6 August 2016 (UTC)reply
Well, no, I could have thought about it more and supplied more details, but that would have taken more time. If you're interested in that sort of detail, it's better to work it out for yourself anyway. --
Trovatore (
talk)
00:26, 7 August 2016 (UTC)reply
As for the Prime Number Theorem, it turned out to be "purely number-theoretical". See our article, the section on
Elementary Proofs and the refs. It's provable in a somewhat weaker system than Peano & later verified by automatic methods even. The Erdos-Selberg "elementary" (in a vaguer mathematical sense meaning no complex analysis) proof was a key step.
John Z (
talk)
02:14, 13 August 2016 (UTC)reply
calculating the average odds?
Hello,
this is not homework, just a problem that has been bothering me.
Say the chance of a hurricane (in the next year) is 10%, the chance of an earthquake is 20% and the chance of a blizzard is 30%.
I want to know the odds that only one of these things is going to happen.
I know how to do that in a very "sisyphean" way:
[(1-0.1)*(1-0.2)*0.3]+[(1-0.1)*0.2*(1-0.3)]+[0.1*(1-0.2)*(1-0.3)]=0.398 so that's 39.8%.
The problem is this way is not practical if you have 20 different factors instead of 3.
Do you know a faster way to solve this sort of problem? maybe some equation that takes the sum of the odds, or the average odds, or the product of the odds, or the number of different factors?
It's just a rewriting of your own calculation (I swapped the order). P × 0.1/(1-0.1) is your 0.1*(1-0.2)*(1-0.3), while P × 0.2/(1-0.2) is your (1-0.1)*0.2*(1-0.3), and P × 0.3/(1-0.3) is your (1-0.1)*(1-0.2)*0.3.
PrimeHunter (
talk)
01:07, 6 August 2016 (UTC)reply
To clarify even further: Suppose you have 5 events, and let be the probability of each of them not happening. The original method involves calculating each of which scales quadratically with the number of events. But all these terms are similar - they can each be found from by eliminating one of the variables. So you calculate once and then instead of calculating , you calculate . This scales linearly. --
Meni Rosenfeld (
talk)
10:27, 7 August 2016 (UTC)reply