# Talk:Commutative property

It's called commutivity, not commutativity, I'm pretty sure. — Preceding unsigned comment added by Phys (talk • contribs) 23:56, 11 August 2003 (UTC)

I came here wondering about which is correct. This very article uses both commutivity and commutativity, which I think is...bad. If you do a Google search with one of the words you'll see that "commutativity" is WAAAAY preferred over "commutivity." But I must say, I've only heard people speak "commutivity" with an emphasis on the first syllable. "Commutativity" would make more sense with the "-ity" being a suffix on "commutative" and I suspect it is the correct version. Now I'm going to start listening closely whenever I have a conversation with somebody about binary operators.... 205.175.97.219 (talk) 08:08, 10 January 2014 (UTC)
The article now has commutativity 11 times and commutivity only once. As you say, Google also greatly prefers commutativity, so I think Wikipedia should consider that the correct term. I will delete the one reference to commutivity as an alternate term. Dirac66 (talk) 01:35, 11 January 2014 (UTC)

This probably needs to be clarified; especially if we are going to use the language of category theory elsewhere. Septentrionalis 01:55, 6 May 2006 (UTC)

"So, subtraction is commutative if and only if x = y and noncommutative for any other pair of real numbers."
Who actually uses that language? Anyone? Melchoir 14:59, 2 June 2006 (UTC)
I say we just x that line.--Emplynx 15:41, 5 June 2006 (UTC)
A more accurate formulation would be The pair (x,y) commutes (or:  x commutes with y) with respect to the operator subtraction, if and only if 
  
    
      
        x
        ≠
        y
      
    
    {\displaystyle x\neq y}
  
.  Since this does not hold for all pairs (x,y), the operation is not commutative.
This is somewhat lengthy, but perhaps worth it.  The point is that the whole operation as one unit is either commutative or non-commutative. JoergenB 11:58, 27 August 2006 (UTC)

Subtraction is of course commutative, because it is simply an addition of a negative number: 3 - 5 = -5 + 3. --2003:C1:4F04:B002:CD37:22CC:28C2:15DA (talk) 16:00, 27 August 2021 (UTC)
Exactly. Subtraction is an addition of a negative number. And Division is a multiplication with an inverse value. --2003:C1:4F06:DB90:9560:B63A:C09F:612B (talk) 16:17, 13 September 2023 (UTC)


### Operator definition
I think that a reference to the difference between function and infix notation for operators might be of use 
for the non-professionals.  I also think restoring a multiplication sign might help.  In my experience, fresh students often find some difficulty in relating the two statements f(y,z) = f(z,y) and yz = zy.
However, I noted that binary operations (as distinguished from operators???) are defined in a more restricted manner in these pages, as operators on one set.  While a commutative operator indeed must be defined on a 'Cartesian square' rather than on an arbitrary Cartesian product, its result may reside in a different set.  After all, an ordinary scalar product (aka 'dot product') on a vector space is ordinarily called a commutative operation. The discussion of `infix' versus `prefix' notation I only found in the `operator' items. JoergenB 12:34, 27 August 2006 (UTC)

Binary operation briefly addresses notation; perhaps too late? It sounds like you have some good ideas for the article. By all means, please take a shot at implementing them! Melchoir 16:33, 27 August 2006 (UTC)
Indeed, it does discuss notation; thanks for pointing it out.  I'll wait with editing, until I have a better grip on the conflicting binary operation definitions issue.  If indeed the range always should be a factor of the domain, for 'binary operations', then several items should be rewritten or removed (e.g., dot_product). JoergenB 01:06, 28 August 2006 (UTC)

this article (and the associativity article) should have some context
describing how the concept is (was) derived. Is it subject to 
proof? Is it axiomatic? drefty.mac 20:32, 26 October 2006 (UTC)

I added a simple example via the "+" operator at the beginning of the page because I think that, for "standard" readers this clarifies commutativity a lot in a sinlge line (see Associativity for comparison).

