# User talk:RDBury

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## Indeterminant

Hello,

You have introduced the word "indeterminant" in polynomial ring, with an edit summary asserting that "variable" is incorrect. I disagree with you. Firstly, "indeterminant" is not used in mathematics, and, in any case, is not a synonymous of "indeterminate" (the first two pages of Google Scholar for "indeterminant" do not link to any mathematical article). "Indeterminate" is clearly the correct and commonly used word. I disagree with you that "variable" is jargon and incorrect here: in modern mathematics (and also in computer science), a "variable" is simply a symbol which may be the name of any mathematical object. The word has lost his original meaning of "representing a number that varies". A witness that "variable" is correct for polynomials is the common terminology of multivariate polynomial. However, although a specialist of polynomials since more than thirty years, I may be wrong. If it is the case, please provide a reliable source.

Sincerely, D.Lazard (talk) 13:45, 30 August 2013 (UTC)

First, sorry I wasn't clear but I meant that the word 'indeterminant' was jargon, not 'variable', but I added a link to it and recently expanded the article. The use of an indeterminant rather than a variable is meant to make that distinction between the polynomial ring over a ring and the ring of polynomial functions over a ring, which is not the same. The second is the homomorphic image of the first but, in particular for finite rings, they are not isomorphic. The term 'indeterminant' is currently used, check the What links here button on the article I just mentioned for examples. I'll try to add some references though, the polynomial article could probably use more in any case. --RDBury (talk) 02:00, 31 August 2013 (UTC)
I should have said 'indeterminate' above and seem to be confusing the 'ate' with 'ant' in my head. I added Herstein (which I own) as a ref and am currently looking for other sources in Google. --RDBury (talk) 05:30, 31 August 2013 (UTC)

## The GCD proof on the maths reference desk

I wanted to reiterate my thanks for your help regarding my recent question on the maths reference desk. I am a PhD student, and have an application of the result in mind, which my thesis supervisor has asked me to write up. I'd like to reference you for the contribution of your argument: is there a way which you would prefer to be referenced? Thanks again, Icthyos (talk) 14:03, 10 October 2013 (UTC)

