Non-associative operations

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There are lots of operations that are not commutative.

I'm looking for striking counter-examples of operations that are not associative.

Or may associativity be genuinely built-in the concept of an operation? May non-associative operations be of genuinely lesser importance?

Which role do algebraic structures with non-associative operations play?

There's a big gap between commutative and non-commuative algebraic structures (e.g. Abelian vs. non-Abelian groups or categories). Both kinds of algebraic structures are of equal importance. Does the same hold for assosiative vs. non-associative algebraic structures?

abstract-algebra examples-counterexamples big-list

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edited Aug 24 at 12:17

Stella Biderman

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asked Aug 23 at 17:07

Hans Stricker

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  Lie algebras$$?
  – Lord Shark the Unknown
  Aug 23 at 17:08
  

  
  
  
  


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  Multiplication in the octonions?
  – Mark S.
  Aug 23 at 17:11
  

  
  
  
  


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  cross product in vector calculus
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  At a pre-school algebra level, the operation of exponentiation is not associative --- $2^(3^2) = 2^9 = 512$ and $(2^3)^2 = 8^2 = 64.$ That is, 2^(3^2) differs from (2^3)^2.
  – Dave L. Renfro
  Aug 23 at 17:33
  
  

  
  
  
  


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  "At a pre-school algebra level" --- Looking at this the next day, I notice that the most natural reading of this has "pre" modifying "school", that is "(pre-school) algebra", rather having "pre" modifying "school algebra", that is "pre-(school algebra)", where the latter is of course what I intended. Incidentally, this distinction gives an example of verbal non-associativity.
  – Dave L. Renfro
  Aug 24 at 12:16
  
  

  
  
  
  

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show 7 more comments

up vote
33
down vote

favorite

17

There are lots of operations that are not commutative.

I'm looking for striking counter-examples of operations that are not associative.

Or may associativity be genuinely built-in the concept of an operation? May non-associative operations be of genuinely lesser importance?

Which role do algebraic structures with non-associative operations play?

There's a big gap between commutative and non-commuative algebraic structures (e.g. Abelian vs. non-Abelian groups or categories). Both kinds of algebraic structures are of equal importance. Does the same hold for assosiative vs. non-associative algebraic structures?

abstract-algebra examples-counterexamples big-list

share|cite|improve this question

edited Aug 24 at 12:17

Stella Biderman

26.2k63175

asked Aug 23 at 17:07

Hans Stricker

4,39613574

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  13
  

  
  
  
  
  
  

  
  Lie algebras$$?
  – Lord Shark the Unknown
  Aug 23 at 17:08
  

  
  
  
  


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  14
  

  
  
  
  
  
  

  
  Multiplication in the octonions?
  – Mark S.
  Aug 23 at 17:11
  

  
  
  
  


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  20
  

  
  
  
  
  
  

  
  cross product in vector calculus
  – trying
  Aug 23 at 17:11
  

  
  
  
  


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  At a pre-school algebra level, the operation of exponentiation is not associative --- $2^(3^2) = 2^9 = 512$ and $(2^3)^2 = 8^2 = 64.$ That is, 2^(3^2) differs from (2^3)^2.
  – Dave L. Renfro
  Aug 23 at 17:33
  
  

  
  
  
  


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  "At a pre-school algebra level" --- Looking at this the next day, I notice that the most natural reading of this has "pre" modifying "school", that is "(pre-school) algebra", rather having "pre" modifying "school algebra", that is "pre-(school algebra)", where the latter is of course what I intended. Incidentally, this distinction gives an example of verbal non-associativity.
  – Dave L. Renfro
  Aug 24 at 12:16
  
  

  
  
  
  

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show 7 more comments

up vote
33
down vote

favorite

17

up vote
33
down vote

favorite

17

17

There are lots of operations that are not commutative.

I'm looking for striking counter-examples of operations that are not associative.

Or may associativity be genuinely built-in the concept of an operation? May non-associative operations be of genuinely lesser importance?

Which role do algebraic structures with non-associative operations play?

There's a big gap between commutative and non-commuative algebraic structures (e.g. Abelian vs. non-Abelian groups or categories). Both kinds of algebraic structures are of equal importance. Does the same hold for assosiative vs. non-associative algebraic structures?

abstract-algebra examples-counterexamples big-list

share|cite|improve this question

edited Aug 24 at 12:17

Stella Biderman

26.2k63175

asked Aug 23 at 17:07

Hans Stricker

4,39613574

There are lots of operations that are not commutative.

I'm looking for striking counter-examples of operations that are not associative.

Or may associativity be genuinely built-in the concept of an operation? May non-associative operations be of genuinely lesser importance?

Which role do algebraic structures with non-associative operations play?

There's a big gap between commutative and non-commuative algebraic structures (e.g. Abelian vs. non-Abelian groups or categories). Both kinds of algebraic structures are of equal importance. Does the same hold for assosiative vs. non-associative algebraic structures?

abstract-algebra examples-counterexamples big-list

share|cite|improve this question

edited Aug 24 at 12:17

Stella Biderman

26.2k63175

asked Aug 23 at 17:07

Hans Stricker

4,39613574

share|cite|improve this question

share|cite|improve this question

edited Aug 24 at 12:17

Stella Biderman

26.2k63175

edited Aug 24 at 12:17

Stella Biderman

26.2k63175

edited Aug 24 at 12:17

Stella Biderman

26.2k63175

26.2k63175

asked Aug 23 at 17:07

Hans Stricker

4,39613574

asked Aug 23 at 17:07

Hans Stricker

4,39613574

asked Aug 23 at 17:07

Hans Stricker

4,39613574

4,39613574

• 
  
  
  

  
  13
  

  
  
  
  
  
  

  
  Lie algebras$$?
  – Lord Shark the Unknown
  Aug 23 at 17:08
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Multiplication in the octonions?
  – Mark S.
  Aug 23 at 17:11
  

  
  
  
  


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  20
  

  
  
  
  
  
  

  
  cross product in vector calculus
  – trying
  Aug 23 at 17:11
  

  
  
  
  


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  42
  

  
  
  
  
  
  

  
  At a pre-school algebra level, the operation of exponentiation is not associative --- $2^(3^2) = 2^9 = 512$ and $(2^3)^2 = 8^2 = 64.$ That is, 2^(3^2) differs from (2^3)^2.
  – Dave L. Renfro
  Aug 23 at 17:33
  
  

  
  
  
  


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  48
  

  
  
  
  
  
  

  
  "At a pre-school algebra level" --- Looking at this the next day, I notice that the most natural reading of this has "pre" modifying "school", that is "(pre-school) algebra", rather having "pre" modifying "school algebra", that is "pre-(school algebra)", where the latter is of course what I intended. Incidentally, this distinction gives an example of verbal non-associativity.
  – Dave L. Renfro
  Aug 24 at 12:16
  
  

  
  
  
  

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show 7 more comments

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  Lie algebras$$?
  – Lord Shark the Unknown
  Aug 23 at 17:08
  

  
  
  
  


• 
  
  
  

  
  14
  

  
  
  
  
  
  

  
  Multiplication in the octonions?
  – Mark S.
  Aug 23 at 17:11
  

  
  
  
  


• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  cross product in vector calculus
  – trying
  Aug 23 at 17:11
  

  
  
  
  


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  42
  

  
  
  
  
  
  

  
  At a pre-school algebra level, the operation of exponentiation is not associative --- $2^(3^2) = 2^9 = 512$ and $(2^3)^2 = 8^2 = 64.$ That is, 2^(3^2) differs from (2^3)^2.
  – Dave L. Renfro
  Aug 23 at 17:33
  
  

  
  
  
  


• 
  
  
  

  
  48
  

  
  
  
  
  
  

  
  "At a pre-school algebra level" --- Looking at this the next day, I notice that the most natural reading of this has "pre" modifying "school", that is "(pre-school) algebra", rather having "pre" modifying "school algebra", that is "pre-(school algebra)", where the latter is of course what I intended. Incidentally, this distinction gives an example of verbal non-associativity.
  – Dave L. Renfro
  Aug 24 at 12:16
  
  

  
  
  
  

13

13

Lie algebras$$?
– Lord Shark the Unknown
Aug 23 at 17:08

Lie algebras$$?
– Lord Shark the Unknown
Aug 23 at 17:08

14

14

Multiplication in the octonions?
– Mark S.
Aug 23 at 17:11

Multiplication in the octonions?
– Mark S.
Aug 23 at 17:11

20

20

cross product in vector calculus
– trying
Aug 23 at 17:11

cross product in vector calculus
– trying
Aug 23 at 17:11

42

42

At a pre-school algebra level, the operation of exponentiation is not associative --- $2^(3^2) = 2^9 = 512$ and $(2^3)^2 = 8^2 = 64.$ That is, 2^(3^2) differs from (2^3)^2.
– Dave L. Renfro
Aug 23 at 17:33

At a pre-school algebra level, the operation of exponentiation is not associative --- $2^(3^2) = 2^9 = 512$ and $(2^3)^2 = 8^2 = 64.$ That is, 2^(3^2) differs from (2^3)^2.
– Dave L. Renfro
Aug 23 at 17:33

48

48

"At a pre-school algebra level" --- Looking at this the next day, I notice that the most natural reading of this has "pre" modifying "school", that is "(pre-school) algebra", rather having "pre" modifying "school algebra", that is "pre-(school algebra)", where the latter is of course what I intended. Incidentally, this distinction gives an example of verbal non-associativity.
– Dave L. Renfro
Aug 24 at 12:16

"At a pre-school algebra level" --- Looking at this the next day, I notice that the most natural reading of this has "pre" modifying "school", that is "(pre-school) algebra", rather having "pre" modifying "school algebra", that is "pre-(school algebra)", where the latter is of course what I intended. Incidentally, this distinction gives an example of verbal non-associativity.
– Dave L. Renfro
Aug 24 at 12:16

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22 Answers
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91
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Subtraction:

$$
(1-2)-3 = -4
$$
$$
1-(2-3) = 2
$$

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edited Aug 23 at 20:53

Martin Argerami

117k1071165

answered Aug 23 at 18:22

Syd Henderson

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  +1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
  – DreamConspiracy
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  Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
  – User123456789
  Aug 24 at 9:29
  
  

  
  
  
  


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  No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
  – Marc van Leeuwen
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  @User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
  – David Z
  Aug 25 at 0:25
  

  
  
  
  


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  I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
  – Adayah
  Aug 25 at 18:56
  
  

  
  
  
  

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up vote
83
down vote

A simple example, and one that even elementary school students should be able to understand, is averaging.

average(average(a,b),c)

and

average(a,average(b,c))

are, generally, not equal to each other.

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answered Aug 23 at 21:49

Acccumulation

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  This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
  – goblin
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  If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
  – 6005
  Aug 25 at 23:16
  

  
  
  
  

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up vote
43
down vote

Division

$$(1div2)div4 = 1/8$$
$$1div(2div4) = 2$$

It is my perception that this is one of the main causes of fractions being difficult to grasp for a lot of people.

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answered Aug 23 at 20:54

rcollyer

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  It's also why that sign ÷ should be banned altogether.
  – Najib Idrissi
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  Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
  – Stella Biderman
  Aug 24 at 12:19
  
  

  
  
  
  


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  Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
  – Marc van Leeuwen
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  @StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
  – mr_e_man
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  Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
  – mr_e_man
  Aug 24 at 21:28
  
  

  
  
  
  

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up vote
39
down vote

The Cross Product

For example let $bf i$, $bf j$ and $bf k$ be the unit vectors. The cross product is the unique bilinear product that satisfies the formulae:

$bf i times j = bf k bf j times k = bf i bf k times i = bf j$

$bf j times i = bf -k bf k times j = bf -i bf i times k = bf -j$

$bf i times i = bf 0 bf j times j = bf 0 bf k times k = bf 0$

Now consider the expressions $(bf i times i) times j$ and $bf i times (i times j)$. The first evaluates to $bf 0 times j = bf 0$ while the second evaluates to $bf i times k = -j$.

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answered Aug 23 at 18:17

Daron

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  This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
  – Chase Ryan Taylor
  Aug 23 at 23:21
  

  
  
  
  


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  Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
  – Adam
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  @ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
  – Toby Bartels
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  @TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
  – Henning Makholm
  Aug 28 at 13:35
  

  
  
  
  


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  @HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
  – Toby Bartels
  Aug 31 at 7:03
  

  
  
  
  

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up vote
37
down vote

Exponentiation:
beginalign*
left(2^2right)^3&=2^6=64\
2^left(2^3right)&=2^8=256
endalign*

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answered Aug 24 at 11:26

Markus Scheuer

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  The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
  – Chase Ryan Taylor
  Aug 26 at 6:39
  

  
  
  
  


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  @ChaseRyanTaylor: Good to see the answer is useful. :-)
  – Markus Scheuer
  Aug 26 at 7:41
  

  
  
  
  

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up vote
28
down vote

In general, we do have:

$$ (A setminus B) setminus C neq A setminus (B setminus C)$$ that is, set difference is non-associative, and it is quite an important elementary operation.

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edited Aug 27 at 7:34

answered Aug 23 at 22:22

Right

1,036213

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up vote
25
down vote

How about the Rock-Paper-Scissors binary relation? It is a (commutative!) binary relation on $r, p, $ and $s$ given by the "winning" relations $rp=p$ (paper covers rock), $rs=r$ (rock smashes scissors), and $sp=s$ (scissors cut paper). For ties, we have $r^2=r, p^2=p$ and $s^2=s$.