I added a whole new introduction. Please check if it is ok. ssepp(talk) 18:25, 31 May 2007 (UTC)

I clarified the introduction a bit and removed the example, it cluttered the lead.  I removed all math from the lead to make it more accessible since this article is in the basics section of math.  I switched the article to use the binary operation notation everywhere, and added the binary function notation in a generalizations section.  I plan to go back through later and add references. -Weston.pace 18:12, 12 July 2007 (UTC)
Does anyone know if multivariate functions are commutative if the order of their parameters doesn't matter? -Weston.pace 20:42, 12 July 2007 (UTC)

Well-written, easy to understand article. Short and to the point. I wish other mathematics articles were this easy to understand! ;-)
There are a few minor issues which need to be addressed before granting GA status. The 'common uses' section could overall be expanded a bit. For one, it mentions that, "the commutativity of well known operations (such as addition and multiplication on real and complex numbers) is often used (or implicitly assumed) in proofs." But there are no references of this, and no examples or wikilinks to proofs to illustrate this. As it stands right now, this is mainly just hearsay -- "Yeah, this is used in all sorts of math proofs,... yadayadayada!" Please provide some actual examples, and possibly a reference or two.
Under 'Examples: Commutative operations in math', it would help if the multiplication example of 2x3 = 3x2 matched up with the graphical example, which is 3x5 = 5x3.
The references to websites (links) need to have dates of retrieval added to them (e.g. when was the URL last accessed/checked? "Retrieved on August 3, 2007"). It would also be helpful if author and publisher (who the website belongs to) information was added to website links as well.
Other than that, this article looks great! Dr. Cash 20:43, 3 August 2007 (UTC)

This article has been reviewed as part of Wikipedia:WikiProject Good articles/Project quality task force. I believe the article currently meets the criteria and should remain listed as a Good article. The article history has been updated to reflect this review. (oldid reference #:152489007) OhanaUnitedTalk page 02:32, 24 August 2007 (UTC)

 This colourful image is used to illustrate the commutivity of multiplication: . The image may illustrate the point for the specific example of 5x3 = 3x5, but it is not persuasive in the general case that AxB = BxA. I think a two-dimensional array of objects (arranged in a grid) would be more appropriate:
                * * *
* * * * *       * * *
* * * * *   =   * * *
* * * * *       * * *
                * * *

This ASCII illustration is more persuasive that M rows containing N objects yields the same total as N columns containing M objects.
Is there someone with the necessay skills willing to create such an illustration? reetep 13:01, 30 October 2007 (UTC)
I'm getting rid of the picture of the grapes because it's just plain confusing. Three does not equal five.
Dithridge (talk) 22:01, 23 October 2008 (UTC)

Well I reverted it; sorry.  I think the image is good and useful: the rectangle of stars is not better, just different.   The grapes image shows clearly that commutativity of multiplication is far from obvious in real-life situations. Robinh 07:10, 24 October 2008 (UTC)
The graph picture is totally unobvious.  I just now realized that the left set had 5 grapes each and the right set had 3 grapes each.  If a picture is going to illustrate an idea it should be obvious from the first moment your eyes see it. Mikemill (talk) 16:35, 4 November 2008 (UTC)
I agree that the grape picture takes a second to realize what's going on. But I think it's great for multiplication because it isn't inherently obvious that multiplication is commutative. In the grape example, as in many applications of multiplication, one operand represents a quantity with units (5 grapes) and the other operand represents a unitless scalar (3). It is not especially intuitive that 3 * (5 grapes) is equal to 5 * (3 grapes) and I think the picture illustrates this well. 205.175.97.219 (talk) 08:12, 10 January 2014 (UTC)

wouldn't it be good to make fact that commutativity is just a kind of combination more visible? could chain some further lookuping by peoples 84.16.123.194 (talk) 05:17, 2 April 2008 (UTC)
{tshepo nobela ul}  —Preceding unsigned comment added by 196.21.218.142 (talk) 00:13, 2 November 2008 (UTC) 

I've restored the sentence

The associative property is closely related to the commutative property.
This was recently removed with "meaningless" as the edit summary.  To the contrary, the associative identity (a•b)•c = a•(b•c) for fixed a and c and arbitrary b is equivalent to the commutative identity (a•—)∘(—•c) = (—•c)∘(a•—).  I found a reference for this in the paper