Actually thank you for the mental workout. I've never thought about help desk answers being referenced. Actually Wikipedia considers itself an unreliable source so I'm not sure that you could use what I wrote up as a reference and still pass peer review. In any case, I assume these results are already well known, it's just easier and more fun to work this kind of thing out for myself than to look it up. I'm not connected with academia so I suppose if you really need to give a source you could just say R.D. Bury (personal communication). --RDBury (talk) 22:33, 10 October 2013 (UTC)
Great, thanks. I was very keen not to reference Wikipedia, for the reason you said, and was planning on just including a version of the proof. Obviously I didn't want to take credit for it, so that's why I asked. Icthyos (talk) 12:05, 11 October 2013 (UTC)
In case you'd like a further mental workout, here's a generalisation of my initial problem. Obviously you retain the right to ignore this, but I thought it might interest you. Take integers A1, B1, A2, B2, ..., An, Bn, with A1 and B2 odd and all the rest even, and gcd(A1, B1, A2, B2, ..., An, Bn) = 1. Does there still exist an integer k such that gcd(B1 + kA1, ..., Bn + kAn) = 1? It seems like this should be easier to prove somehow, because it feels somehow more difficult for longer lists of integers to not be coprime, but the 'rank 2' case which we were tackling before can't be directly applied, as far as I can see. While A1*B2 - B1*A2 is non-zero, it is no longer the case that gcd(A1,B1,A2,B2) need be 1. Any thoughts? Icthyos (talk) 14:25, 14 October 2013 (UTC)
I'm thinking the n=2 case does generalize, but you have to look at the problem in a different way. The way I'm looking at it that you're given two points v and w in the lattice Zn. We're interested in when the affine sublattice generated by v and w (i.e. {..., 2v-w, v, w, 2w-v ...}) has any points with relatively prime coordinates. A point has relatively prime coordinates iff it isn't a multiple other than ±1 of some other point. The first necessary condition is that v and w aren't the same multiple of other points, which is the same as saying the gcd of all the coordinates is 1. A condition that is necessary with a few exceptions is that the line containing v and w does not pass through the origin, since if it did the v and w would be multiples of the same point x and so would every other point on the affine lattice. This implies the gcd of the coordinates is >1 unless one of the points was ±x. Conversely, if v and w are not the same multiple of other points and the line through v and w does not contain the origin then the affine sublattice contains a point that is not a multiple of another point. The n=2 case has been done. But n=2 is actually the only case, just restrict attention to Zn∩<v,w>, the lattice of points in Zn which are in the linear subspace generated by v and w. This is a lattice of dimension 2 so it's isomorphic to Z2 and any statement about Z2 applies to this lattice as well. --RDBury (talk) 23:56, 14 October 2013 (UTC)
I'm almost with you. I considered this more geometric approach a couple of weeks ago, but didn't push it hard enough apparently! A geometric interpretation of gcd is that gcd(A1, ..., An) is one less than the number of points in Zn lying on the line joining the origin to the point (A1, ..., An) (i.e. if the gcd is 1, the number of points is 2: the endpoints of the line). It's your final sentence that I'm having the trouble with. There is certainly an isomorphism between the two lattices of dimension 2, but one of these lattices is sitting inside Zn -- call this one Y. I agree that in the n=2 case we can find a point 'with gcd 1', and that when we map this into Y, the line joining that point to the origin will still only intersect the other points of Y at the line's endpoints. But isn't it possible for that line to intersect some other lattice point in Zn which isn't a member of Y, and thus for the point in question to not have gcd 1? I do feel like this is the right way to show it, but I feel like I'm missing something. Icthyos (talk) 14:04, 15 October 2013 (UTC)
Good point; I was relying on intuition bit much there. There is a lattice Y embedded in Zn and the question is whether y∈Y is "prime" in Y iff it's "prime" in Zn, where "prime" means there is no k≠±1 and point x so that y=kx. This is false in general but in this case Y is Zn∩L where L is a linear subspace of Qn. If y is not prime in Y then y=kx with x in Y⊆Zn and y is not prime in Zn. So being prime in Zn implies being prime in Y without any extra conditions on Y. Suppose Y is not prime in Zn, so y=kx with k≠±1, x in Zn. Since y∈Y⊆L, x=(1/k)y∈L and x∈Zn so x∈Y and y is not prime in Y. I should have specified why I chose Y the way I did rather than the sublattice generated by v and w. --RDBury (talk) 03:20, 16 October 2013 (UTC)

## Nomination of Amanda Brooks for deletion

A discussion is taking place as to whether the article Amanda Brooks is suitable for inclusion in Wikipedia according to Wikipedia's policies and guidelines or whether it should be deleted.

The article will be discussed at Wikipedia:Articles for deletion/Amanda Brooks until a consensus is reached, and anyone is welcome to contribute to the discussion. The nomination will explain the policies and guidelines which are of concern. The discussion focuses on high-quality evidence and our policies and guidelines.

Users may edit the article during the discussion, including to improve the article to address concerns raised in the discussion. However, do not remove the article-for-deletion notice from the top of the article. Orange Mike | Talk 17:50, 30 November 2013 (UTC)

## thanks for the reference desk response on spinoza and axiomatic math - I've left you a couple of more questions :)

Hi there. Thanks for your response at the reference desk on spinoza and axiomatic mathematics. I've left you a couple of more questions there. 213.246.165.17 (talk) 13:41, 12 September 2014 (UTC)

You might want to put your answer to this question at Math.SE. Mark Hurd (talk) 08:20, 30 September 2014 (UTC)

## Binomial coefficient identity

I answered your question here. (It is unclear to me whether you will receive notification automatically, which is why I leave this note here.) --JBL (talk) 17:27, 2 June 2015 (UTC)