In it, we have the expressions $r(ps)$ and $(rp)s$, which simplify as follows:
$$r(ps)=rs=r$$
and
$$(rp)s=ps=s$$

Here, non-associativity and the lack of an always winning position are closely related.

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answered Aug 26 at 0:54

Adam

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  Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
  – leftaroundabout
  Aug 26 at 21:04
  
  

  
  
  
  

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up vote
16
down vote

The Cartesian product is actually not associative, since if $A, B, C$ are sets, then

$$(A times B) times C = ((a, b), c) : a in A, b in B, c in C \ A times (B times C) = (a, (b, c)) : a in A, b in B, c in C $$

so $(A times B) times C neq A times (B times C)$.

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answered Aug 23 at 19:23

Adayah

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  There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
  – md2perpe
  Aug 23 at 20:18
  

  
  
  
  


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  While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
  – Stella Biderman
  Aug 24 at 12:17
  

  
  
  
  


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  While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
  – Frenzy Li
  Aug 24 at 18:35
  

  
  
  
  


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  @StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
  – Anton Fetisov
  Aug 25 at 1:17
  
  

  
  
  
  

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up vote
14
down vote

Take any abelian group $G$ and define $x * y := x - y$. Now,

$$
(x*y)*z = (x-y) - z = x - y - z
$$

and

$$
x * (y * z) = x - (y - z) = x - y + z
$$

So $*$ will be associative if and only if $2z = 0$ for all $z in G$. Thus, any abelian group with no $2$-torsion will give rise to such an example.

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answered Aug 23 at 17:18

Guido A.

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  How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
  – Marc van Leeuwen
  Aug 24 at 13:02
  
  

  
  
  
  


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  I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
  – Guido A.
  Aug 24 at 18:03
  

  
  
  
  

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up vote
11
down vote

Here's the simple example:

In $BbbR$, define $$a*b=2a+b$$ where the $+$ in the RHS is the usual addition in $BbbR$

Then $*$ is not associative

$2*(0*1)=2*(2(0)+1)=2*1=2(2)+1=5$

whereas

$(2*0)*1=(2(2)+0)*1=4*1=2(4)+1=9$

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edited Aug 23 at 17:31

answered Aug 23 at 17:19

Chinnapparaj R

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  Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
  – Hans Stricker
  Aug 23 at 17:23
  

  
  
  
  


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  2
  

  
  
  
  
  
  

  
  @HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
  – Christopher
  Aug 24 at 14:59
  

  
  
  
  

add a comment | 

up vote
11
down vote

This one may be a stretch, but hey: what about the English language!

Consider the constituent "The guardian of the king's throne".

That could mean "the throne of the guardian of the king" if we associate the words of the original -repeated genitive?- constituent as "(The guardian of the king)'s throne".
(I've put the words involving "genitive operation" in bold)

But it could also mean "the guardian of the throne of the king" if we associate the words of the original constituent as "The guardian of (the king's throne)".

This illustrates that in English the genitive operation may not always be nicely associative. I'm sure there are many other good (funnier) examples and better accounts/overviews of (non)associativity in language may be out there!

share|cite|improve this answer

edited Aug 25 at 8:59

answered Aug 24 at 22:42

Thibaut Demaerel

476311

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  +1 for the creativity, but I am not so sure that this can be strictly considered an operation
  – Ender Wiggins
  Aug 26 at 8:05
  

  
  
  
  

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up vote
10
down vote

Take the space $M_ntimes n(K)$ of all $ntimes n$ matrices over a field $K$ and consider the operation $[M,N]=M.N-N.M$. That operation is non-associative. That's a very natural example. But since an operation on a set $A$ is simply any map from $Atimes A$ into $A$, you can easily built lots of examples. For instance, in $mathbb R$, you define, say, $xodot y=x+e^y$. It is not associative, of course.

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answered Aug 23 at 17:12

José Carlos Santos

119k16101182

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  This being the commutator.
  – leftaroundabout
  Aug 26 at 21:01
  

  
  
  
  


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  @leftaroundabout Indeed.
  – José Carlos Santos
  Aug 26 at 21:06
  

  
  
  
  

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up vote
10
down vote

Which role do algebraic structures with non-associative operations
play?

Probably the most important such structure is a Lie Algebra. Lie algebras are fundamental to the study of Lie groups, and appear in many other areas of mathematics.

The “product” operation of Lie algebras is called the Lie bracket $[x,y]$, and it is non-associative except for rare, degenerate circumstances. It does satisfy the Jacobi identity

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$

Non-associative products with this property arise very naturally, often as commutators $[x,y]=xy-yx$ of some other associative, non-commutative operation.

share|cite|improve this answer

answered Aug 25 at 18:51

Alon Amit

10.5k3766

• 
  
  
  

  
  

  
  
  
  
  
  

  
  While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
  – Theo Bendit
  Aug 28 at 8:49
  

  
  
  
  


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  The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
  – Alon Amit
  Aug 28 at 8:50
  
  

  
  
  
  


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  Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
  – Theo Bendit
  Aug 28 at 8:55
  

  
  
  
  


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  We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
  – Alon Amit
  Aug 28 at 15:45
  

  
  
  
  

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up vote
9
down vote

Take a Steiner Triple System STS(n). It is a set $S$ of n elements with a set of subsets (called blocks) of $3$ elements from $S$ such that every pair of elements in $S$ is in exactly one of the blocks.

For example, the unique (up to isomorphism) STS(7) is
$a,b,c a,d,e a,f,g b,d,f b,e,g c,d,g c,e,f$

Now, given an STS define an operation $*$ on S:
$x*x:=x$ for all $x$ in $S$, and $x*y:=z$ where $x,y,z$ is a block. This operation is not associative.

https://en.wikipedia.org/wiki/Steiner_system

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answered Aug 23 at 21:12

Josh B.

2,40511323

add a comment | 

up vote
9
down vote

Similar to the case of set differences and exponentiation, implication is not associative:

$$ARightarrow (B Rightarrow C) notequiv (ARightarrow B) Rightarrow C$$

In fact

$$ARightarrow (BRightarrow C) equiv Awedge B Rightarrow C$$

The "usual" way of looking at implication is not in vitro, but in relation to $wedge$ (although alternatively there is something called an "implication algebra"). So it is a non-associative operation intimately related to an associative operation (this can be made precise).

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answered Aug 25 at 11:33

Stefan Perko

8,31611641

add a comment | 

up vote
8
down vote

Tensor product of (bi)modules over a quasi-bialgebra.

Let $Bbbk$ be a field and let $(A,m,u,Delta,varepsilon,Phi)$ be a quasi-bialgebra over $Bbbk$. The category $_AmathfrakM$ of left $A$-modules is a monoidal category with tensor product $otimes:=otimes_Bbbk$ and unit $Bbbk$ (the action on $Motimes N$ is the diagonal one given by $Delta$). In $_AmathfrakM$, $(Motimes N)otimes P$ and $Motimes (Notimes P)$ are different $A$-modules (this is due to the non-coassociativity of $Delta$) and you just have an isomorphism
$$alpha_M,N,P:(Motimes N)otimes P to Motimes (Notimes P),(motimes n)otimes pmapsto Phicdot(motimes (notimes p))$$
(the associativity constraint, in fact).

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answered Aug 24 at 11:41

Ender Wiggins

755320

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  Quasi bi-algebras must be terrible.
  – Allawonder
  Aug 25 at 20:35
  

  
  
  
  


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  Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
  – Ender Wiggins
  Aug 26 at 8:07
  

  
  
  
  


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  What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
  – Allawonder
  Aug 26 at 18:58
  

  
  
  
  


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  @Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
  – Ender Wiggins
  Aug 27 at 5:03
  

  
  
  
  


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  If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
  – Allawonder
  Aug 27 at 6:00
  
  

  
  
  
  

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up vote
8
down vote

Quite similar to the "cartesian product" example : composition of paths !

Let $X$ be a topological space, $alpha, beta : [0,1]to X$ continuous maps with $alpha(1)=beta(0)$, then we may define $alphastarbeta : [0,1]to X$ by concatenating in the obvious way.

However, $star$ is not associative (on the nose). This is very interesting because while $alphastar(betastargamma)neq (alphastarbeta)stargamma$ in general, this equation holds up to path homotopy.

It's very similar to the cartesian product example, because though $Atimes (Btimes C)neq (Atimes B)times C$ in general, this equality holds up to natural isomorphism.

As mentioned in the comments below the answer about cartesian products, studying structures that "should" be associative, but aren't associative "on the nose" is a big part of algebraic topology

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answered Aug 25 at 12:50

Max

10.8k1836

add a comment | 

up vote
7
down vote

Given an associative (but possibly non-commutative) algebra (over a field with characteristic other than $2$, e.g. $mathbb R$) we can make it into a so called Jordan algebra by equipping it with the symmetrized product

$$xcirc y := frac 1 2(xy + yx).$$

Being a Jordan algebra means that $circ$ is commutative and satisfies the Jordan identity

$$(xcirc y)circ(xcirc x) = xcirc (ycirc (xcirc x))$$

They have applications to quantum mechanics as (c.f. Jordan operator algebras) but are also studied for their one sake.

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answered Aug 25 at 20:37

Stefan Perko

8,31611641

add a comment | 

up vote
6
down vote

The multiplication of octonions is not associative, and they have many applications in mathematics and physics.

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edited Aug 25 at 15:09

answered Aug 25 at 12:41

mo-user

1812

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  I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
  – wgrenard
  Aug 25 at 15:05
  

  
  
  
  


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  Thanks for the comment, fixed.
  – mo-user
  Aug 25 at 15:10
  

  
  
  
  


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  More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
  – Ender Wiggins
  Aug 26 at 9:54
  

  
  
  
  

add a comment | 

up vote
4
down vote

Malcev algebras

Malcev algebras are also an example. A Malcev algebra $A$ over a filed $Bbbk$ is a vector space with an internal composition law $cdot$ such that
$$x^2=0,\
xcdot y=-ycdot x,\
J(x,y,z)cdot x = J(x,y,xcdot z),$$
where $J(x,y,z)=(xcdot y)cdot z+(ycdot z)cdot x+(zcdot x)cdot y$ (see Sagle, Malcev Algebras, §2).

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answered Aug 26 at 9:47

Ender Wiggins

755320

add a comment | 

up vote
2
down vote

Neither the Sheffer Stroke (NAND), nor the Pierce arrow (NOR) associate. Both have an important status in logic, as does implication. Reverse implication also does not associate.

I asked somewhat of a related question a few years ago: Is the Ratio of Associative Binary Operations to All Binary Operations on a Set of $n$ Elements Generally Small? Associative binary operations on a finite set are rare in comparison to the class of most binary operations on a finite set (statistically speaking). Additionally, there exist all of the counterxamples in these answers. So, we can't build in the concept of associativity to a general concept of an operation. Doing so would lead to many contradictions.

Given that a goal of the study of abstract algebra lies in studying all concrete algebras by abstract means, the study of non-associative algebras is more important than associative algebras, since non-associative algebras are vastly more common than associative algebras.

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answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

add a comment | 

up vote
1
down vote

Subtraction / Division have been mentoined of course, but let me explain why they are as important or even more important than Addition / Multiplication and arguably only less talked about since we are more comfortable working with associativity.

You can explain what a group is (almost) solely by its division (subtraction in the Abelian case).

A (division) group is thus a set $G$ with a binary operation $/ : Gtimes Gto G$ s.t.

• $a/a = b/b$,


• $fraca/a(a/a)/a = a$,


• $fracab/c = fraca(c/c)/c)/b$.

(I used "big fractions" for increased readibility).

A group morphism $f : Gto H$ is then just a function $Gto H$ s.t. $f(a/b) = f(a)/f(b)$.

There is a small "caveat". The "division groups" almost coincide with ordinary groups; just pick $gin G$ and let $e_G := g/g$, $acdot b := a/(e_G/b)$ and $a^-1 = e_G/a$. This leaves open the possibility of $G$ being empty. So in this theory, there is an "empty group".

If you don't want an "empty group", then we can simply insist on the existence of an identity $e_G$ s.t. $a/a = e_G$ for all $ain G$.