Došen, Kosta; Petrić, Zoran (2006), "Associativity as Commutativity", J. Symbolic Logic, 71 (1): 217–226, doi:10.2178/jsl/1140641170 (arXiv:math/0506600)
I'd like to check whether additional references are available.  In particular, is anyone aware of an earlier reference for the claim that associativity is a kind of commutativity (and not just that associativity can be defined by a commutative diagram)? Michael Slone (talk) 01:12, 17 April 2009 (UTC)

This article could do with an explanation or brief discussion, or even a link to such, of non-commuting linear operators as used in quantum mechanics.  There is a link to particle statistics, labelled as a link to a discussion of commutativity in physics, but that page does not in fact discuss commutativity at all.  Nathaniel Virgo (talk) 11:30, 24 June 2010 (UTC)

I have now added a new section on commutativity and the uncertainty principle in quantum mechanics, along with a link to the main article on the Heisenberg uncertainty principle for further explanations. Dirac66 (talk) 01:20, 9 July 2010 (UTC)

Am I the only one who does not understand the diagram purporting to show that concatenation is non-commutative? (in the section "Noncommutative operations in everyday life")? What is the meaning of EA + ≠ + TTEA followed by ( ) ≠ ( ) EATTEA ??
A simpler and clearer example might be EA + TT = EATT ≠ TTEA = TT + EA. Dirac66 (talk) 02:42, 10 July 2010 (UTC)

Yes, despite my best efforts I cannot understand it either, your example is clearer.  —Preceding unsigned comment added by 80.169.145.30 (talk) 09:35, 12 August 2010 (UTC)
Thanks for the support. I have now changed the article to use the simpler example, although with one T because EAT and TEA are English words. I can't do a nice multicolour graphic like the previous one, but clearly it is more important to be understandable and mathematically correct. Dirac66 (talk) 02:25, 13 August 2010 (UTC)

In the Related Concepts section.  I think the image used to show the symmetry of addition isn't really clear to someone who doesn't know what they're looking for.  I think explicitly drawing the y=x line and pointing out that squares opposite each other from the line are the same shade would help clarify things in this way. 71.208.180.110 (talk) 19:41, 11 November 2010 (UTC)

New page title should be Commutative property. I object to the 2006 move on similar grounds to the objection voiced at #Old requested move, December 2005. Commutativity is not a word, at least not one commonly used by reliable sources in the mathematics community. I will suggest moving all the other -ivity words to their correct locations also. Cliff (talk) 21:28, 1 April 2011 (UTC)

Attempted move since no discussion for over a month, however was unable to perform the move because page currently exists as redirect here. Requesting administrator assistance below. Cliff (talk) 23:11, 5 May 2011 (UTC)


### Duh
Of course commutativity is a word and it's even a noun http://www.merriam-webster.com/dictionary/commutativity. As for the extremely dubious claim that "commutativity" as a word is not used by mathematicians try searching for it in Google Books. This article is written by the clueless, (GA) reviewed by the clueless and maintained by the even more clueless... Which is the essence of Wikipedia. JMP EAX (talk) 01:01, 26 August 2014 (UTC)
In fact if I search for "commutative property" in GB I find mostly texts aimed at high school and below, at least on the 1st page of hits. Contrast with the same search for commutativity. So the argument(s) really are "difficult word for American kids". JMP EAX (talk) 03:45, 26 August 2014 (UTC)

There is no excuse for using a more difficult word in a general encyclopaedia when a more easily understood one is available.  We are not writing papers for peer reviewed journals here and there should be no assumption of background knowledge.  If you have such a low opinion of Wikipedia what on earth are you doing here?  If you think everyone who writes for Wikipedia is clueless, what does that say about you?  But in any case there are, in fact, several notable mathematicians writing articles on Wikipedia. SpinningSpark 14:59, 5 November 2014 (UTC)

In this section we have;

