Is there ANY context in which f(x,x) is noncommutative?

Question

Stupid question, but one occasionally reads such things as "the operation $\ast$ is noncommutative for all $x,y$ such that $x\neq y$" or "$x\ast y$ is commutative iff $x=y$". These statements bother me, because they imply that there is some operation $\cdot$ for which $x=y\not\implies x\cdot y=y\cdot x$ which in turn implies $x\cdot x\neq x\cdot x$.

Is this the result of poor writing, or is there some legitimate reason to call an operation between an element and itself "commutative".

Answer 1

The issue isn't talking about when they do commute but when they don't.

There are three options.

1) For every $x\ne y $, $f (x,y)=f (y,x) $. Thus we say $f$ always commute. (It's commutative)

2) For $x\ne y$ sometimes $f(x,y)\ne f (y,x) $. Thus it's not always commutative. (It's not commutative.)

3) for $x\ne y $ we always have $f (x,y)\ne f (y,x) $.

We'd like to say of 3) that 3) is never commutative.

But we can't say that. We can't say that because all functions have to commute when $x=y $.

So for 3) or only options are to state either it never commutes when $x\ne y $ or, equivalently, the only time $f $ commutes is if $x=y $.

Is this the result of poor writing, or is there some legitimate reason to call an operation between an element and itself "commutative".

It's not poor writing. Just the opposite. Proper writing requires that that case $x=y $ does commute. Even if all others dont.

or is there some legitimate reason to call an operation between an element and itself "commutative".

Well if $x=y $ then $f (x,y)=f (y,x) $. That's a legitimate reason, isn't it.

Your seem to be saying it needn't be stated as it is always true. Well, fair enough but the texts are stating we need to always make an exception because we are not allowed to say a function never commutes. We must in those cases point out that $x=y $ is the only case the do.

Answer 2

I think the statement is simply saying "$x * y \ne y * x$ for all $x \ne y$." Note that this is stronger than saying "there exists $x$ and $y$ such that $x * y \ne y * x$." You are correct that $x * x = x * x$ always, but this is not the point of the statement.

Answer 3

What you are reading looks to me like sloppy writing. If $x = y$, then $x * y = y * x$ for any operator $*$ (for which $x *x$ is defined). An operator $*$ is commutative iff $x * y = y *x$ for all $x$ and $y$ (for which $x * y$ is defined). To prove this property, you only have to consider the case when $x \neq y$, but it is pointless and confusing to exclude this special case from the definition and it is wrong to say "$x * y$ is commutative": $x *y$ is a value in the algebraic structure and not an operator.

It does make sense to talk about commutativity on subsets of the domain of definition of an operator. E.g., you can say "multiplication is not commutative on the quaternions $\Bbb{H}$, but is commutative on the complex numbers $\Bbb{C} \subseteq \Bbb{H}$". You can say "in the quaternions, commutativity of multiplications fails for the elements $\mathbf{i}$ and $\mathbf{j}$". However, you don't say "$\mathbf{i}\mathbf{j}$ is non-commutative" or "$\mathbf{i}\mathbf{i}$ is commutative"($\mathbf{i}\mathbf{j}$ and $\mathbf{i}\mathbf{i}$" are not operations, but rather quaternions, namely $\mathbf{k}$ and $-1$).

Note: in the above, I am not attacking the usual abuse of notation whereby we write formulas with free variables and use them to denote functions (like "the function $f(x, y) = x + y + x*y$"). What you shouldn't do is talk about properties of the function in a context where you have constrained the free variables: to talk about properties (such as commutativity) of "the function $f(x, y)$ where $x = y$" is poor writing.