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answered Aug 25 at 20:13

Stefan Perko

8,31611641

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up vote
91
down vote

Subtraction:

$$
(1-2)-3 = -4
$$
$$
1-(2-3) = 2
$$

share|cite|improve this answer

edited Aug 23 at 20:53

Martin Argerami

117k1071165

answered Aug 23 at 18:22

Syd Henderson

1,55668

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  +1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
  – DreamConspiracy
  Aug 23 at 22:21
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
  – User123456789
  Aug 24 at 9:29
  
  

  
  
  
  


• 
  
  
  

  
  18
  

  
  
  
  
  
  

  
  No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
  – Marc van Leeuwen
  Aug 24 at 12:47
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
  – David Z
  Aug 25 at 0:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
  – Adayah
  Aug 25 at 18:56
  
  

  
  
  
  

 | 
show 5 more comments

up vote
91
down vote

Subtraction:

$$
(1-2)-3 = -4
$$
$$
1-(2-3) = 2
$$

share|cite|improve this answer

edited Aug 23 at 20:53

Martin Argerami

117k1071165

answered Aug 23 at 18:22

Syd Henderson

1,55668

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  +1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
  – DreamConspiracy
  Aug 23 at 22:21
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
  – User123456789
  Aug 24 at 9:29
  
  

  
  
  
  


• 
  
  
  

  
  18
  

  
  
  
  
  
  

  
  No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
  – Marc van Leeuwen
  Aug 24 at 12:47
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
  – David Z
  Aug 25 at 0:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
  – Adayah
  Aug 25 at 18:56
  
  

  
  
  
  

 | 
show 5 more comments

up vote
91
down vote

up vote
91
down vote

Subtraction:

$$
(1-2)-3 = -4
$$
$$
1-(2-3) = 2
$$

share|cite|improve this answer

edited Aug 23 at 20:53

Martin Argerami

117k1071165

answered Aug 23 at 18:22

Syd Henderson

1,55668

Subtraction:

$$
(1-2)-3 = -4
$$
$$
1-(2-3) = 2
$$

share|cite|improve this answer

edited Aug 23 at 20:53

Martin Argerami

117k1071165

answered Aug 23 at 18:22

Syd Henderson

1,55668

share|cite|improve this answer

share|cite|improve this answer

edited Aug 23 at 20:53

Martin Argerami

117k1071165

edited Aug 23 at 20:53

Martin Argerami

117k1071165

edited Aug 23 at 20:53

Martin Argerami

117k1071165

117k1071165

answered Aug 23 at 18:22

Syd Henderson

1,55668

answered Aug 23 at 18:22

Syd Henderson

1,55668

answered Aug 23 at 18:22

Syd Henderson

1,55668

1,55668

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  +1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
  – DreamConspiracy
  Aug 23 at 22:21
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
  – User123456789
  Aug 24 at 9:29
  
  

  
  
  
  


• 
  
  
  

  
  18
  

  
  
  
  
  
  

  
  No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
  – Marc van Leeuwen
  Aug 24 at 12:47
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
  – David Z
  Aug 25 at 0:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
  – Adayah
  Aug 25 at 18:56
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  +1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
  – DreamConspiracy
  Aug 23 at 22:21
  

  
  
  
  


• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
  – User123456789
  Aug 24 at 9:29
  
  

  
  
  
  


• 
  
  
  

  
  18
  

  
  
  
  
  
  

  
  No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
  – Marc van Leeuwen
  Aug 24 at 12:47
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
  – David Z
  Aug 25 at 0:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
  – Adayah
  Aug 25 at 18:56
  
  

  
  
  
  

8

8

+1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
– DreamConspiracy
Aug 23 at 22:21

+1, however this gets a little sketchy once you consider various possible definitions of subtraction and how it should be treated as an operation
– DreamConspiracy
Aug 23 at 22:21

6

6

Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
– User123456789
Aug 24 at 9:29

Also, this no longer holds if you consider subtraction as the adding of negative numbers, (so $1-2-3=1+(-2)+(-3)$).
– User123456789
Aug 24 at 9:29

18

18

No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
– Marc van Leeuwen
Aug 24 at 12:47

No matter how one defines subtraction (but assuming its usual relation to addition, which is associative), it is not associative (unless $x=0-x$ holds for all $x$, in which case subtraction is the same as addition), since it obeys the laws $(x-y)-z=x+(0-y)+(0-z)$ and $x-(y-z)=x+(0-y)+z$, and after some cancellation the two are equal iff $0-z=z$. (Here I avoided possible confusion by not using the negation operation $-x$.) The comment by @User123456789 only shows that you can work around the lack of associativity of subtraction by rewriting in terms of addition; this is beside the point.
– Marc van Leeuwen
Aug 24 at 12:47

3

3

@User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
– David Z
Aug 25 at 0:25

@User123456789 I think the relevant calculations would actually be $(1 - 2) - 3 = (1 + (-2)) + (-3) = (-1) + (-3) = -4$ and $1 - (2 - 3) = 1 + (-(2 + (-3))) = 1 + (-(-1)) = 1 + 1 = 2$.
– David Z
Aug 25 at 0:25

1

1

I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
– Adayah
Aug 25 at 18:56

I somehow find it hilarious that you can get 67 upvotes on a professional math website for something that looks like a solution to a 1st grader's exercise. (Though I'm not implying they are undeserved, for that is a perfectly correct answer to a not very much 1st grade question. )
– Adayah
Aug 25 at 18:56

 | 
show 5 more comments

up vote
83
down vote

A simple example, and one that even elementary school students should be able to understand, is averaging.

average(average(a,b),c)

and

average(a,average(b,c))

are, generally, not equal to each other.

share|cite|improve this answer

answered Aug 23 at 21:49

Acccumulation

5,2742515

• 
  
  
  

  
  29
  

  
  
  
  
  
  

  
  This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
  – goblin
  Aug 24 at 6:31
  
  

  
  
  
  


• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
  – 6005
  Aug 25 at 23:16
  

  
  
  
  

add a comment | 

up vote
83
down vote

A simple example, and one that even elementary school students should be able to understand, is averaging.

average(average(a,b),c)

and

average(a,average(b,c))

are, generally, not equal to each other.

share|cite|improve this answer

answered Aug 23 at 21:49

Acccumulation

5,2742515

• 
  
  
  

  
  29
  

  
  
  
  
  
  

  
  This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
  – goblin
  Aug 24 at 6:31
  
  

  
  
  
  


• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
  – 6005
  Aug 25 at 23:16
  

  
  
  
  

add a comment | 

up vote
83
down vote

up vote
83
down vote

A simple example, and one that even elementary school students should be able to understand, is averaging.

average(average(a,b),c)

and

average(a,average(b,c))

are, generally, not equal to each other.

share|cite|improve this answer

answered Aug 23 at 21:49

Acccumulation

5,2742515

A simple example, and one that even elementary school students should be able to understand, is averaging.

average(average(a,b),c)

and

average(a,average(b,c))

are, generally, not equal to each other.

share|cite|improve this answer

answered Aug 23 at 21:49

Acccumulation

5,2742515

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 21:49

Acccumulation

5,2742515

answered Aug 23 at 21:49

Acccumulation

5,2742515

answered Aug 23 at 21:49

Acccumulation

5,2742515

5,2742515

• 
  
  
  

  
  29
  

  
  
  
  
  
  

  
  This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
  – goblin
  Aug 24 at 6:31
  
  

  
  
  
  


• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
  – 6005
  Aug 25 at 23:16
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  29
  

  
  
  
  
  
  

  
  This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
  – goblin
  Aug 24 at 6:31
  
  

  
  
  
  


• 
  
  
  

  
  10
  

  
  
  
  
  
  

  
  If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
  – 6005
  Aug 25 at 23:16
  

  
  
  
  

29

29

This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
– goblin
Aug 24 at 6:31

This is an especially interesting example because averaging is commutative, idempotent, and medial. But it's not associative, of course.
– goblin
Aug 24 at 6:31

10

10

If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
– 6005
Aug 25 at 23:16

If anyone is looking for a definition of "medial": en.wikipedia.org/wiki/Medial_magma
– 6005
Aug 25 at 23:16

add a comment | 

up vote
43
down vote

Division

$$(1div2)div4 = 1/8$$
$$1div(2div4) = 2$$

It is my perception that this is one of the main causes of fractions being difficult to grasp for a lot of people.

share|cite|improve this answer

answered Aug 23 at 20:54

rcollyer

1,52211426

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  It's also why that sign ÷ should be banned altogether.
  – Najib Idrissi
  Aug 24 at 11:09
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
  – Stella Biderman
  Aug 24 at 12:19
  
  

  
  
  
  


• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
  – Marc van Leeuwen
  Aug 24 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
  – mr_e_man
  Aug 24 at 19:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
  – mr_e_man
  Aug 24 at 21:28
  
  

  
  
  
  

 | 
show 5 more comments

up vote
43
down vote

Division

$$(1div2)div4 = 1/8$$
$$1div(2div4) = 2$$

It is my perception that this is one of the main causes of fractions being difficult to grasp for a lot of people.

share|cite|improve this answer

answered Aug 23 at 20:54

rcollyer

1,52211426

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  It's also why that sign ÷ should be banned altogether.
  – Najib Idrissi
  Aug 24 at 11:09
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
  – Stella Biderman
  Aug 24 at 12:19
  
  

  
  
  
  


• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
  – Marc van Leeuwen
  Aug 24 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
  – mr_e_man
  Aug 24 at 19:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
  – mr_e_man
  Aug 24 at 21:28
  
  

  
  
  
  

 | 
show 5 more comments

up vote
43
down vote

up vote
43
down vote

Division

$$(1div2)div4 = 1/8$$
$$1div(2div4) = 2$$

It is my perception that this is one of the main causes of fractions being difficult to grasp for a lot of people.

share|cite|improve this answer

answered Aug 23 at 20:54

rcollyer

1,52211426

Division

$$(1div2)div4 = 1/8$$
$$1div(2div4) = 2$$

It is my perception that this is one of the main causes of fractions being difficult to grasp for a lot of people.

share|cite|improve this answer

answered Aug 23 at 20:54

rcollyer

1,52211426

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 20:54

rcollyer

1,52211426

answered Aug 23 at 20:54

rcollyer

1,52211426

answered Aug 23 at 20:54

rcollyer

1,52211426

1,52211426

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  It's also why that sign ÷ should be banned altogether.
  – Najib Idrissi
  Aug 24 at 11:09
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
  – Stella Biderman
  Aug 24 at 12:19
  
  

  
  
  
  


• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
  – Marc van Leeuwen
  Aug 24 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
  – mr_e_man
  Aug 24 at 19:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
  – mr_e_man
  Aug 24 at 21:28
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  It's also why that sign ÷ should be banned altogether.
  – Najib Idrissi
  Aug 24 at 11:09
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
  – Stella Biderman
  Aug 24 at 12:19
  
  

  
  
  
  


• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
  – Marc van Leeuwen
  Aug 24 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
  – mr_e_man
  Aug 24 at 19:05
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
  – mr_e_man
  Aug 24 at 21:28
  
  

  
  
  
  

6

6

It's also why that sign ÷ should be banned altogether.
– Najib Idrissi
Aug 24 at 11:09

It's also why that sign ÷ should be banned altogether.
– Najib Idrissi
Aug 24 at 11:09

4

4

Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
– Stella Biderman
Aug 24 at 12:19

Like the subtraction example, this can be explained away by defining $a div b = acdot b^-1$. In fact, I view this more as a problem with the symbol $div$ more than anything else.
– Stella Biderman
Aug 24 at 12:19

19

19

Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
– Marc van Leeuwen
Aug 24 at 12:57

Come one, there is nothing wrong with the symbol '$div$' (apart from forcing you to lift your pen from the paper twice) although in programming it is usually written '$/$'. It obeys similar laws to '$-$' (though has higher precedence, the same as '$times$'), in particular it associates to the left, so $adiv bdiv c$ is interpreted as $(adiv b)div c$ (or $a/b/c$ is interpreted as $(a/b)/c$, check with your favourite programming language).
– Marc van Leeuwen
Aug 24 at 12:57

4

4

@StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
– mr_e_man
Aug 24 at 19:05

@StellaBiderman -- That doesn't "explain it away", it's still non-associative: $$(acdot b^-1)cdot c^-1neq acdot(bcdot c^-1)^-1$$ I guess your point is that multiplication is more fundamental... I agree.
– mr_e_man
Aug 24 at 19:05

2

2

Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
– mr_e_man
Aug 24 at 21:28

Not directly related: I once guessed that $div$ was a fraction bar with placeholder dots, like the inner product $langlecdot, ,cdotrangle$. divisbyzero.com/2017/09/15/the-division-symbol-goes-viral
– mr_e_man
Aug 24 at 21:28

 | 
show 5 more comments

up vote
39
down vote

The Cross Product

For example let $bf i$, $bf j$ and $bf k$ be the unit vectors. The cross product is the unique bilinear product that satisfies the formulae:

$bf i times j = bf k bf j times k = bf i bf k times i = bf j$

$bf j times i = bf -k bf k times j = bf -i bf i times k = bf -j$

$bf i times i = bf 0 bf j times j = bf 0 bf k times k = bf 0$

Now consider the expressions $(bf i times i) times j$ and $bf i times (i times j)$. The first evaluates to $bf 0 times j = bf 0$ while the second evaluates to $bf i times k = -j$.

share|cite|improve this answer

answered Aug 23 at 18:17

Daron

4,87811024

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
  – Chase Ryan Taylor
  Aug 23 at 23:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
  – Adam
  Aug 26 at 0:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
  – Toby Bartels
  Aug 26 at 6:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
  – Henning Makholm
  Aug 28 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
  – Toby Bartels
  Aug 31 at 7:03
  

  
  
  
  

add a comment | 

up vote
39
down vote

The Cross Product

For example let $bf i$, $bf j$ and $bf k$ be the unit vectors. The cross product is the unique bilinear product that satisfies the formulae:

$bf i times j = bf k bf j times k = bf i bf k times i = bf j$

$bf j times i = bf -k bf k times j = bf -i bf i times k = bf -j$

$bf i times i = bf 0 bf j times j = bf 0 bf k times k = bf 0$

Now consider the expressions $(bf i times i) times j$ and $bf i times (i times j)$. The first evaluates to $bf 0 times j = bf 0$ while the second evaluates to $bf i times k = -j$.