Some truth functions are noncommutative, since the truth tables for the functions are different when one changes the order of the operands.
For example, Bpq = Cqp; Cpq = Bqp; Fpq = Gqp; Gpq = Fqp; Hpq = Iqp; Ipq = Hqp; Lpq = Mqp; Mpq = Lqp.
It may be that I am just not familiar with the notation being used here, but it seems pretty meaningless to me.  I especially don't understand why it is necessary to repeat the same thing over and over with different letters.  In the end, there is no statement or proof that Bpq != Bqp so I fail to see how this necessariy amounts to noncommutative.  Why not simply give an example of a truth table that does not commute such as,

SpinningSpark 15:25, 5 November 2014 (UTC)

Since there was no response, I have made the change. SpinningSpark 18:56, 30 November 2014 (UTC)
I notice that the Bpq, Cpq, etc. notation refers to the list of functions at Table of binary truth functions. However I suspects most non-specialists do not know this notation, so your version is clearly preferable.
We could also note that your truth tables are equivalent to the functions (A AND NOT B) and (B AND NOT A). (Write in proper notation if you wish; I haven't figured out yet how to parse these expressions.) Dirac66 (talk) 01:49, 2 December 2014 (UTC)
Ah, thanks for the link.  But even after I now understand the notation, I still maintain that the statement does not demonstrate non-commutativity.  The statement Bpq = Cqp; Cpq = Bqp does not exclude commutativity since Bpq = Bqp is still a possible solution.  The whole thing relies on the reader already knowing that Bpq does not commute, in which case the author may as well have stopped writing after "Some truth functions are noncommutative". At the very least some explanation or link for the notation should have been provided. I suspect that even a lot of specialists have not come across this notation, I spent years on this stuff as an undegraduate EE and never saw it.  I'll put your suggestion in the article, along with the link. SpinningSpark 10:01, 7 January 2015 (UTC)

As a general principle, any technical article should start with a simple introduction and simple example(s), with more complex material further down. This enables beginners to understand something at least. This article is viewed several hundred times a day (see Page View Statistics on history page) so it is reasonable to assume that some readers are beginners. 
Earlier today Bloodholds added an intro section Simply put to explain why addition and multiplication are commutative while subtraction and division are not. These operations are familiar to everyone and so form a better starting point than formal propositional logic which were the previous starting point.
However this new section was deleted by Wcherowi with the edit summary Article is not only about arithmetic operations. This was too simple. It is true that the deleted material was not perfect: the title Simply put might suggest that it includes everything in a simple way, and should be revised to Simple example. Also the table could be improved. But I do not think these are reasons to delete the material entirely; instead it should be used as a starting point and improved.
Also today was not the first attempt to add a simple intro. This talk page includes a section new introduction with a mention of an intro added in 2007. I checked back and found this which seems nice and clear, again based on simple arithmetic operations. I have not checked when it was removed. Dirac66 (talk) 18:33, 21 May 2015 (UTC)

I essentially agree with what you have written. I do not revert casually and I stewed over this one for quite a while. This was a judgement call as to whether or not this was an improvement to the page. As you noted, this entry needed some work and lately I've been letting "how much work is needed to fix an entry" be a factor in those decisions. What finally swayed me was the fact that this has a GA rating and in my mind a GA rating means that this type of addition should not be necessary. I have to admit however that I did not compare the current article to the one at the last time it was reviewed. Having done so now, I can see that the article has seriously drifted away from what it was. I still don't think that Bloodholds addition is what is needed — it belabors an obvious point and the tables are just too much — that content can be conveyed in the lead in one or two simple sentences. A restructuring of the page would be in order. This page should not start out with logic and advanced topics before stating and generalizing the simple arithmetic examples. The lead you mentioned was gone by July 2007, before the article achieved GA status. I think that with just a little effort this page can be put back into GA shape. Bill Cherowitzo  (talk) 21:09, 21 May 2015 (UTC)
I've given a first pass at rearranging the sections of this article. This might (or might not) be sufficient to address your concerns. Bill Cherowitzo  (talk) 21:49, 23 May 2015 (UTC)
Yes, that seems much better now. Beginners will now be able to read the first few sections without being lost. Thank you. Dirac66 (talk) 22:44, 23 May 2015 (UTC)

