Is there ANY context in which f(x,x) is noncommutative?
Stupid question, but one occasionally reads such things as "the operation $ast$ is noncommutative for all $x,y$ such that $xneq y$" or "$xast y$ is commutative iff $x=y$". These statements bother me, because they imply that there is some operation $cdot$ for which $x=ynotimplies xcdot y=ycdot x$ which in turn implies $xcdot xneq xcdot x$.
Is this the result of poor writing, or is there some legitimate reason to call an operation between an element and itself "commutative".
abstract-algebra group-theory binary-operations
|
show 2 more comments
Stupid question, but one occasionally reads such things as "the operation $ast$ is noncommutative for all $x,y$ such that $xneq y$" or "$xast y$ is commutative iff $x=y$". These statements bother me, because they imply that there is some operation $cdot$ for which $x=ynotimplies xcdot y=ycdot x$ which in turn implies $xcdot xneq xcdot x$.
Is this the result of poor writing, or is there some legitimate reason to call an operation between an element and itself "commutative".
abstract-algebra group-theory binary-operations
The statements you give about non-commutativity fail if $y=x$
– aidangallagher4
Nov 22 '18 at 20:36
There are three options and we need to distinguish between them 1) They commute always; that's easy to express 2) they don't always commute or they sometimes don't commute. That's easy to express 3) the third option is harder to express, they absolutely never commute.... except in the cases where they act on themselves, cause... you know if x=x you can't not switch them.. but that's the exception... all real cases they don't commute. That third one requires some form af adress.
– fleablood
Nov 22 '18 at 20:45
@fleablood: I have never encountered the third case arising "in nature" or seen any writing that needs to talk about operations that are "nowhere commutative". If the OP is reading material that needs to deal with that case, then he or she should definitely provide a reference, as the writing style used seems to be very poor from the OP's quotations.
– Rob Arthan
Nov 22 '18 at 21:03
@Rob Arthan that's why it isn't sloppy writing. To describe a function that never (non-trivially) commutes we must [ne way or another that it does trivially commute.
– fleablood
Nov 22 '18 at 21:05
I've never seen it either but clearly the OP has. And in such a case we'd have to make an exception for the trivial case, wouldn't we.
– fleablood
Nov 22 '18 at 21:07
|
show 2 more comments
Stupid question, but one occasionally reads such things as "the operation $ast$ is noncommutative for all $x,y$ such that $xneq y$" or "$xast y$ is commutative iff $x=y$". These statements bother me, because they imply that there is some operation $cdot$ for which $x=ynotimplies xcdot y=ycdot x$ which in turn implies $xcdot xneq xcdot x$.
Is this the result of poor writing, or is there some legitimate reason to call an operation between an element and itself "commutative".
abstract-algebra group-theory binary-operations
Stupid question, but one occasionally reads such things as "the operation $ast$ is noncommutative for all $x,y$ such that $xneq y$" or "$xast y$ is commutative iff $x=y$". These statements bother me, because they imply that there is some operation $cdot$ for which $x=ynotimplies xcdot y=ycdot x$ which in turn implies $xcdot xneq xcdot x$.
Is this the result of poor writing, or is there some legitimate reason to call an operation between an element and itself "commutative".
abstract-algebra group-theory binary-operations
abstract-algebra group-theory binary-operations
asked Nov 22 '18 at 20:30
R. Burton
34619
34619
The statements you give about non-commutativity fail if $y=x$
– aidangallagher4
Nov 22 '18 at 20:36
There are three options and we need to distinguish between them 1) They commute always; that's easy to express 2) they don't always commute or they sometimes don't commute. That's easy to express 3) the third option is harder to express, they absolutely never commute.... except in the cases where they act on themselves, cause... you know if x=x you can't not switch them.. but that's the exception... all real cases they don't commute. That third one requires some form af adress.