share|cite|improve this answer

answered Aug 23 at 18:17

Daron

4,87811024

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
  – Chase Ryan Taylor
  Aug 23 at 23:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
  – Adam
  Aug 26 at 0:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
  – Toby Bartels
  Aug 26 at 6:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
  – Henning Makholm
  Aug 28 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
  – Toby Bartels
  Aug 31 at 7:03
  

  
  
  
  

add a comment | 

up vote
39
down vote

up vote
39
down vote

The Cross Product

For example let $bf i$, $bf j$ and $bf k$ be the unit vectors. The cross product is the unique bilinear product that satisfies the formulae:

$bf i times j = bf k bf j times k = bf i bf k times i = bf j$

$bf j times i = bf -k bf k times j = bf -i bf i times k = bf -j$

$bf i times i = bf 0 bf j times j = bf 0 bf k times k = bf 0$

Now consider the expressions $(bf i times i) times j$ and $bf i times (i times j)$. The first evaluates to $bf 0 times j = bf 0$ while the second evaluates to $bf i times k = -j$.

share|cite|improve this answer

answered Aug 23 at 18:17

Daron

4,87811024

The Cross Product

For example let $bf i$, $bf j$ and $bf k$ be the unit vectors. The cross product is the unique bilinear product that satisfies the formulae:

$bf i times j = bf k bf j times k = bf i bf k times i = bf j$

$bf j times i = bf -k bf k times j = bf -i bf i times k = bf -j$

$bf i times i = bf 0 bf j times j = bf 0 bf k times k = bf 0$

Now consider the expressions $(bf i times i) times j$ and $bf i times (i times j)$. The first evaluates to $bf 0 times j = bf 0$ while the second evaluates to $bf i times k = -j$.

share|cite|improve this answer

answered Aug 23 at 18:17

Daron

4,87811024

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 18:17

Daron

4,87811024

answered Aug 23 at 18:17

Daron

4,87811024

answered Aug 23 at 18:17

Daron

4,87811024

4,87811024

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
  – Chase Ryan Taylor
  Aug 23 at 23:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
  – Adam
  Aug 26 at 0:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
  – Toby Bartels
  Aug 26 at 6:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
  – Henning Makholm
  Aug 28 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
  – Toby Bartels
  Aug 31 at 7:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
  – Chase Ryan Taylor
  Aug 23 at 23:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
  – Adam
  Aug 26 at 0:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
  – Toby Bartels
  Aug 26 at 6:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
  – Henning Makholm
  Aug 28 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
  – Toby Bartels
  Aug 31 at 7:03
  

  
  
  
  

This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
– Chase Ryan Taylor
Aug 23 at 23:21

This would’ve been an awesome time to use the MathJax trick newcommandimathbf i
– Chase Ryan Taylor
Aug 23 at 23:21

1

1

Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
– Adam
Aug 26 at 0:59

Worth mentioning that the cross product can be thought of as the Lie bracket of the Lie algebra $mathfraksu(2)$ -- tying this to another answer.
– Adam
Aug 26 at 0:59

1

1

@ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
– Toby Bartels
Aug 26 at 6:26

@ChaseRyanTaylor : OMG, MathJax allows that now? And it even works in one LaTeX environment if defined in another!? You are my new best friend for telling me about this!
– Toby Bartels
Aug 26 at 6:26

@TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
– Henning Makholm
Aug 28 at 13:35

@TobyBartels: It is not recommended, because it affects all math on the same page. If someone else writes an answer where they do newcommandiimath instead, you get dueling definitions and random results as answers move around according to vote totals.
– Henning Makholm
Aug 28 at 13:35

@HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
– Toby Bartels
Aug 31 at 7:03

@HenningMakholm : OK, maybe that it is a little too magic. I will endeavour to be responsible!
– Toby Bartels
Aug 31 at 7:03

add a comment | 

up vote
37
down vote

Exponentiation:
beginalign*
left(2^2right)^3&=2^6=64\
2^left(2^3right)&=2^8=256
endalign*

share|cite|improve this answer

answered Aug 24 at 11:26

Markus Scheuer

57k452136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
  – Chase Ryan Taylor
  Aug 26 at 6:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor: Good to see the answer is useful. :-)
  – Markus Scheuer
  Aug 26 at 7:41
  

  
  
  
  

add a comment | 

up vote
37
down vote

Exponentiation:
beginalign*
left(2^2right)^3&=2^6=64\
2^left(2^3right)&=2^8=256
endalign*

share|cite|improve this answer

answered Aug 24 at 11:26

Markus Scheuer

57k452136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
  – Chase Ryan Taylor
  Aug 26 at 6:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor: Good to see the answer is useful. :-)
  – Markus Scheuer
  Aug 26 at 7:41
  

  
  
  
  

add a comment | 

up vote
37
down vote

up vote
37
down vote

Exponentiation:
beginalign*
left(2^2right)^3&=2^6=64\
2^left(2^3right)&=2^8=256
endalign*

share|cite|improve this answer

answered Aug 24 at 11:26

Markus Scheuer

57k452136

Exponentiation:
beginalign*
left(2^2right)^3&=2^6=64\
2^left(2^3right)&=2^8=256
endalign*

share|cite|improve this answer

answered Aug 24 at 11:26

Markus Scheuer

57k452136

share|cite|improve this answer

share|cite|improve this answer

answered Aug 24 at 11:26

Markus Scheuer

57k452136

answered Aug 24 at 11:26

Markus Scheuer

57k452136

answered Aug 24 at 11:26

Markus Scheuer

57k452136

57k452136

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
  – Chase Ryan Taylor
  Aug 26 at 6:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor: Good to see the answer is useful. :-)
  – Markus Scheuer
  Aug 26 at 7:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
  – Chase Ryan Taylor
  Aug 26 at 6:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ChaseRyanTaylor: Good to see the answer is useful. :-)
  – Markus Scheuer
  Aug 26 at 7:41
  

  
  
  
  

The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
– Chase Ryan Taylor
Aug 26 at 6:39

The Wikipedia hyperlink was a, uh, nice touch; I’d actually never heard of “exponentiation” before....
– Chase Ryan Taylor
Aug 26 at 6:39

1

1

@ChaseRyanTaylor: Good to see the answer is useful. :-)
– Markus Scheuer
Aug 26 at 7:41

@ChaseRyanTaylor: Good to see the answer is useful. :-)
– Markus Scheuer
Aug 26 at 7:41

add a comment | 

up vote
28
down vote

In general, we do have:

$$ (A setminus B) setminus C neq A setminus (B setminus C)$$ that is, set difference is non-associative, and it is quite an important elementary operation.

share|cite|improve this answer

edited Aug 27 at 7:34

answered Aug 23 at 22:22

Right

1,036213

add a comment | 

up vote
28
down vote

In general, we do have:

$$ (A setminus B) setminus C neq A setminus (B setminus C)$$ that is, set difference is non-associative, and it is quite an important elementary operation.

share|cite|improve this answer

edited Aug 27 at 7:34

answered Aug 23 at 22:22

Right

1,036213

add a comment | 

up vote
28
down vote

up vote
28
down vote

In general, we do have:

$$ (A setminus B) setminus C neq A setminus (B setminus C)$$ that is, set difference is non-associative, and it is quite an important elementary operation.

share|cite|improve this answer

edited Aug 27 at 7:34

answered Aug 23 at 22:22

Right

1,036213

In general, we do have:

$$ (A setminus B) setminus C neq A setminus (B setminus C)$$ that is, set difference is non-associative, and it is quite an important elementary operation.

share|cite|improve this answer

edited Aug 27 at 7:34

answered Aug 23 at 22:22

Right

1,036213

share|cite|improve this answer

share|cite|improve this answer

edited Aug 27 at 7:34

edited Aug 27 at 7:34

edited Aug 27 at 7:34

answered Aug 23 at 22:22

Right

1,036213

answered Aug 23 at 22:22

Right

1,036213

answered Aug 23 at 22:22

Right

1,036213

1,036213

add a comment | 

add a comment | 

up vote
25
down vote

How about the Rock-Paper-Scissors binary relation? It is a (commutative!) binary relation on $r, p, $ and $s$ given by the "winning" relations $rp=p$ (paper covers rock), $rs=r$ (rock smashes scissors), and $sp=s$ (scissors cut paper). For ties, we have $r^2=r, p^2=p$ and $s^2=s$.

In it, we have the expressions $r(ps)$ and $(rp)s$, which simplify as follows:
$$r(ps)=rs=r$$
and
$$(rp)s=ps=s$$

Here, non-associativity and the lack of an always winning position are closely related.

share|cite|improve this answer

answered Aug 26 at 0:54

Adam

1,342915

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
  – leftaroundabout
  Aug 26 at 21:04
  
  

  
  
  
  

add a comment | 

up vote
25
down vote

How about the Rock-Paper-Scissors binary relation? It is a (commutative!) binary relation on $r, p, $ and $s$ given by the "winning" relations $rp=p$ (paper covers rock), $rs=r$ (rock smashes scissors), and $sp=s$ (scissors cut paper). For ties, we have $r^2=r, p^2=p$ and $s^2=s$.

In it, we have the expressions $r(ps)$ and $(rp)s$, which simplify as follows:
$$r(ps)=rs=r$$
and
$$(rp)s=ps=s$$

Here, non-associativity and the lack of an always winning position are closely related.

share|cite|improve this answer

answered Aug 26 at 0:54

Adam

1,342915

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
  – leftaroundabout
  Aug 26 at 21:04
  
  

  
  
  
  

add a comment | 

up vote
25
down vote

up vote
25
down vote

How about the Rock-Paper-Scissors binary relation? It is a (commutative!) binary relation on $r, p, $ and $s$ given by the "winning" relations $rp=p$ (paper covers rock), $rs=r$ (rock smashes scissors), and $sp=s$ (scissors cut paper). For ties, we have $r^2=r, p^2=p$ and $s^2=s$.

In it, we have the expressions $r(ps)$ and $(rp)s$, which simplify as follows:
$$r(ps)=rs=r$$
and
$$(rp)s=ps=s$$

Here, non-associativity and the lack of an always winning position are closely related.

share|cite|improve this answer

answered Aug 26 at 0:54

Adam

1,342915

How about the Rock-Paper-Scissors binary relation? It is a (commutative!) binary relation on $r, p, $ and $s$ given by the "winning" relations $rp=p$ (paper covers rock), $rs=r$ (rock smashes scissors), and $sp=s$ (scissors cut paper). For ties, we have $r^2=r, p^2=p$ and $s^2=s$.

In it, we have the expressions $r(ps)$ and $(rp)s$, which simplify as follows:
$$r(ps)=rs=r$$
and
$$(rp)s=ps=s$$

Here, non-associativity and the lack of an always winning position are closely related.

share|cite|improve this answer

answered Aug 26 at 0:54

Adam

1,342915

share|cite|improve this answer

share|cite|improve this answer

answered Aug 26 at 0:54

Adam

1,342915

answered Aug 26 at 0:54

Adam

1,342915

answered Aug 26 at 0:54

Adam

1,342915

1,342915

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
  – leftaroundabout
  Aug 26 at 21:04
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
  – leftaroundabout
  Aug 26 at 21:04
  
  

  
  
  
  

Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
– leftaroundabout
Aug 26 at 21:04

Interesting way to look at it. I would have rather described Rock-Paper-Scissors as an anticommutative relation $mathrmr,p,stimesmathrmr,p,sto -1,0,1$.
– leftaroundabout
Aug 26 at 21:04

add a comment | 

up vote
16
down vote

The Cartesian product is actually not associative, since if $A, B, C$ are sets, then

$$(A times B) times C = ((a, b), c) : a in A, b in B, c in C \ A times (B times C) = (a, (b, c)) : a in A, b in B, c in C $$

so $(A times B) times C neq A times (B times C)$.

share|cite|improve this answer

answered Aug 23 at 19:23

Adayah

6,771823

• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
  – md2perpe
  Aug 23 at 20:18
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
  – Stella Biderman
  Aug 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
  – Frenzy Li
  Aug 24 at 18:35
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  @StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
  – Anton Fetisov
  Aug 25 at 1:17
  
  

  
  
  
  

add a comment | 

up vote
16
down vote

The Cartesian product is actually not associative, since if $A, B, C$ are sets, then

$$(A times B) times C = ((a, b), c) : a in A, b in B, c in C \ A times (B times C) = (a, (b, c)) : a in A, b in B, c in C $$

so $(A times B) times C neq A times (B times C)$.

share|cite|improve this answer

answered Aug 23 at 19:23

Adayah

6,771823

• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
  – md2perpe
  Aug 23 at 20:18
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
  – Stella Biderman
  Aug 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
  – Frenzy Li
  Aug 24 at 18:35
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  @StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
  – Anton Fetisov
  Aug 25 at 1:17
  
  

  
  
  
  

add a comment | 

up vote
16
down vote

up vote
16
down vote

The Cartesian product is actually not associative, since if $A, B, C$ are sets, then

$$(A times B) times C = ((a, b), c) : a in A, b in B, c in C \ A times (B times C) = (a, (b, c)) : a in A, b in B, c in C $$

so $(A times B) times C neq A times (B times C)$.