The article tries to show some everyday uses of communativity, but it is unclear.  specifically: 

    Putting on socks resembles a commutative operation since which sock is put on first is unimportant. Either way, the result (having both socks on), is the same. In contrast, putting on underwear and trousers is not commutative.

putting on socks would be commutative in terms of time: it doesn't matter whether you put on the right sock or the left sock first.  Most socks would also be commutative in space: putting the right sock on the left foot and the left on the right foot wouldn't matter.
But this can cause a confusion for some readers if they try to apply the analogy elsewhere.  For example: Shoes.
Shoes are time-commutative, it won't matter which shoe you put on when you do it correctly.
But they are not space-commutative:  if you put your right shoe on your left foot, your feet will hurt at the end of the day.  — Preceding unsigned comment added by 216.81.81.83 (talk) 13:01, 17 December 2015 (UTC) 

I concur with this.  It bothered me enough to attempt to learn how to contact an editor here on the backend just to bring it up, but I see it's already been mentioned. Rannison (talk) 08:29, 26 March 2019 (UTC)

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These statements from the article (the first two from the lede the third from the "Common Uses" section) needed reconciling:

"In mathematics, a binary operation is commutative if changing the order of the operands does not change the result."

"A corresponding property exists for binary relations; a binary relation is said to be symmetric if the relation applies regardless of the order of its operands;"

"The commutative property (or commutative law) is a property generally associated with binary operations and functions."

Assuming the statements are correct, I therefore changed the opening sentence of the article to make it clear that the two terms, commutative and symmetric, at least for binary mathematics, are synonymous.
LookingGlass (talk) 10:34, 12 November 2022 (UTC)

Although commutativity of binary operations is strongly related to symmetry of functions and relations, the term are not synonymous. So, I have reverted your edits, removed section "Common uses", and replaced "corresponding" with analogous". This should resolve your concerns. D.Lazard (talk) 11:44, 12 November 2022 (UTC)
@D.Lazard The key issue remains, so it doesn't; with regard to the minor edit to the semantics of the opening sentence: I hesitated, anticipating a reaction such as yours.  In the end though I found the existing too 'jarring' to leave.  Why you consider the word "analogous" an improvement to "corresponding" I cannot tell.  Whichever word word seems redundant as the sentence can be stripped of the prefixed clause entirely to say simply: "a binary relation is said to be symmetric if the relation applies regardless of the order of its operands" - which the lede says is what the word commutative means, and that THIS is the word that is used for binary etc etc!  If you have the time to consider more carefully, perhaps you will both see and address the issue raised.  Although no doubt the text you protect is clear to you, Wiki is intended for a wide audience - perhaps even one wide enough to embrace people as stupid as myself.  But thank-you!  You remind me why I abandoned editing here, and in doing so have saved me the time I was just about to invest - leaving me instead to waste it in this! LookingGlass (talk) 15:02, 12 November 2022 (UTC)
I agree that "analogous" is not better than "corresponding". The right word is "similar". You say "is what the word commutative mean". This is not true: "commutative" is an adjective whose meaning may depends on the noun it qualifies. Wikipedia must not establish any new meaning for a word. It must report on the common meanings. You may consider as an issue that nobody uses  "commutative" for relations, but you must live with that. If there is another issue in the article, you must say more precisely what it is. D.Lazard (talk) 15:40, 12 November 2022 (UTC)

We didn't mention: commutative property in filters and quantum entanglement. — Preceding unsigned comment added by LookingGlass (talk • contribs) 11:36, 12 November 2022 (UTC)

Pinging @Jacobolus and @D.Lazard. I have done the improvement in my sandbox before implementing to this article for a long time. However, does it actually helpful to readers? I was trying to bundling into a single section, which provides the conformtable of glancing the article, and add some sources. Dedhert.Jr (talk) 08:19, 9 March 2024 (UTC)