– fleablood
Nov 22 '18 at 20:45
@fleablood: I have never encountered the third case arising "in nature" or seen any writing that needs to talk about operations that are "nowhere commutative". If the OP is reading material that needs to deal with that case, then he or she should definitely provide a reference, as the writing style used seems to be very poor from the OP's quotations.
– Rob Arthan
Nov 22 '18 at 21:03
@Rob Arthan that's why it isn't sloppy writing. To describe a function that never (non-trivially) commutes we must [ne way or another that it does trivially commute.
– fleablood
Nov 22 '18 at 21:05
I've never seen it either but clearly the OP has. And in such a case we'd have to make an exception for the trivial case, wouldn't we.
– fleablood
Nov 22 '18 at 21:07
|
show 2 more comments
The statements you give about non-commutativity fail if $y=x$
– aidangallagher4
Nov 22 '18 at 20:36
There are three options and we need to distinguish between them 1) They commute always; that's easy to express 2) they don't always commute or they sometimes don't commute. That's easy to express 3) the third option is harder to express, they absolutely never commute.... except in the cases where they act on themselves, cause... you know if x=x you can't not switch them.. but that's the exception... all real cases they don't commute. That third one requires some form af adress.
– fleablood
Nov 22 '18 at 20:45
@fleablood: I have never encountered the third case arising "in nature" or seen any writing that needs to talk about operations that are "nowhere commutative". If the OP is reading material that needs to deal with that case, then he or she should definitely provide a reference, as the writing style used seems to be very poor from the OP's quotations.
– Rob Arthan
Nov 22 '18 at 21:03
@Rob Arthan that's why it isn't sloppy writing. To describe a function that never (non-trivially) commutes we must [ne way or another that it does trivially commute.
– fleablood
Nov 22 '18 at 21:05
I've never seen it either but clearly the OP has. And in such a case we'd have to make an exception for the trivial case, wouldn't we.
– fleablood
Nov 22 '18 at 21:07
The statements you give about non-commutativity fail if $y=x$
– aidangallagher4
Nov 22 '18 at 20:36
The statements you give about non-commutativity fail if $y=x$
– aidangallagher4
Nov 22 '18 at 20:36
There are three options and we need to distinguish between them 1) They commute always; that's easy to express 2) they don't always commute or they sometimes don't commute. That's easy to express 3) the third option is harder to express, they absolutely never commute.... except in the cases where they act on themselves, cause... you know if x=x you can't not switch them.. but that's the exception... all real cases they don't commute. That third one requires some form af adress.
– fleablood
Nov 22 '18 at 20:45
There are three options and we need to distinguish between them 1) They commute always; that's easy to express 2) they don't always commute or they sometimes don't commute. That's easy to express 3) the third option is harder to express, they absolutely never commute.... except in the cases where they act on themselves, cause... you know if x=x you can't not switch them.. but that's the exception... all real cases they don't commute. That third one requires some form af adress.
– fleablood
Nov 22 '18 at 20:45
@fleablood: I have never encountered the third case arising "in nature" or seen any writing that needs to talk about operations that are "nowhere commutative". If the OP is reading material that needs to deal with that case, then he or she should definitely provide a reference, as the writing style used seems to be very poor from the OP's quotations.
– Rob Arthan
Nov 22 '18 at 21:03
@fleablood: I have never encountered the third case arising "in nature" or seen any writing that needs to talk about operations that are "nowhere commutative". If the OP is reading material that needs to deal with that case, then he or she should definitely provide a reference, as the writing style used seems to be very poor from the OP's quotations.
– Rob Arthan
Nov 22 '18 at 21:03
@Rob Arthan that's why it isn't sloppy writing. To describe a function that never (non-trivially) commutes we must [ne way or another that it does trivially commute.
– fleablood
Nov 22 '18 at 21:05
@Rob Arthan that's why it isn't sloppy writing. To describe a function that never (non-trivially) commutes we must [ne way or another that it does trivially commute.