share|cite|improve this answer

answered Aug 23 at 19:23

Adayah

6,771823

The Cartesian product is actually not associative, since if $A, B, C$ are sets, then

$$(A times B) times C = ((a, b), c) : a in A, b in B, c in C \ A times (B times C) = (a, (b, c)) : a in A, b in B, c in C $$

so $(A times B) times C neq A times (B times C)$.

share|cite|improve this answer

answered Aug 23 at 19:23

Adayah

6,771823

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 19:23

Adayah

6,771823

answered Aug 23 at 19:23

Adayah

6,771823

answered Aug 23 at 19:23

Adayah

6,771823

6,771823

• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
  – md2perpe
  Aug 23 at 20:18
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
  – Stella Biderman
  Aug 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
  – Frenzy Li
  Aug 24 at 18:35
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  @StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
  – Anton Fetisov
  Aug 25 at 1:17
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  20
  

  
  
  
  
  
  

  
  There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
  – md2perpe
  Aug 23 at 20:18
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
  – Stella Biderman
  Aug 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
  – Frenzy Li
  Aug 24 at 18:35
  

  
  
  
  


• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  @StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
  – Anton Fetisov
  Aug 25 at 1:17
  
  

  
  
  
  

20

20

There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
– md2perpe
Aug 23 at 20:18

There is however an obvious bijection between $(A times B) times C$ and $A times (B times C)$. Therefore they are practically identical and are also normally treated as such.
– md2perpe
Aug 23 at 20:18

11

11

While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
– Stella Biderman
Aug 24 at 12:17

While this might be technically true, no one cares and putting it forth as an example is bad mathematics. You might as well argue that $5inmathbb R$ and $5+ 0iinmathbb C$ are different numbers.
– Stella Biderman
Aug 24 at 12:17

9

9

While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
– Frenzy Li
Aug 24 at 18:35

While the comments above have merit, it's still important to notice the technical differences between $=$ (equality) and $cong$ (isomorphism), especially for learners of algebra.
– Frenzy Li
Aug 24 at 18:35

11

11

@StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
– Anton Fetisov
Aug 25 at 1:17

@StellaBiderman "no one cares" - the entire field of algebraic topology begs to differ. The techniques of working with structures which "should" be associative but actually aren't have been a central point of study for the last 30 years, and the non-associativity of cartesian product is a prototypical example of such operations: the notion of $E_infty$-algebras is defined to mirror it (or rather the set coproduct - which seems even more obviously associative). Suffices to say that this structure is the raison d'etre for homotopy groups of spheres and Steenrod algebras.
– Anton Fetisov
Aug 25 at 1:17

add a comment | 

up vote
14
down vote

Take any abelian group $G$ and define $x * y := x - y$. Now,

$$
(x*y)*z = (x-y) - z = x - y - z
$$

and

$$
x * (y * z) = x - (y - z) = x - y + z
$$

So $*$ will be associative if and only if $2z = 0$ for all $z in G$. Thus, any abelian group with no $2$-torsion will give rise to such an example.

share|cite|improve this answer

answered Aug 23 at 17:18

Guido A.

4,562726

• 
  
  
  

  
  13
  

  
  
  
  
  
  

  
  How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
  – Marc van Leeuwen
  Aug 24 at 13:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
  – Guido A.
  Aug 24 at 18:03
  

  
  
  
  

add a comment | 

up vote
14
down vote

Take any abelian group $G$ and define $x * y := x - y$. Now,

$$
(x*y)*z = (x-y) - z = x - y - z
$$

and

$$
x * (y * z) = x - (y - z) = x - y + z
$$

So $*$ will be associative if and only if $2z = 0$ for all $z in G$. Thus, any abelian group with no $2$-torsion will give rise to such an example.

share|cite|improve this answer

answered Aug 23 at 17:18

Guido A.

4,562726

• 
  
  
  

  
  13
  

  
  
  
  
  
  

  
  How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
  – Marc van Leeuwen
  Aug 24 at 13:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
  – Guido A.
  Aug 24 at 18:03
  

  
  
  
  

add a comment | 

up vote
14
down vote

up vote
14
down vote

Take any abelian group $G$ and define $x * y := x - y$. Now,

$$
(x*y)*z = (x-y) - z = x - y - z
$$

and

$$
x * (y * z) = x - (y - z) = x - y + z
$$

So $*$ will be associative if and only if $2z = 0$ for all $z in G$. Thus, any abelian group with no $2$-torsion will give rise to such an example.

share|cite|improve this answer

answered Aug 23 at 17:18

Guido A.

4,562726

Take any abelian group $G$ and define $x * y := x - y$. Now,

$$
(x*y)*z = (x-y) - z = x - y - z
$$

and

$$
x * (y * z) = x - (y - z) = x - y + z
$$

So $*$ will be associative if and only if $2z = 0$ for all $z in G$. Thus, any abelian group with no $2$-torsion will give rise to such an example.

share|cite|improve this answer

answered Aug 23 at 17:18

Guido A.

4,562726

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 17:18

Guido A.

4,562726

answered Aug 23 at 17:18

Guido A.

4,562726

answered Aug 23 at 17:18

Guido A.

4,562726

4,562726

• 
  
  
  

  
  13
  

  
  
  
  
  
  

  
  How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
  – Marc van Leeuwen
  Aug 24 at 13:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
  – Guido A.
  Aug 24 at 18:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  13
  

  
  
  
  
  
  

  
  How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
  – Marc van Leeuwen
  Aug 24 at 13:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
  – Guido A.
  Aug 24 at 18:03
  

  
  
  
  

13

13

How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
– Marc van Leeuwen
Aug 24 at 13:02

How does writing the operator as $*$ rather than $-$ make any difference? In other words, this is the same answer as the one saying "subtraction". Which admittedly seems to be more recent than this answer. Nice to spot the $2$-torsion though, even if you could have said "any group that is not an elementary $2$-group".
– Marc van Leeuwen
Aug 24 at 13:02

2

2

I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
– Guido A.
Aug 24 at 18:03

I agree that the extra notation does not add much, it just felt clearer to me. As for the latter, I didn't know the definition for elementary $p$-groups, but I see how it conveys precisely what fails here (I don't know much abstract algebra). This is essentially subtraction, yes, but that answer was posterior and I figured I could point out this generalization since it is virtually effortless.
– Guido A.
Aug 24 at 18:03

add a comment | 

up vote
11
down vote

Here's the simple example:

In $BbbR$, define $$a*b=2a+b$$ where the $+$ in the RHS is the usual addition in $BbbR$

Then $*$ is not associative

$2*(0*1)=2*(2(0)+1)=2*1=2(2)+1=5$

whereas

$(2*0)*1=(2(2)+0)*1=4*1=2(4)+1=9$

share|cite|improve this answer

edited Aug 23 at 17:31

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
  – Hans Stricker
  Aug 23 at 17:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
  – Christopher
  Aug 24 at 14:59
  

  
  
  
  

add a comment | 

up vote
11
down vote

Here's the simple example:

In $BbbR$, define $$a*b=2a+b$$ where the $+$ in the RHS is the usual addition in $BbbR$

Then $*$ is not associative

$2*(0*1)=2*(2(0)+1)=2*1=2(2)+1=5$

whereas

$(2*0)*1=(2(2)+0)*1=4*1=2(4)+1=9$

share|cite|improve this answer

edited Aug 23 at 17:31

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
  – Hans Stricker
  Aug 23 at 17:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
  – Christopher
  Aug 24 at 14:59
  

  
  
  
  

add a comment | 

up vote
11
down vote

up vote
11
down vote

Here's the simple example:

In $BbbR$, define $$a*b=2a+b$$ where the $+$ in the RHS is the usual addition in $BbbR$

Then $*$ is not associative

$2*(0*1)=2*(2(0)+1)=2*1=2(2)+1=5$

whereas

$(2*0)*1=(2(2)+0)*1=4*1=2(4)+1=9$

share|cite|improve this answer

edited Aug 23 at 17:31

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

Here's the simple example:

In $BbbR$, define $$a*b=2a+b$$ where the $+$ in the RHS is the usual addition in $BbbR$

Then $*$ is not associative

$2*(0*1)=2*(2(0)+1)=2*1=2(2)+1=5$

whereas

$(2*0)*1=(2(2)+0)*1=4*1=2(4)+1=9$

share|cite|improve this answer

edited Aug 23 at 17:31

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

share|cite|improve this answer

share|cite|improve this answer

edited Aug 23 at 17:31

edited Aug 23 at 17:31

edited Aug 23 at 17:31

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

answered Aug 23 at 17:19

Chinnapparaj R

1,924318

1,924318

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
  – Hans Stricker
  Aug 23 at 17:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
  – Christopher
  Aug 24 at 14:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
  – Hans Stricker
  Aug 23 at 17:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
  – Christopher
  Aug 24 at 14:59
  

  
  
  
  

1

1

Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
– Hans Stricker
Aug 23 at 17:23

Thanks for that. But may be there some reason why this operation didn't gain much attention? Might it have to do with its non-associativity?
– Hans Stricker
Aug 23 at 17:23

2

2

@HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
– Christopher
Aug 24 at 14:59

@HansStricker no, it's because it's pretty much useless - in general, we only study mathematical objects (particularly algebraic ones, I think) just because they're there - they have to have some kind of external motivation or a lot of internal elegance. And this operation has neither.
– Christopher
Aug 24 at 14:59

add a comment | 

up vote
11
down vote

This one may be a stretch, but hey: what about the English language!

Consider the constituent "The guardian of the king's throne".

That could mean "the throne of the guardian of the king" if we associate the words of the original -repeated genitive?- constituent as "(The guardian of the king)'s throne".
(I've put the words involving "genitive operation" in bold)

But it could also mean "the guardian of the throne of the king" if we associate the words of the original constituent as "The guardian of (the king's throne)".

This illustrates that in English the genitive operation may not always be nicely associative. I'm sure there are many other good (funnier) examples and better accounts/overviews of (non)associativity in language may be out there!

share|cite|improve this answer

edited Aug 25 at 8:59

answered Aug 24 at 22:42

Thibaut Demaerel

476311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 for the creativity, but I am not so sure that this can be strictly considered an operation
  – Ender Wiggins
  Aug 26 at 8:05
  

  
  
  
  

add a comment | 

up vote
11
down vote

This one may be a stretch, but hey: what about the English language!

Consider the constituent "The guardian of the king's throne".

That could mean "the throne of the guardian of the king" if we associate the words of the original -repeated genitive?- constituent as "(The guardian of the king)'s throne".
(I've put the words involving "genitive operation" in bold)

But it could also mean "the guardian of the throne of the king" if we associate the words of the original constituent as "The guardian of (the king's throne)".

This illustrates that in English the genitive operation may not always be nicely associative. I'm sure there are many other good (funnier) examples and better accounts/overviews of (non)associativity in language may be out there!

share|cite|improve this answer

edited Aug 25 at 8:59

answered Aug 24 at 22:42

Thibaut Demaerel

476311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 for the creativity, but I am not so sure that this can be strictly considered an operation
  – Ender Wiggins
  Aug 26 at 8:05
  

  
  
  
  

add a comment | 

up vote
11
down vote

up vote
11
down vote

This one may be a stretch, but hey: what about the English language!

Consider the constituent "The guardian of the king's throne".

That could mean "the throne of the guardian of the king" if we associate the words of the original -repeated genitive?- constituent as "(The guardian of the king)'s throne".
(I've put the words involving "genitive operation" in bold)

But it could also mean "the guardian of the throne of the king" if we associate the words of the original constituent as "The guardian of (the king's throne)".

This illustrates that in English the genitive operation may not always be nicely associative. I'm sure there are many other good (funnier) examples and better accounts/overviews of (non)associativity in language may be out there!

share|cite|improve this answer

edited Aug 25 at 8:59

answered Aug 24 at 22:42

Thibaut Demaerel

476311

This one may be a stretch, but hey: what about the English language!

Consider the constituent "The guardian of the king's throne".

That could mean "the throne of the guardian of the king" if we associate the words of the original -repeated genitive?- constituent as "(The guardian of the king)'s throne".
(I've put the words involving "genitive operation" in bold)

But it could also mean "the guardian of the throne of the king" if we associate the words of the original constituent as "The guardian of (the king's throne)".