I don't understand clearly the point solved by this draft, and the advantage with respect to incremental improvements. Neverthetless,
I see in the draft: "In contrast, the commutative property states that the order of the terms does not affect the final result." This is wrong when there are more than three terms. A true version woud consist of replacing "in contrast" with "when the associative property holds".
In both versions, the etymology is unsourced and blatanly wrong: the suffix -atif, -ative is a standard way to built an adjective from a noun with the suffix -ation, which, itself, means "action of". So "commutative" means litteraly "relative to the action of exchanging".
D.Lazard (talk) 14:13, 9 March 2024 (UTC)
@D.Lazard Thank you for the comments here. All of these changes were copied from the article, and I prefer to tidy up the sections. However, it seems I could not find the sources about them, including the definition in general and related properties. Can you help me in this case? Dedhert.Jr (talk) 12:56, 10 March 2024 (UTC)
I am not good for providing citations. However, I rewrote the etymology sentence for keeping only evidences that can be verified in any good French dictionary. I do not know, and it is not important here, whether commutatif is derived directly from commuter or through commutation. Both are compatible with my formulation. D.Lazard (talk) 18:06, 10 March 2024 (UTC)

@D.Lazard, I see you have partially restored the edit in the definition. My question is, are those two totally different interpretations? To me, it seems the same thing, but each of them describes the operation and the elements under operation, or basically just confusing to read? Dedhert.Jr (talk) 10:05, 3 May 2025 (UTC)

There is only one definition, and no interpretation. By this edit, you removed "Mathematical" in the heading, which was a good thing, but your rewrite of the section content suffers of many issues: "without noticing the order of elements" is non sensical in a definition; "Equivalently, the first element commutes with another ..." is gramatically incorrect as the "first element" is quantified with "every"; also, "to commute" requires a definition that was in the preceding version, but was removed by this edit; "Mathematically speaking" is confusing by suggesting that the remainder of the article is not about mathematics; the remainder of the sentence is logically incorrect as fixing elements with "let" before quantifying them with "every".
On the opposite, the previous definition was perfectly correct, except for the order of the sentences ("to commute" was used (just) before being defined, and its definition is interleaved between the definition of "noncommutative" and the explanation of its meaning (this explanation may fundamental for some readers, and its removal is an other issue of your edit). D.Lazard (talk) 11:04, 4 May 2025 (UTC)
@D.Lazard Thank you for the explanation here. I have forgotten the main reason here, but for now, I am trying to WP:ONEDOWN target the school audience because the elementary school studies this. Luckily, I have some sources for the other definition. And in my head, I am thinking roughly in the following, but I am not sure if this is another disagreement:  Let 
  
    
      
        x
      
    
    {\displaystyle x}
  
 and 
  
    
      
        y
      
    
    {\displaystyle y}
  
 be element of a set 
  
    
      
        S
      
    
    {\displaystyle S}
  
. A binary operation 
  
    
      
        ∗
      
    
    {\displaystyle *}
  
 on a set S is commutative, or one says that 
  
    
      
        x
      
    
    {\displaystyle x}
  
 commutes with 
  
    
      
        y
      
    
    {\displaystyle y}
  
, or that 
  
    
      
        x
      
    
    {\displaystyle x}
  
 and 
  
    
      
        y
      
    
    {\displaystyle y}
  
 commute under 
  
    
      
        ∗
      
    
    {\displaystyle *}
  
 if:
  
    
      
        x
        ∗
        y
        =
        y
        ∗
        x
      
    
    {\displaystyle x*y=y*x}
  
 for all 
  
    
      
        x
        ,
        y
        ∈
        S
      
    
    {\displaystyle x,y\in S}
  
. So, an operation is commutative if every two elements commute. Likewise, an operation that is not commutative is said to be noncommutative. An operation is noncommutative if there are two elements such that 
  
    
      
        x
        ∗
        y
        ≠
        y
        ∗
        x
        .
      