## "Least Symmetric Triangle"

FYI: https://en.wikipedia.org/w/index.php?title=Wikipedia%3AReference_desk%2FArchives%2FMathematics%2F2015_June_30&type=revision&diff=671046944&oldid=670300447 — Preceding unsigned comment added by 109.153.232.17 (talk) 01:41, 12 July 2015 (UTC)

Thanks, I looked at it. Sounds like the full solution will be a project in both geometry and computer programming. --RDBury (talk) 01:57, 12 July 2015 (UTC)

## Ref desk

Hi!Thanks for the answer at refdesk about percent change.--85.121.32.1 (talk) 11:49, 17 May 2016 (UTC)

## ArbCom Elections 2016: Voting now open!

 Hello, RDBury. Voting in the 2016 Arbitration Committee elections is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate in the 2016 election, please review the candidates' statements and submit your choices on the voting page. MediaWiki message delivery (talk) 22:08, 21 November 2016 (UTC)

## Happy Birthday

This is Chuck. You aren't answering your phone. Call Ann&Ernie right away if you are because Ann is going to drive out to check on you. Then call me.

Chuck — Preceding unsigned comment added by Cy Guy (talkcontribs) 22:56, 12 September 2017 (UTC)

Wikipedia:Reference desk/Mathematics#Curvature_from_partial_derivatives

Did you find my answer useful ?
If not, are you upset that I supplied a numerical methods answer ?

Thanks, StuRat (talk) 21:28, 23 October 2017 (UTC)

All I expect is people to make their best effort to provide some useful information on the question, since that's all anyone can expect from me when I answer a question. I don't expect answers to be useful or even correct as long as as that was the intent. I'm not sure what happened to make this an issue between you and JBL, and a certainly hope no one is expecting me to take sides, but it seems to me that further discussion or argument serves little purpose. --RDBury (talk) 00:27, 24 October 2017 (UTC)
Hi RDBury, apologies for dragging you into this, it was poorly considered. All the best, JBL (talk) 00:31, 24 October 2017 (UTC)
Thanks, RDBury. I was attempting to provide useful info, as I will in the future. If you don't find it useful, you have my apologies. All the best, StuRat (talk) 01:47, 24 October 2017 (UTC)

## ArbCom 2017 election voter message

 Hello, RDBury. Voting in the 2017 Arbitration Committee elections is now open until 23.59 on Sunday, 10 December. All users who registered an account before Saturday, 28 October 2017, made at least 150 mainspace edits before Wednesday, 1 November 2017 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate in the 2017 election, please review the candidates and submit your choices on the voting page. MediaWiki message delivery (talk) 18:42, 3 December 2017 (UTC)

## Are you a dictator?!

Hello, why did you delete my question?

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Mathematics&diff=843010554&oldid=843010117

I am solving some open problems in number theory which for centuries nobody could have any claim about them!

sorry, you can not be a monarch! — Preceding unsigned comment added by 89.45.57.67 (talk) 07:45, 26 May 2018 (UTC)

and this question is also from me, maybe you will be glad if delete it!

https://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Collatz_conjecture_is_almost_done! — Preceding unsigned comment added by 188.212.49.189 (talk) 10:09, 26 May 2018 (UTC)

I believe I was following the policy outlined under WP:EVADE. If you disagree, feel free to take the matter up with an admin or get a third opinion via WP:THIRD; I'll abide by the decision of a neutral third party. --RDBury (talk) 13:19, 26 May 2018 (UTC)
Thank you for guidance, okay let me do it! and I believe toleration on opinions of other people will make a person greater than past! — Preceding unsigned comment added by 94.101.247.97 (talk) 05:30, 27 May 2018 (UTC)
1. I want improve my theories on open problems in number theory by advice from specialists and also I can not have access academic circles but apparently @RDBury prevents me do it please express your opinion, thanks! 14:20, 27 May 2018 (UTC) — Preceding unsigned comment added by 188.212.58.42 (talk)
Okay, nobody typed hence you won and I have to leave here forever though I am used to profiting Wikipedia.org and I also deliver my best thanks to the USA. — Preceding unsigned comment added by 188.212.62.143 (talk) 15:55, 9 June 2018 (UTC)