This illustrates that in English the genitive operation may not always be nicely associative. I'm sure there are many other good (funnier) examples and better accounts/overviews of (non)associativity in language may be out there!

share|cite|improve this answer

edited Aug 25 at 8:59

answered Aug 24 at 22:42

Thibaut Demaerel

476311

share|cite|improve this answer

share|cite|improve this answer

edited Aug 25 at 8:59

edited Aug 25 at 8:59

edited Aug 25 at 8:59

answered Aug 24 at 22:42

Thibaut Demaerel

476311

answered Aug 24 at 22:42

Thibaut Demaerel

476311

answered Aug 24 at 22:42

Thibaut Demaerel

476311

476311

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 for the creativity, but I am not so sure that this can be strictly considered an operation
  – Ender Wiggins
  Aug 26 at 8:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  +1 for the creativity, but I am not so sure that this can be strictly considered an operation
  – Ender Wiggins
  Aug 26 at 8:05
  

  
  
  
  

1

1

+1 for the creativity, but I am not so sure that this can be strictly considered an operation
– Ender Wiggins
Aug 26 at 8:05

+1 for the creativity, but I am not so sure that this can be strictly considered an operation
– Ender Wiggins
Aug 26 at 8:05

add a comment | 

up vote
10
down vote

Take the space $M_ntimes n(K)$ of all $ntimes n$ matrices over a field $K$ and consider the operation $[M,N]=M.N-N.M$. That operation is non-associative. That's a very natural example. But since an operation on a set $A$ is simply any map from $Atimes A$ into $A$, you can easily built lots of examples. For instance, in $mathbb R$, you define, say, $xodot y=x+e^y$. It is not associative, of course.

share|cite|improve this answer

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This being the commutator.
  – leftaroundabout
  Aug 26 at 21:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @leftaroundabout Indeed.
  – José Carlos Santos
  Aug 26 at 21:06
  

  
  
  
  

add a comment | 

up vote
10
down vote

Take the space $M_ntimes n(K)$ of all $ntimes n$ matrices over a field $K$ and consider the operation $[M,N]=M.N-N.M$. That operation is non-associative. That's a very natural example. But since an operation on a set $A$ is simply any map from $Atimes A$ into $A$, you can easily built lots of examples. For instance, in $mathbb R$, you define, say, $xodot y=x+e^y$. It is not associative, of course.

share|cite|improve this answer

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This being the commutator.
  – leftaroundabout
  Aug 26 at 21:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @leftaroundabout Indeed.
  – José Carlos Santos
  Aug 26 at 21:06
  

  
  
  
  

add a comment | 

up vote
10
down vote

up vote
10
down vote

Take the space $M_ntimes n(K)$ of all $ntimes n$ matrices over a field $K$ and consider the operation $[M,N]=M.N-N.M$. That operation is non-associative. That's a very natural example. But since an operation on a set $A$ is simply any map from $Atimes A$ into $A$, you can easily built lots of examples. For instance, in $mathbb R$, you define, say, $xodot y=x+e^y$. It is not associative, of course.

share|cite|improve this answer

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

Take the space $M_ntimes n(K)$ of all $ntimes n$ matrices over a field $K$ and consider the operation $[M,N]=M.N-N.M$. That operation is non-associative. That's a very natural example. But since an operation on a set $A$ is simply any map from $Atimes A$ into $A$, you can easily built lots of examples. For instance, in $mathbb R$, you define, say, $xodot y=x+e^y$. It is not associative, of course.

share|cite|improve this answer

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

answered Aug 23 at 17:12

José Carlos Santos

119k16101182

119k16101182

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This being the commutator.
  – leftaroundabout
  Aug 26 at 21:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @leftaroundabout Indeed.
  – José Carlos Santos
  Aug 26 at 21:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This being the commutator.
  – leftaroundabout
  Aug 26 at 21:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @leftaroundabout Indeed.
  – José Carlos Santos
  Aug 26 at 21:06
  

  
  
  
  

This being the commutator.
– leftaroundabout
Aug 26 at 21:01

This being the commutator.
– leftaroundabout
Aug 26 at 21:01

@leftaroundabout Indeed.
– José Carlos Santos
Aug 26 at 21:06

@leftaroundabout Indeed.
– José Carlos Santos
Aug 26 at 21:06

add a comment | 

up vote
10
down vote

Which role do algebraic structures with non-associative operations
play?

Probably the most important such structure is a Lie Algebra. Lie algebras are fundamental to the study of Lie groups, and appear in many other areas of mathematics.

The “product” operation of Lie algebras is called the Lie bracket $[x,y]$, and it is non-associative except for rare, degenerate circumstances. It does satisfy the Jacobi identity

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$

Non-associative products with this property arise very naturally, often as commutators $[x,y]=xy-yx$ of some other associative, non-commutative operation.

share|cite|improve this answer

answered Aug 25 at 18:51

Alon Amit

10.5k3766

• 
  
  
  

  
  

  
  
  
  
  
  

  
  While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
  – Theo Bendit
  Aug 28 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
  – Alon Amit
  Aug 28 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
  – Theo Bendit
  Aug 28 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
  – Alon Amit
  Aug 28 at 15:45
  

  
  
  
  

add a comment | 

up vote
10
down vote

Which role do algebraic structures with non-associative operations
play?

Probably the most important such structure is a Lie Algebra. Lie algebras are fundamental to the study of Lie groups, and appear in many other areas of mathematics.

The “product” operation of Lie algebras is called the Lie bracket $[x,y]$, and it is non-associative except for rare, degenerate circumstances. It does satisfy the Jacobi identity

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$

Non-associative products with this property arise very naturally, often as commutators $[x,y]=xy-yx$ of some other associative, non-commutative operation.

share|cite|improve this answer

answered Aug 25 at 18:51

Alon Amit

10.5k3766

• 
  
  
  

  
  

  
  
  
  
  
  

  
  While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
  – Theo Bendit
  Aug 28 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
  – Alon Amit
  Aug 28 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
  – Theo Bendit
  Aug 28 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
  – Alon Amit
  Aug 28 at 15:45
  

  
  
  
  

add a comment | 

up vote
10
down vote

up vote
10
down vote

Which role do algebraic structures with non-associative operations
play?

Probably the most important such structure is a Lie Algebra. Lie algebras are fundamental to the study of Lie groups, and appear in many other areas of mathematics.

The “product” operation of Lie algebras is called the Lie bracket $[x,y]$, and it is non-associative except for rare, degenerate circumstances. It does satisfy the Jacobi identity

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$

Non-associative products with this property arise very naturally, often as commutators $[x,y]=xy-yx$ of some other associative, non-commutative operation.

share|cite|improve this answer

answered Aug 25 at 18:51

Alon Amit

10.5k3766

Which role do algebraic structures with non-associative operations
play?

Probably the most important such structure is a Lie Algebra. Lie algebras are fundamental to the study of Lie groups, and appear in many other areas of mathematics.

The “product” operation of Lie algebras is called the Lie bracket $[x,y]$, and it is non-associative except for rare, degenerate circumstances. It does satisfy the Jacobi identity

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$

Non-associative products with this property arise very naturally, often as commutators $[x,y]=xy-yx$ of some other associative, non-commutative operation.

share|cite|improve this answer

answered Aug 25 at 18:51

Alon Amit

10.5k3766

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 18:51

Alon Amit

10.5k3766

answered Aug 25 at 18:51

Alon Amit

10.5k3766

answered Aug 25 at 18:51

Alon Amit

10.5k3766

10.5k3766

• 
  
  
  

  
  

  
  
  
  
  
  

  
  While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
  – Theo Bendit
  Aug 28 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
  – Alon Amit
  Aug 28 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
  – Theo Bendit
  Aug 28 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
  – Alon Amit
  Aug 28 at 15:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
  – Theo Bendit
  Aug 28 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
  – Alon Amit
  Aug 28 at 8:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
  – Theo Bendit
  Aug 28 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
  – Alon Amit
  Aug 28 at 15:45
  

  
  
  
  

While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
– Theo Bendit
Aug 28 at 8:49

While I don't doubt the importance of Lie Algebras, I wouldn't consider them more important than subtraction, division, or exponentiation!
– Theo Bendit
Aug 28 at 8:49

The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
– Alon Amit
Aug 28 at 8:50

The difference is that none of these operations is the primary defining operation of any algebraic structure. The Lie bracket certainly is.
– Alon Amit
Aug 28 at 8:50

Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
– Theo Bendit
Aug 28 at 8:55

Well, there is Stefan Perko's answer below, but I'd more argue that subtraction/division and exponentiation are inextricable parts of groups, which truly are the most important algebraic structure. It may not be explicitly listed in the axioms, but it's a part of all groups.
– Theo Bendit
Aug 28 at 8:55

We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
– Alon Amit
Aug 28 at 15:45

We are past splitting hairs here, but: Subtraction is a fine example of a non-associative operation. Groups are not a good example of a non-associative algebraic structure.
– Alon Amit
Aug 28 at 15:45

add a comment | 

up vote
9
down vote

Take a Steiner Triple System STS(n). It is a set $S$ of n elements with a set of subsets (called blocks) of $3$ elements from $S$ such that every pair of elements in $S$ is in exactly one of the blocks.

For example, the unique (up to isomorphism) STS(7) is
$a,b,c a,d,e a,f,g b,d,f b,e,g c,d,g c,e,f$

Now, given an STS define an operation $*$ on S:
$x*x:=x$ for all $x$ in $S$, and $x*y:=z$ where $x,y,z$ is a block. This operation is not associative.

https://en.wikipedia.org/wiki/Steiner_system

share|cite|improve this answer

answered Aug 23 at 21:12

Josh B.

2,40511323

add a comment | 

up vote
9
down vote

Take a Steiner Triple System STS(n). It is a set $S$ of n elements with a set of subsets (called blocks) of $3$ elements from $S$ such that every pair of elements in $S$ is in exactly one of the blocks.

For example, the unique (up to isomorphism) STS(7) is
$a,b,c a,d,e a,f,g b,d,f b,e,g c,d,g c,e,f$

Now, given an STS define an operation $*$ on S:
$x*x:=x$ for all $x$ in $S$, and $x*y:=z$ where $x,y,z$ is a block. This operation is not associative.

https://en.wikipedia.org/wiki/Steiner_system

share|cite|improve this answer

answered Aug 23 at 21:12

Josh B.

2,40511323

add a comment | 

up vote
9
down vote

up vote
9
down vote

Take a Steiner Triple System STS(n). It is a set $S$ of n elements with a set of subsets (called blocks) of $3$ elements from $S$ such that every pair of elements in $S$ is in exactly one of the blocks.

For example, the unique (up to isomorphism) STS(7) is
$a,b,c a,d,e a,f,g b,d,f b,e,g c,d,g c,e,f$

Now, given an STS define an operation $*$ on S:
$x*x:=x$ for all $x$ in $S$, and $x*y:=z$ where $x,y,z$ is a block. This operation is not associative.

https://en.wikipedia.org/wiki/Steiner_system

share|cite|improve this answer

answered Aug 23 at 21:12

Josh B.

2,40511323

Take a Steiner Triple System STS(n). It is a set $S$ of n elements with a set of subsets (called blocks) of $3$ elements from $S$ such that every pair of elements in $S$ is in exactly one of the blocks.

For example, the unique (up to isomorphism) STS(7) is
$a,b,c a,d,e a,f,g b,d,f b,e,g c,d,g c,e,f$

Now, given an STS define an operation $*$ on S:
$x*x:=x$ for all $x$ in $S$, and $x*y:=z$ where $x,y,z$ is a block. This operation is not associative.

https://en.wikipedia.org/wiki/Steiner_system

share|cite|improve this answer

answered Aug 23 at 21:12

Josh B.

2,40511323

share|cite|improve this answer

share|cite|improve this answer

answered Aug 23 at 21:12

Josh B.

2,40511323

answered Aug 23 at 21:12

Josh B.

2,40511323

answered Aug 23 at 21:12

Josh B.

2,40511323

2,40511323

add a comment | 

add a comment | 

up vote
9
down vote

Similar to the case of set differences and exponentiation, implication is not associative:

$$ARightarrow (B Rightarrow C) notequiv (ARightarrow B) Rightarrow C$$

In fact

$$ARightarrow (BRightarrow C) equiv Awedge B Rightarrow C$$

The "usual" way of looking at implication is not in vitro, but in relation to $wedge$ (although alternatively there is something called an "implication algebra"). So it is a non-associative operation intimately related to an associative operation (this can be made precise).

share|cite|improve this answer

answered Aug 25 at 11:33

Stefan Perko

8,31611641

add a comment | 

up vote
9
down vote

Similar to the case of set differences and exponentiation, implication is not associative:

$$ARightarrow (B Rightarrow C) notequiv (ARightarrow B) Rightarrow C$$

In fact

$$ARightarrow (BRightarrow C) equiv Awedge B Rightarrow C$$

The "usual" way of looking at implication is not in vitro, but in relation to $wedge$ (although alternatively there is something called an "implication algebra"). So it is a non-associative operation intimately related to an associative operation (this can be made precise).

share|cite|improve this answer

answered Aug 25 at 11:33

Stefan Perko

8,31611641

add a comment | 

up vote
9
down vote

up vote
9
down vote

Similar to the case of set differences and exponentiation, implication is not associative:

$$ARightarrow (B Rightarrow C) notequiv (ARightarrow B) Rightarrow C$$

In fact

$$ARightarrow (BRightarrow C) equiv Awedge B Rightarrow C$$

The "usual" way of looking at implication is not in vitro, but in relation to $wedge$ (although alternatively there is something called an "implication algebra"). So it is a non-associative operation intimately related to an associative operation (this can be made precise).

share|cite|improve this answer

answered Aug 25 at 11:33

Stefan Perko

8,31611641

Similar to the case of set differences and exponentiation, implication is not associative:

$$ARightarrow (B Rightarrow C) notequiv (ARightarrow B) Rightarrow C$$

In fact

$$ARightarrow (BRightarrow C) equiv Awedge B Rightarrow C$$

The "usual" way of looking at implication is not in vitro, but in relation to $wedge$ (although alternatively there is something called an "implication algebra"). So it is a non-associative operation intimately related to an associative operation (this can be made precise).

share|cite|improve this answer

answered Aug 25 at 11:33

Stefan Perko

8,31611641

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 11:33

Stefan Perko

8,31611641

answered Aug 25 at 11:33

Stefan Perko

8,31611641

answered Aug 25 at 11:33

Stefan Perko

8,31611641

8,31611641

add a comment | 

add a comment | 

up vote
8
down vote

Tensor product of (bi)modules over a quasi-bialgebra.