    
    {\displaystyle x*y\neq y*x.}
  
 This does not exclude the possibility that some pairs of elements commute.
 Perhaps the definition in the body seems to have a repetitive formula, which is why I am trying to rewrite again, but maintaining the body. Speaking of non-commutative, I completely forgot to add it to be honest. Dedhert.Jr (talk) 13:39, 4 May 2025 (UTC)
Again, your suggesion is incorrect: one can not fix 
  
    
      
        x
      
    
    {\displaystyle x}
  
 and 
  
    
      
        y
      
    
    {\displaystyle y}
  
 with "Let 
  
    
      
        x
      
    
    {\displaystyle x}
  
 and 
  
    
      
        y
      
    
    {\displaystyle y}
  
", and then apply "for all" to them.
Again, There is only one definition of a commutative operation and one definition of the verb "to commute". These are not the same, and this makes your "So" very confusing. A reasonable question is the order in which the two concepts are defined. In an elementary textbook for young people, it may reasonable to define "to commute first". But a Wikipedia is not a textbook, and the audience of this article is not limited to students of elementary courses. Since "commutative operation" is much more commonly used than "two specific elements commute", it is better for WP to define "commutative operation" first. D.Lazard (talk) 14:41, 4 May 2025 (UTC)

My edits (mentioning the non-commutativity of addition in the context of infinite series) were reverted by @D.Lazard with the comment "Changing the term order in a series is not an operation", which I find confusing. What does that mean, and why does that justify the revert? From my POV, the operation is the sum, and the example shows how the sum is not commutative. In other words, changing the term order in a series is commutation; commutation may not be an operation (I would argue that it is), but a sum certainly is. — Guillaume Pelletier ~ 15:41, 4 October 2025 (UTC)

There is a clear definition of an operation in Operation (mathematics) and yor edit was placed in section § Noncommutative operations. "Changing the order of terms in a series" does satisfy this definition, and thus does not belong to this section. Moreover, the signs plus that appear in the notation of a series do no represent an addition individually but are a part of the notation of the series. So, introducing this example in this article would be confusing for readers who are not expert in series. In any case there are many more elementary examples that are less controversial; so, it is not helpful to consider series in this article. D.Lazard (talk) 17:15, 4 October 2025 (UTC)
Thank you for your response, and I am certainly prepared to let the matter rest... but I am just as confused as I was before.
When you write "'Changing the order of terms in a series' does satisfy this definition", do you mean "does not satisfy this definition"?
What do you mean by "the signs plus that appear in the notation of a series do no represent an addition individually but are a part of the notation of the series"? I have always understood summation to be repeated addition, so I'm really not sure what you mean here.
It seems to me like your response ignores what I am saying in my previous comment: changing term order is commutation, and the operation is the sum. If your response addresses this, I do not understand how.
Mainly in regards to #3, I am not sure why we are discussing whether commutation is an operation, it seems like an entirely separate matter. Would appreciate any further clarification for my own benefit. — Guillaume Pelletier ~ 15:01, 5 October 2025 (UTC)
Everything what you say may be useful for explaining what is a series (see the first lines of Series), but becomes incorrect if one wants using it in a rigorous reasonning. I acknowledge that the difference is rather subtle. This is the main reason for not introducing series in this elementary article. About your points
Yes, I omitted "not"
A (finite) summation is indeed a repeated addition. The sum of a series is the limit of its partial sums. So, the sum of a series is not exactly a repeated addition, and the sigh plus are here only for emphasizing that one talks of a series and not of a sequence.
Changing the term order in not commutation since the series is not an operation; it is a permutation.
D.Lazard (talk) 17:46, 5 October 2025 (UTC)
That is helpful, thank you. — Guillaume Pelletier ~ 14:33, 6 October 2025 (UTC)

As I understand the terms, these are the most elementary algebraic properties:
The reflexive property (of equality): a = a.  a = b if and only if b = a.
The symmetric property (of operation R): aRb = bRa.  (Notice the visible symmetry.)
The associative property (of operation R): (aRb)Rc = aR(bRc).
The commutative property obtains when both the symmetric and associative properties obtain.

 Because the commutative property presumes associativity, parentheses are no longer needed, allowing:
   (aRb)Rc = aRbRc.
 Because the commutative property presumes symmetry, the order of arguments is irrelevant, allowing:
   (aRb)Rc = aRcRb = bRaRc = bRcRa = cRaRb = cRbRa.

