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Isn't rule #6 wrong?
Do all the brackets here represent the Legendre symbol? If so, it should say so!
No, only the ones in the definition (and it's stated there). The other brackets represent Jacobi symbols.
is it just me or rule #10 is wrong since it should be stated that m,n must be relativly prime? —The preceding unsigned comment was added by 217.132.13.228 ( talk • contribs) 06:15, 21 February 2007 (UTC)
I rewrote the page, added references, explained in more detail just what Jacobi added to teh calculateon of (a|p) Virginia-American ( talk) 21:38, 2 March 2008 (UTC)
I just cut huge rambling sections of this page, including all the calculations. My objection is that the whole point of the Jacobi symbol is that it is polynomial time computable, and the calculations shown used factorization. —Preceding unsigned comment added by 18.51.7.209 ( talk) 00:47, 28 October 2008 (UTC)
Ok, I made an account and finished writing up how the Jacobi symbol is calculated. The page actually already had this embedded in the 'motivation' section; it's just difficult to read. At this point, I think my little section supersedes both 'further properties' and 'motivation', but I'll leave it to the wiki consensus....
I would also be interested in seeing areas such as the 'motivation' section being dissolved. Currently, it reads too much like a math lecture, which I think is at odds with WP:NOTTEXTBOOK. It's trying to be more educational, but I think it makes it less readily informative. CountingPine ( talk) 09:42, 28 October 2008 (UTC)
I put in the "motivation" section to show how the Legendre symbol can require factorization to calculate, whereas the Jacobi symbol can be calculated as quickly as a gcd.
In fact, I wanted it to be *less* like a typical math lecture, in that most math lectures would simply define the symbol and prove the formulas without giving any reason for doing so.
Virginia-American ( talk) 20:04, 28 October 2008 (UTC)
I've cleaned it up well enough to remove the "story" tag. (I think!) Virginia-American ( talk) 16:24, 5 November 2008 (UTC)
Rule #6 is incorrect, quadratic reciprocity should be something like..
If both n and m are odd, then (n/m) = (m/n) unless both n and m are congruent to 3 mod 4. In this case, (n/m) = -(m/n). —Preceding unsigned comment added by 97.77.52.94 ( talk) 02:48, 9 March 2011 (UTC)
It said on the image caption that any non-zero residue m mod n must have (m|n)=1. I initially couldn't figure out how to prove this, then I found a counterexample: m=10 is not congruent to 0 mod n=15, and it is congruent to 5^2=25 mod 15, so it is a non-zero residue. However, (10|15)=(10|3)(10|5)=1*0=0, not 1. Thus, I'm removing the claim. Any objections? David815 ( talk) 15:29, 22 February 2015 (UTC)
So 9 has no quadratic residues aside from 0? 2² = 4 doesn't make 4 a quadratic residue? Am I missing something? Quite misleading... 2804:14D:BAA1:3E5:51B2:48F8:7C8A:685A ( talk) 00:16, 5 May 2016 (UTC)
The worked problem begins with "Given that 9907 is prime," but that is unnecessary. The calculation will work merely given that 9907 is odd and positive. Colin McLarty ( talk) 13:20, 14 January 2017 (UTC)
I've noticed that there might be a mistake in the formula presented, when I tried to use this formula to calculate (2/9) in a Jocobi Symbol it equals (2/3)^2 = 1, which is clearly incorrect. You can see it by squaring each of the numbers 0 through 8, you should notice that the only possible remainders mod 9 are 0,1,4,7 which implies 2 is not a quadratic residue, however according to formula presented on the site it would be. I hope it gets fixed soon. 62.122.115.147 ( talk) 15:12, 12 June 2022 (UTC)
I'd like to remove the first one. Any problems? James in dc ( talk) 02:49, 7 December 2023 (UTC)
I have rewritten this article the main changes are
Please visit my sandbox User:James in dc/sandbox/Jacobi symbol and comment. I will replace the article in a week or so, after I finish with the notes.
James in dc ( talk) 23:13, 18 December 2023 (UTC)
This
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
Isn't rule #6 wrong?