Let $Bbbk$ be a field and let $(A,m,u,Delta,varepsilon,Phi)$ be a quasi-bialgebra over $Bbbk$. The category $_AmathfrakM$ of left $A$-modules is a monoidal category with tensor product $otimes:=otimes_Bbbk$ and unit $Bbbk$ (the action on $Motimes N$ is the diagonal one given by $Delta$). In $_AmathfrakM$, $(Motimes N)otimes P$ and $Motimes (Notimes P)$ are different $A$-modules (this is due to the non-coassociativity of $Delta$) and you just have an isomorphism
$$alpha_M,N,P:(Motimes N)otimes P to Motimes (Notimes P),(motimes n)otimes pmapsto Phicdot(motimes (notimes p))$$
(the associativity constraint, in fact).

share|cite|improve this answer

answered Aug 24 at 11:41

Ender Wiggins

755320

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quasi bi-algebras must be terrible.
  – Allawonder
  Aug 25 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
  – Ender Wiggins
  Aug 26 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
  – Allawonder
  Aug 26 at 18:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
  – Ender Wiggins
  Aug 27 at 5:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
  – Allawonder
  Aug 27 at 6:00
  
  

  
  
  
  

add a comment | 

up vote
8
down vote

Tensor product of (bi)modules over a quasi-bialgebra.

Let $Bbbk$ be a field and let $(A,m,u,Delta,varepsilon,Phi)$ be a quasi-bialgebra over $Bbbk$. The category $_AmathfrakM$ of left $A$-modules is a monoidal category with tensor product $otimes:=otimes_Bbbk$ and unit $Bbbk$ (the action on $Motimes N$ is the diagonal one given by $Delta$). In $_AmathfrakM$, $(Motimes N)otimes P$ and $Motimes (Notimes P)$ are different $A$-modules (this is due to the non-coassociativity of $Delta$) and you just have an isomorphism
$$alpha_M,N,P:(Motimes N)otimes P to Motimes (Notimes P),(motimes n)otimes pmapsto Phicdot(motimes (notimes p))$$
(the associativity constraint, in fact).

share|cite|improve this answer

answered Aug 24 at 11:41

Ender Wiggins

755320

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quasi bi-algebras must be terrible.
  – Allawonder
  Aug 25 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
  – Ender Wiggins
  Aug 26 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
  – Allawonder
  Aug 26 at 18:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
  – Ender Wiggins
  Aug 27 at 5:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
  – Allawonder
  Aug 27 at 6:00
  
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

Tensor product of (bi)modules over a quasi-bialgebra.

Let $Bbbk$ be a field and let $(A,m,u,Delta,varepsilon,Phi)$ be a quasi-bialgebra over $Bbbk$. The category $_AmathfrakM$ of left $A$-modules is a monoidal category with tensor product $otimes:=otimes_Bbbk$ and unit $Bbbk$ (the action on $Motimes N$ is the diagonal one given by $Delta$). In $_AmathfrakM$, $(Motimes N)otimes P$ and $Motimes (Notimes P)$ are different $A$-modules (this is due to the non-coassociativity of $Delta$) and you just have an isomorphism
$$alpha_M,N,P:(Motimes N)otimes P to Motimes (Notimes P),(motimes n)otimes pmapsto Phicdot(motimes (notimes p))$$
(the associativity constraint, in fact).

share|cite|improve this answer

answered Aug 24 at 11:41

Ender Wiggins

755320

Tensor product of (bi)modules over a quasi-bialgebra.

Let $Bbbk$ be a field and let $(A,m,u,Delta,varepsilon,Phi)$ be a quasi-bialgebra over $Bbbk$. The category $_AmathfrakM$ of left $A$-modules is a monoidal category with tensor product $otimes:=otimes_Bbbk$ and unit $Bbbk$ (the action on $Motimes N$ is the diagonal one given by $Delta$). In $_AmathfrakM$, $(Motimes N)otimes P$ and $Motimes (Notimes P)$ are different $A$-modules (this is due to the non-coassociativity of $Delta$) and you just have an isomorphism
$$alpha_M,N,P:(Motimes N)otimes P to Motimes (Notimes P),(motimes n)otimes pmapsto Phicdot(motimes (notimes p))$$
(the associativity constraint, in fact).

share|cite|improve this answer

answered Aug 24 at 11:41

Ender Wiggins

755320

share|cite|improve this answer

share|cite|improve this answer

answered Aug 24 at 11:41

Ender Wiggins

755320

answered Aug 24 at 11:41

Ender Wiggins

755320

answered Aug 24 at 11:41

Ender Wiggins

755320

755320

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quasi bi-algebras must be terrible.
  – Allawonder
  Aug 25 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
  – Ender Wiggins
  Aug 26 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
  – Allawonder
  Aug 26 at 18:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
  – Ender Wiggins
  Aug 27 at 5:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
  – Allawonder
  Aug 27 at 6:00
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Quasi bi-algebras must be terrible.
  – Allawonder
  Aug 25 at 20:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
  – Ender Wiggins
  Aug 26 at 8:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
  – Allawonder
  Aug 26 at 18:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
  – Ender Wiggins
  Aug 27 at 5:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
  – Allawonder
  Aug 27 at 6:00
  
  

  
  
  
  

2

2

Quasi bi-algebras must be terrible.
– Allawonder
Aug 25 at 20:35

Quasi bi-algebras must be terrible.
– Allawonder
Aug 25 at 20:35

Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
– Ender Wiggins
Aug 26 at 8:07

Only if you perform computations with elements, otherwise (categorically speaking) they behave even better than bialgebras
– Ender Wiggins
Aug 26 at 8:07

What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
– Allawonder
Aug 26 at 18:58

What would be the point of developing a theory one could not feel? How could one ever do without computation, no matter how minimal?
– Allawonder
Aug 26 at 18:58

2

2

@Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
– Ender Wiggins
Aug 27 at 5:03

@Allawonder I am not saying that you have to do without computations, I am just saying that they behave nicely from a categorical point of view. You can do computations, in fact. And with some practice in Sweedler Notation, they are not more difficult than ordinary bialgebra's ones. Simply, they are longer. But mathematics is not simply computing, in my opinion: which computations do you find in Category Theory or Logic? Quasi-bialgebras are nice eg because they are exactly those algebras whose category of modules is monoidal with tensor product the one over $Bbbk$ and unit object $Bbbk$
– Ender Wiggins
Aug 27 at 5:03

If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
– Allawonder
Aug 27 at 6:00

If one has to perform some computation, then they must be terrible, as I first said. Note that this says nothing about their niceness from other perspectives; neither does it mean I'm reducing math to computation. I only make the remark that these bi-algebras must be terrible, at least computationally. This, we both agree, is true. This does not make them any less interesting or worthy as an answer to the question in OP, to be even clearer.
– Allawonder
Aug 27 at 6:00

add a comment | 

up vote
8
down vote

Quite similar to the "cartesian product" example : composition of paths !

Let $X$ be a topological space, $alpha, beta : [0,1]to X$ continuous maps with $alpha(1)=beta(0)$, then we may define $alphastarbeta : [0,1]to X$ by concatenating in the obvious way.

However, $star$ is not associative (on the nose). This is very interesting because while $alphastar(betastargamma)neq (alphastarbeta)stargamma$ in general, this equation holds up to path homotopy.

It's very similar to the cartesian product example, because though $Atimes (Btimes C)neq (Atimes B)times C$ in general, this equality holds up to natural isomorphism.

As mentioned in the comments below the answer about cartesian products, studying structures that "should" be associative, but aren't associative "on the nose" is a big part of algebraic topology

share|cite|improve this answer

answered Aug 25 at 12:50

Max

10.8k1836

add a comment | 

up vote
8
down vote

Quite similar to the "cartesian product" example : composition of paths !

Let $X$ be a topological space, $alpha, beta : [0,1]to X$ continuous maps with $alpha(1)=beta(0)$, then we may define $alphastarbeta : [0,1]to X$ by concatenating in the obvious way.

However, $star$ is not associative (on the nose). This is very interesting because while $alphastar(betastargamma)neq (alphastarbeta)stargamma$ in general, this equation holds up to path homotopy.

It's very similar to the cartesian product example, because though $Atimes (Btimes C)neq (Atimes B)times C$ in general, this equality holds up to natural isomorphism.

As mentioned in the comments below the answer about cartesian products, studying structures that "should" be associative, but aren't associative "on the nose" is a big part of algebraic topology

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answered Aug 25 at 12:50

Max

10.8k1836

add a comment | 

up vote
8
down vote

up vote
8
down vote

Quite similar to the "cartesian product" example : composition of paths !

Let $X$ be a topological space, $alpha, beta : [0,1]to X$ continuous maps with $alpha(1)=beta(0)$, then we may define $alphastarbeta : [0,1]to X$ by concatenating in the obvious way.

However, $star$ is not associative (on the nose). This is very interesting because while $alphastar(betastargamma)neq (alphastarbeta)stargamma$ in general, this equation holds up to path homotopy.

It's very similar to the cartesian product example, because though $Atimes (Btimes C)neq (Atimes B)times C$ in general, this equality holds up to natural isomorphism.

As mentioned in the comments below the answer about cartesian products, studying structures that "should" be associative, but aren't associative "on the nose" is a big part of algebraic topology

share|cite|improve this answer

answered Aug 25 at 12:50

Max

10.8k1836

Quite similar to the "cartesian product" example : composition of paths !

Let $X$ be a topological space, $alpha, beta : [0,1]to X$ continuous maps with $alpha(1)=beta(0)$, then we may define $alphastarbeta : [0,1]to X$ by concatenating in the obvious way.

However, $star$ is not associative (on the nose). This is very interesting because while $alphastar(betastargamma)neq (alphastarbeta)stargamma$ in general, this equation holds up to path homotopy.

It's very similar to the cartesian product example, because though $Atimes (Btimes C)neq (Atimes B)times C$ in general, this equality holds up to natural isomorphism.

As mentioned in the comments below the answer about cartesian products, studying structures that "should" be associative, but aren't associative "on the nose" is a big part of algebraic topology

share|cite|improve this answer

answered Aug 25 at 12:50

Max

10.8k1836

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 12:50

Max

10.8k1836

answered Aug 25 at 12:50

Max

10.8k1836

answered Aug 25 at 12:50

Max

10.8k1836

10.8k1836

add a comment | 

add a comment | 

up vote
7
down vote

Given an associative (but possibly non-commutative) algebra (over a field with characteristic other than $2$, e.g. $mathbb R$) we can make it into a so called Jordan algebra by equipping it with the symmetrized product

$$xcirc y := frac 1 2(xy + yx).$$

Being a Jordan algebra means that $circ$ is commutative and satisfies the Jordan identity

$$(xcirc y)circ(xcirc x) = xcirc (ycirc (xcirc x))$$

They have applications to quantum mechanics as (c.f. Jordan operator algebras) but are also studied for their one sake.

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answered Aug 25 at 20:37

Stefan Perko

8,31611641

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up vote
7
down vote

Given an associative (but possibly non-commutative) algebra (over a field with characteristic other than $2$, e.g. $mathbb R$) we can make it into a so called Jordan algebra by equipping it with the symmetrized product

$$xcirc y := frac 1 2(xy + yx).$$

Being a Jordan algebra means that $circ$ is commutative and satisfies the Jordan identity

$$(xcirc y)circ(xcirc x) = xcirc (ycirc (xcirc x))$$

They have applications to quantum mechanics as (c.f. Jordan operator algebras) but are also studied for their one sake.

share|cite|improve this answer

answered Aug 25 at 20:37

Stefan Perko

8,31611641

add a comment | 

up vote
7
down vote

up vote
7
down vote

Given an associative (but possibly non-commutative) algebra (over a field with characteristic other than $2$, e.g. $mathbb R$) we can make it into a so called Jordan algebra by equipping it with the symmetrized product

$$xcirc y := frac 1 2(xy + yx).$$

Being a Jordan algebra means that $circ$ is commutative and satisfies the Jordan identity

$$(xcirc y)circ(xcirc x) = xcirc (ycirc (xcirc x))$$

They have applications to quantum mechanics as (c.f. Jordan operator algebras) but are also studied for their one sake.

share|cite|improve this answer

answered Aug 25 at 20:37

Stefan Perko

8,31611641

Given an associative (but possibly non-commutative) algebra (over a field with characteristic other than $2$, e.g. $mathbb R$) we can make it into a so called Jordan algebra by equipping it with the symmetrized product

$$xcirc y := frac 1 2(xy + yx).$$

Being a Jordan algebra means that $circ$ is commutative and satisfies the Jordan identity

$$(xcirc y)circ(xcirc x) = xcirc (ycirc (xcirc x))$$

They have applications to quantum mechanics as (c.f. Jordan operator algebras) but are also studied for their one sake.

share|cite|improve this answer

answered Aug 25 at 20:37

Stefan Perko

8,31611641

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 20:37

Stefan Perko

8,31611641

answered Aug 25 at 20:37

Stefan Perko

8,31611641

answered Aug 25 at 20:37

Stefan Perko

8,31611641

8,31611641

add a comment | 

add a comment | 

up vote
6
down vote

The multiplication of octonions is not associative, and they have many applications in mathematics and physics.