This is correct and substantially at odds with the contents of this article because the article does not address the associativity requirement for commutativity.
I also note that there is no Wikipedia entry for the symmetric property.
Correcting this mistake would not take too much work, but considering the widespread acceptance of the erroneous understanding of what commutativity is, I would like to ensure the debate is over before I invest time in this (I have other math to do). KikiLamb5 (talk) 21:03, 6 November 2025 (UTC)

You are confusing operation and relation. The property of freely exchange the arguments of a relation is called symmetry and is described in symmetric relation. There are many other sorts of symmetric  properties, including the property of a square to have many symmetries (see symmetric property) The property of freely exchange the operands of an operation is called commutative property, and is never called "symmetry".
Also, the commutative  and the associative propeties are independent: many associative operations are not commutative, and it exists commutative operations that are not associative, even if they are not commonly encountered. Parentheses are required for a commutative operation that is not associative.
So, the article is perfectly correct and does not need to be corrected. D.Lazard (talk) 22:28, 6 November 2025 (UTC)
“You are confusing operation and relation.”
All operators imply a mapping of all possible inputs to some outputs (here, allowing a null output if the operation is not defined for the input set, though we usually restrict (and I am, and have been, restricting my) conversation to a discussion of non-null outputs).  A mapping can be represented as a set of tuples.  Any set of tuples can be called a finitary relation.  So, every operation implies a finitary relation.
Yes, I used “R” in place of the operator, and an R, in such a glyphic context, often indicates an finitary relation in the process of being defined.  However, I specifically stated, “operation R” more than once.
So I am not confusing operations and relations beyond their natural coincidence.
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“The property of freely exchange the arguments of a relation is called symmetry …”
I am talking about algebraic symmetry*, and I assume that is what you mean.  I am not sure what you mean by “freely”.  “Arguments” is from the vocabulary of operations and functions, not finitary relations (though functions are mappings, mappings are sets of tuples, and any set of tuples is a finitary relation).  I think it is you who is confused about operations and relations.
By “algebraic symmetry”, I mean the symmetry in a written mathematical expression, not symmetry in mathematical expressions that are restricted to operations valid for algebraic numbers.
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“…and is described in symmetric relation.”
Thanks for the citation. What is stated on the Symmetric Relation page is what I understand as the “symmetric property” that some operations (not operators) have.
This operation/operator distinction refers to cases where it is not only the choice of operator that affects which representational properties obtain for a given valid mathematical expression, because the classes of the operands affect the properties as well, as can be seen in linear algebra, for example, where square diagonal matrices enjoy algebraic properties under matrix multiplication that other kinds of matrices do not.
Also, “operators” have generic operands, which is to say an operand of an operator has a specific data-type but no implied data.  “Operations” are instances of the computation that an operator encapsulates, and they have operands that represent data of the data-type specified by the operator.
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“There are many other sorts of symmetric properties…”
In my first line, I indicated I was talking about elementary algebraic properties.  Mentioning other meanings of “symmetry” without any specific correlation to algebraic expressions or their properties does not seem relevant to this discussion.
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“The property of freely exchange the operands of an operation is called commutative property…”
You use “commutative property” here, and refer to “symmetric relation” above.  I think anyone would have to admit that, as statements, they are remarkably similar and should both have the same class designation.  “Relation” already has a heavily defended mathematical meaning that is not well-aligned with the meaning we are writing about, and it is already easy to confuse the current mathematical meaning of “relation” with other meanings in casual conversation.  Giving “relation” another mathematical meaning would make writing more complicated. I think “property” is a better choice for the class of statements, and that choice has history.
In your cited statement above, again, it is unclear what “freely” means.  “Exchange” is also somewhat problematic.  [From this point I made some mistakes; so I am removing the rest of the original response and will correct.]
OriginalKikiLamb5 (talk) 22:14, 8 November 2025 (UTC)
DeletesKikiLamb5 (talk) 05:05, 9 November 2025 (UTC)