– fleablood
Nov 22 '18 at 21:05
I've never seen it either but clearly the OP has. And in such a case we'd have to make an exception for the trivial case, wouldn't we.
– fleablood
Nov 22 '18 at 21:07
I've never seen it either but clearly the OP has. And in such a case we'd have to make an exception for the trivial case, wouldn't we.
– fleablood
Nov 22 '18 at 21:07
|
show 2 more comments
3 Answers
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The issue isn't talking about when they do commute but when they don't.
There are three options.
1) For every $xne y $, $f (x,y)=f (y,x) $. Thus we say $f$ always commute. (It's commutative)
2) For $xne y$ sometimes $f(x,y)ne f (y,x) $. Thus it's not always commutative. (It's not commutative.)
3) for $xne 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 $xne 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.
add a comment |
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.
add a comment |
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.
add a comment |
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3 Answers
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3 Answers
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The issue isn't talking about when they do commute but when they don't.
There are three options.
1) For every $xne y $, $f (x,y)=f (y,x) $. Thus we say $f$ always commute. (It's commutative)
2) For $xne y$ sometimes $f(x,y)ne f (y,x) $. Thus it's not always commutative. (It's not commutative.)
3) for $xne 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 $xne 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.
add a comment |
The issue isn't talking about when they do commute but when they don't.
There are three options.
1) For every $xne y $, $f (x,y)=f (y,x) $. Thus we say $f$ always commute. (It's commutative)
2) For $xne y$ sometimes $f(x,y)ne f (y,x) $. Thus it's not always commutative. (It's not commutative.)
3) for $xne 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 $xne 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.
add a comment |
The issue isn't talking about when they do commute but when they don't.
There are three options.
1) For every $xne y $, $f (x,y)=f (y,x) $. Thus we say $f$ always commute. (It's commutative)
2) For $xne y$ sometimes $f(x,y)ne f (y,x) $. Thus it's not always commutative. (It's not commutative.)
3) for $xne 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 $xne 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.
The issue isn't talking about when they do commute but when they don't.
There are three options.
1) For every $xne y $, $f (x,y)=f (y,x) $. Thus we say $f$ always commute. (It's commutative)
2) For $xne y$ sometimes $f(x,y)ne f (y,x) $. Thus it's not always commutative. (It's not commutative.)
3) for $xne 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 $xne 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.
edited Nov 22 '18 at 21:02
answered Nov 22 '18 at 20:55
fleablood
68.2k22684
68.2k22684
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
edited Nov 22 '18 at 21:15
answered Nov 22 '18 at 20:56
Rob Arthan
29k42866
29k42866
add a comment |
add a comment |
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.
add a comment |
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.
add a comment |
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.
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.
answered Nov 22 '18 at 20:36
angryavian
39k23180
39k23180
add a comment |
add a comment |
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The statements you give about non-commutativity fail if $y=x$
– aidangallagher4
Nov 22 '18 at 20:36
There are three options and we need to distinguish between them 1) They commute always; that's easy to express 2) they don't always commute or they sometimes don't commute. That's easy to express 3) the third option is harder to express, they absolutely never commute.... except in the cases where they act on themselves, cause... you know if x=x you can't not switch them.. but that's the exception... all real cases they don't commute. That third one requires some form af adress.
– fleablood
Nov 22 '18 at 20:45
@fleablood: I have never encountered the third case arising "in nature" or seen any writing that needs to talk about operations that are "nowhere commutative". If the OP is reading material that needs to deal with that case, then he or she should definitely provide a reference, as the writing style used seems to be very poor from the OP's quotations.
– Rob Arthan
Nov 22 '18 at 21:03
@Rob Arthan that's why it isn't sloppy writing. To describe a function that never (non-trivially) commutes we must [ne way or another that it does trivially commute.
– fleablood
Nov 22 '18 at 21:05
I've never seen it either but clearly the OP has. And in such a case we'd have to make an exception for the trivial case, wouldn't we.
– fleablood
Nov 22 '18 at 21:07