share|cite|improve this answer

edited Aug 25 at 15:09

answered Aug 25 at 12:41

mo-user

1812

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
  – wgrenard
  Aug 25 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the comment, fixed.
  – mo-user
  Aug 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
  – Ender Wiggins
  Aug 26 at 9:54
  

  
  
  
  

add a comment | 

up vote
6
down vote

The multiplication of octonions is not associative, and they have many applications in mathematics and physics.

share|cite|improve this answer

edited Aug 25 at 15:09

answered Aug 25 at 12:41

mo-user

1812

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
  – wgrenard
  Aug 25 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the comment, fixed.
  – mo-user
  Aug 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
  – Ender Wiggins
  Aug 26 at 9:54
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

The multiplication of octonions is not associative, and they have many applications in mathematics and physics.

share|cite|improve this answer

edited Aug 25 at 15:09

answered Aug 25 at 12:41

mo-user

1812

The multiplication of octonions is not associative, and they have many applications in mathematics and physics.

share|cite|improve this answer

edited Aug 25 at 15:09

answered Aug 25 at 12:41

mo-user

1812

share|cite|improve this answer

share|cite|improve this answer

edited Aug 25 at 15:09

edited Aug 25 at 15:09

edited Aug 25 at 15:09

answered Aug 25 at 12:41

mo-user

1812

answered Aug 25 at 12:41

mo-user

1812

answered Aug 25 at 12:41

mo-user

1812

1812

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
  – wgrenard
  Aug 25 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the comment, fixed.
  – mo-user
  Aug 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
  – Ender Wiggins
  Aug 26 at 9:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
  – wgrenard
  Aug 25 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the comment, fixed.
  – mo-user
  Aug 25 at 15:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
  – Ender Wiggins
  Aug 26 at 9:54
  

  
  
  
  

I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
– wgrenard
Aug 25 at 15:05

I think you mean to say that multiplication of octonions is not associative. The octonions themselves are not an operation, so it doesn’t make sense to talk about their associativity or non-associativity.
– wgrenard
Aug 25 at 15:05

Thanks for the comment, fixed.
– mo-user
Aug 25 at 15:10

Thanks for the comment, fixed.
– mo-user
Aug 25 at 15:10

More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
– Ender Wiggins
Aug 26 at 9:54

More in general, this property is true for any algebra obtained via the Cayley-Dickson process after the octonions.
– Ender Wiggins
Aug 26 at 9:54

add a comment | 

up vote
4
down vote

Malcev algebras

Malcev algebras are also an example. A Malcev algebra $A$ over a filed $Bbbk$ is a vector space with an internal composition law $cdot$ such that
$$x^2=0,\
xcdot y=-ycdot x,\
J(x,y,z)cdot x = J(x,y,xcdot z),$$
where $J(x,y,z)=(xcdot y)cdot z+(ycdot z)cdot x+(zcdot x)cdot y$ (see Sagle, Malcev Algebras, §2).

share|cite|improve this answer

answered Aug 26 at 9:47

Ender Wiggins

755320

add a comment | 

up vote
4
down vote

Malcev algebras

Malcev algebras are also an example. A Malcev algebra $A$ over a filed $Bbbk$ is a vector space with an internal composition law $cdot$ such that
$$x^2=0,\
xcdot y=-ycdot x,\
J(x,y,z)cdot x = J(x,y,xcdot z),$$
where $J(x,y,z)=(xcdot y)cdot z+(ycdot z)cdot x+(zcdot x)cdot y$ (see Sagle, Malcev Algebras, §2).

share|cite|improve this answer

answered Aug 26 at 9:47

Ender Wiggins

755320

add a comment | 

up vote
4
down vote

up vote
4
down vote

Malcev algebras

Malcev algebras are also an example. A Malcev algebra $A$ over a filed $Bbbk$ is a vector space with an internal composition law $cdot$ such that
$$x^2=0,\
xcdot y=-ycdot x,\
J(x,y,z)cdot x = J(x,y,xcdot z),$$
where $J(x,y,z)=(xcdot y)cdot z+(ycdot z)cdot x+(zcdot x)cdot y$ (see Sagle, Malcev Algebras, §2).

share|cite|improve this answer

answered Aug 26 at 9:47

Ender Wiggins

755320

Malcev algebras

Malcev algebras are also an example. A Malcev algebra $A$ over a filed $Bbbk$ is a vector space with an internal composition law $cdot$ such that
$$x^2=0,\
xcdot y=-ycdot x,\
J(x,y,z)cdot x = J(x,y,xcdot z),$$
where $J(x,y,z)=(xcdot y)cdot z+(ycdot z)cdot x+(zcdot x)cdot y$ (see Sagle, Malcev Algebras, §2).

share|cite|improve this answer

answered Aug 26 at 9:47

Ender Wiggins

755320

share|cite|improve this answer

share|cite|improve this answer

answered Aug 26 at 9:47

Ender Wiggins

755320

answered Aug 26 at 9:47

Ender Wiggins

755320

answered Aug 26 at 9:47

Ender Wiggins

755320

755320

add a comment | 

add a comment | 

up vote
2
down vote

Neither the Sheffer Stroke (NAND), nor the Pierce arrow (NOR) associate. Both have an important status in logic, as does implication. Reverse implication also does not associate.

I asked somewhat of a related question a few years ago: Is the Ratio of Associative Binary Operations to All Binary Operations on a Set of $n$ Elements Generally Small? Associative binary operations on a finite set are rare in comparison to the class of most binary operations on a finite set (statistically speaking). Additionally, there exist all of the counterxamples in these answers. So, we can't build in the concept of associativity to a general concept of an operation. Doing so would lead to many contradictions.

Given that a goal of the study of abstract algebra lies in studying all concrete algebras by abstract means, the study of non-associative algebras is more important than associative algebras, since non-associative algebras are vastly more common than associative algebras.

share|cite|improve this answer

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

add a comment | 

up vote
2
down vote

Neither the Sheffer Stroke (NAND), nor the Pierce arrow (NOR) associate. Both have an important status in logic, as does implication. Reverse implication also does not associate.

I asked somewhat of a related question a few years ago: Is the Ratio of Associative Binary Operations to All Binary Operations on a Set of $n$ Elements Generally Small? Associative binary operations on a finite set are rare in comparison to the class of most binary operations on a finite set (statistically speaking). Additionally, there exist all of the counterxamples in these answers. So, we can't build in the concept of associativity to a general concept of an operation. Doing so would lead to many contradictions.

Given that a goal of the study of abstract algebra lies in studying all concrete algebras by abstract means, the study of non-associative algebras is more important than associative algebras, since non-associative algebras are vastly more common than associative algebras.

share|cite|improve this answer

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

add a comment | 

up vote
2
down vote

up vote
2
down vote

Neither the Sheffer Stroke (NAND), nor the Pierce arrow (NOR) associate. Both have an important status in logic, as does implication. Reverse implication also does not associate.

I asked somewhat of a related question a few years ago: Is the Ratio of Associative Binary Operations to All Binary Operations on a Set of $n$ Elements Generally Small? Associative binary operations on a finite set are rare in comparison to the class of most binary operations on a finite set (statistically speaking). Additionally, there exist all of the counterxamples in these answers. So, we can't build in the concept of associativity to a general concept of an operation. Doing so would lead to many contradictions.

Given that a goal of the study of abstract algebra lies in studying all concrete algebras by abstract means, the study of non-associative algebras is more important than associative algebras, since non-associative algebras are vastly more common than associative algebras.

share|cite|improve this answer

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

Neither the Sheffer Stroke (NAND), nor the Pierce arrow (NOR) associate. Both have an important status in logic, as does implication. Reverse implication also does not associate.

I asked somewhat of a related question a few years ago: Is the Ratio of Associative Binary Operations to All Binary Operations on a Set of $n$ Elements Generally Small? Associative binary operations on a finite set are rare in comparison to the class of most binary operations on a finite set (statistically speaking). Additionally, there exist all of the counterxamples in these answers. So, we can't build in the concept of associativity to a general concept of an operation. Doing so would lead to many contradictions.

Given that a goal of the study of abstract algebra lies in studying all concrete algebras by abstract means, the study of non-associative algebras is more important than associative algebras, since non-associative algebras are vastly more common than associative algebras.

share|cite|improve this answer

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

answered Aug 25 at 18:27

Doug Spoonwood

7,82312043

7,82312043

add a comment | 

add a comment | 

up vote
1
down vote

Subtraction / Division have been mentoined of course, but let me explain why they are as important or even more important than Addition / Multiplication and arguably only less talked about since we are more comfortable working with associativity.

You can explain what a group is (almost) solely by its division (subtraction in the Abelian case).

A (division) group is thus a set $G$ with a binary operation $/ : Gtimes Gto G$ s.t.

• $a/a = b/b$,


• $fraca/a(a/a)/a = a$,


• $fracab/c = fraca(c/c)/c)/b$.

(I used "big fractions" for increased readibility).

A group morphism $f : Gto H$ is then just a function $Gto H$ s.t. $f(a/b) = f(a)/f(b)$.

There is a small "caveat". The "division groups" almost coincide with ordinary groups; just pick $gin G$ and let $e_G := g/g$, $acdot b := a/(e_G/b)$ and $a^-1 = e_G/a$. This leaves open the possibility of $G$ being empty. So in this theory, there is an "empty group".

If you don't want an "empty group", then we can simply insist on the existence of an identity $e_G$ s.t. $a/a = e_G$ for all $ain G$.

share|cite|improve this answer

answered Aug 25 at 20:13

Stefan Perko

8,31611641

add a comment | 

up vote
1
down vote

Subtraction / Division have been mentoined of course, but let me explain why they are as important or even more important than Addition / Multiplication and arguably only less talked about since we are more comfortable working with associativity.

You can explain what a group is (almost) solely by its division (subtraction in the Abelian case).

A (division) group is thus a set $G$ with a binary operation $/ : Gtimes Gto G$ s.t.

• $a/a = b/b$,


• $fraca/a(a/a)/a = a$,


• $fracab/c = fraca(c/c)/c)/b$.

(I used "big fractions" for increased readibility).

A group morphism $f : Gto H$ is then just a function $Gto H$ s.t. $f(a/b) = f(a)/f(b)$.

There is a small "caveat". The "division groups" almost coincide with ordinary groups; just pick $gin G$ and let $e_G := g/g$, $acdot b := a/(e_G/b)$ and $a^-1 = e_G/a$. This leaves open the possibility of $G$ being empty. So in this theory, there is an "empty group".

If you don't want an "empty group", then we can simply insist on the existence of an identity $e_G$ s.t. $a/a = e_G$ for all $ain G$.

share|cite|improve this answer

answered Aug 25 at 20:13

Stefan Perko

8,31611641

add a comment | 

up vote
1
down vote

up vote
1
down vote

Subtraction / Division have been mentoined of course, but let me explain why they are as important or even more important than Addition / Multiplication and arguably only less talked about since we are more comfortable working with associativity.

You can explain what a group is (almost) solely by its division (subtraction in the Abelian case).

A (division) group is thus a set $G$ with a binary operation $/ : Gtimes Gto G$ s.t.

• $a/a = b/b$,


• $fraca/a(a/a)/a = a$,


• $fracab/c = fraca(c/c)/c)/b$.

(I used "big fractions" for increased readibility).

A group morphism $f : Gto H$ is then just a function $Gto H$ s.t. $f(a/b) = f(a)/f(b)$.

There is a small "caveat". The "division groups" almost coincide with ordinary groups; just pick $gin G$ and let $e_G := g/g$, $acdot b := a/(e_G/b)$ and $a^-1 = e_G/a$. This leaves open the possibility of $G$ being empty. So in this theory, there is an "empty group".

If you don't want an "empty group", then we can simply insist on the existence of an identity $e_G$ s.t. $a/a = e_G$ for all $ain G$.

share|cite|improve this answer

answered Aug 25 at 20:13

Stefan Perko

8,31611641

Subtraction / Division have been mentoined of course, but let me explain why they are as important or even more important than Addition / Multiplication and arguably only less talked about since we are more comfortable working with associativity.

You can explain what a group is (almost) solely by its division (subtraction in the Abelian case).

A (division) group is thus a set $G$ with a binary operation $/ : Gtimes Gto G$ s.t.

• $a/a = b/b$,


• $fraca/a(a/a)/a = a$,


• $fracab/c = fraca(c/c)/c)/b$.

(I used "big fractions" for increased readibility).

A group morphism $f : Gto H$ is then just a function $Gto H$ s.t. $f(a/b) = f(a)/f(b)$.

There is a small "caveat". The "division groups" almost coincide with ordinary groups; just pick $gin G$ and let $e_G := g/g$, $acdot b := a/(e_G/b)$ and $a^-1 = e_G/a$. This leaves open the possibility of $G$ being empty. So in this theory, there is an "empty group".

If you don't want an "empty group", then we can simply insist on the existence of an identity $e_G$ s.t. $a/a = e_G$ for all $ain G$.

share|cite|improve this answer

answered Aug 25 at 20:13

Stefan Perko

8,31611641

share|cite|improve this answer

share|cite|improve this answer

answered Aug 25 at 20:13

Stefan Perko

8,31611641

answered Aug 25 at 20:13

Stefan Perko

8,31611641

answered Aug 25 at 20:13

Stefan Perko

8,31611641

8,31611641

add a comment | 

add a comment | 

 

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