Do all the brackets here represent the Legendre symbol? If so, it should say so!
No, only the ones in the definition (and it's stated there). The other brackets represent Jacobi symbols.
is it just me or rule #10 is wrong since it should be stated that m,n must be relativly prime? —The preceding unsigned comment was added by 217.132.13.228 ( talk • contribs) 06:15, 21 February 2007 (UTC)
I rewrote the page, added references, explained in more detail just what Jacobi added to teh calculateon of (a|p) Virginia-American ( talk) 21:38, 2 March 2008 (UTC)
I just cut huge rambling sections of this page, including all the calculations. My objection is that the whole point of the Jacobi symbol is that it is polynomial time computable, and the calculations shown used factorization. —Preceding unsigned comment added by 18.51.7.209 ( talk) 00:47, 28 October 2008 (UTC)
Ok, I made an account and finished writing up how the Jacobi symbol is calculated. The page actually already had this embedded in the 'motivation' section; it's just difficult to read. At this point, I think my little section supersedes both 'further properties' and 'motivation', but I'll leave it to the wiki consensus....
I would also be interested in seeing areas such as the 'motivation' section being dissolved. Currently, it reads too much like a math lecture, which I think is at odds with WP:NOTTEXTBOOK. It's trying to be more educational, but I think it makes it less readily informative. CountingPine ( talk) 09:42, 28 October 2008 (UTC)
I put in the "motivation" section to show how the Legendre symbol can require factorization to calculate, whereas the Jacobi symbol can be calculated as quickly as a gcd.
In fact, I wanted it to be *less* like a typical math lecture, in that most math lectures would simply define the symbol and prove the formulas without giving any reason for doing so.
Virginia-American ( talk) 20:04, 28 October 2008 (UTC)
I've cleaned it up well enough to remove the "story" tag. (I think!) Virginia-American ( talk) 16:24, 5 November 2008 (UTC)
Rule #6 is incorrect, quadratic reciprocity should be something like..
If both n and m are odd, then (n/m) = (m/n) unless both n and m are congruent to 3 mod 4. In this case, (n/m) = -(m/n). —Preceding unsigned comment added by 97.77.52.94 ( talk) 02:48, 9 March 2011 (UTC)
It said on the image caption that any non-zero residue m mod n must have (m|n)=1. I initially couldn't figure out how to prove this, then I found a counterexample: m=10 is not congruent to 0 mod n=15, and it is congruent to 5^2=25 mod 15, so it is a non-zero residue. However, (10|15)=(10|3)(10|5)=1*0=0, not 1. Thus, I'm removing the claim. Any objections? David815 ( talk) 15:29, 22 February 2015 (UTC)
So 9 has no quadratic residues aside from 0? 2² = 4 doesn't make 4 a quadratic residue? Am I missing something? Quite misleading... 2804:14D:BAA1:3E5:51B2:48F8:7C8A:685A ( talk) 00:16, 5 May 2016 (UTC)
The worked problem begins with "Given that 9907 is prime," but that is unnecessary. The calculation will work merely given that 9907 is odd and positive. Colin McLarty ( talk) 13:20, 14 January 2017 (UTC)
I've noticed that there might be a mistake in the formula presented, when I tried to use this formula to calculate (2/9) in a Jocobi Symbol it equals (2/3)^2 = 1, which is clearly incorrect. You can see it by squaring each of the numbers 0 through 8, you should notice that the only possible remainders mod 9 are 0,1,4,7 which implies 2 is not a quadratic residue, however according to formula presented on the site it would be. I hope it gets fixed soon. 62.122.115.147 ( talk) 15:12, 12 June 2022 (UTC)
I'd like to remove the first one. Any problems? James in dc ( talk) 02:49, 7 December 2023 (UTC)
I have rewritten this article the main changes are
Please visit my sandbox User:James in dc/sandbox/Jacobi symbol and comment. I will replace the article in a week or so, after I finish with the notes.
James in dc ( talk) 23:13, 18 December 2023 (UTC)