ASSOCIATIVE OPERATOR

[Dr.Peterson]

As it was originally this a*(b*c)*d so i thought what could be the representation which is fully parenthesized(FPE NOT SURE ) keeping b*c inside one parenthesis ; either a*((b*c) *d) or (a(b*c))*d But A FPE has exactly one set of Parentheses enclosing each operator and its operands.

abcd for this to be FPE this can be --> ( a*( (b*c)*d ) ) or ( (a*(b*c) )*d ) these are perfectly matching with the def of FPE. "exactly one set of Parentheses enclosing each operator and its operands" but taking any two of five forms a((bc)*d) or (a*b) * (c*d) these are not fully parentisesized.

fpe version of a((bc)*d --> ( a*( (b*c)*d ) )
fpe version of (a*b) * (c*d) -->( (a*b) * (c*d) )
What is FPE has to do with associative property?

I don't think anyone cares what constitutes a "fully parenthesized expression", except perhaps computer scientists. Where did you get the "definition", and why does it matter to you? I didn't use the phrase as a formally defined term, and did not intend to give you something new to obsess over.

If the operator is associative, then all that matters is that you can insert parentheses anywhere you want and the value doesn't change. The five cases in the Wikipedia page under "Generalized associative law" all include enough parentheses to make the order in which operations are to be performed clear, without additional rules such as "left-associativity" [which I now see is discussed further down the page under "Notation for non-associative operations", and which corresponds to my "left-to-right evaluation"]. There is no need for an additional pair of parentheses around the entire expression.

As for your last question, did someone tell you that "FPE" is important? I didn't. The only place it would matter is with a non-associative operation.

 

[Saumyojit]

You could also write a*b*(c*d) if you want (which would literally mean (a*b)*(c*d) ).

post no 14 why does a*b*(c*d) literally mean (a*b)*(c*d) but not a*(b*(c*d) ) ?

all include enough parentheses to make the order in which operations are to be performed clear

Out of 5 forms , one form (a*b)*(c*d) is there which kind of gives same precedence ( DO me first) to both sides creates a confusion.

AND U SAID THAT

a*(b*c*d) is perfectly legal, and has the same value as all the others.

then why these two forms a*(b*c*d) & (a*b*c)*d are not included in the 5 form list making it total 7 forms of applying ASSOCIAITIVITY without changing the order of factors ---> thats bcoz these 2 extra FORMS does not include enough parentheses to make the order in which operations are to be performed clear . For a*(b*c*d) the order goes like this
1: first the order of operation goes to (b*c*d) part and then gets confused which one to execute (b*c) or (c*d). I HOPE MY REASON IS CORRECT.

without additional rules such as "left-associativity"

Out of 5 forms these 2 forms
(a*(b*c))*d and ( (a*b)*c)*d is expressing left - associativity implicitly but also a associative operation does not require left associativity or right.

There is no need for an additional pair of parentheses around the entire expression

this i was doing to meet the demands of defination of FPE.

 

[Dr.Peterson]

post no 14 why does a*b*(c*d) literally mean (a*b)*(c*d) but not a*(b*(c*d) ) ?
Out of 5 forms , one form (a*b)*(c*d) is there which kind of gives same precedence ( DO me first) to both sides creates a confusion.

When I said "literally", I meant, if you apply a left-to-right rule (left associativity). In that case, as you read a*b*(c*d) after performing what is in parentheses, you first encounter the * between a and b, so you would evaluate that first, making the expression behave like (a*b)*(c*d).

To put it another way, under the left-to right rule, the operations are performed in the order shown here by subscripts:

• a*₂b*₃(c*₁d)
• (a*₁b)*₂(c*₁d) -- here the two 1's are independent, and can be done in either order
• a₃*(b₂*(c*₁d) )

The first two of these are equivalent. But, again, this only matters when * is not associative, so you can really ignore it.

My point about independence seems to be one you are missing. Whether we do (a*₁b)*₃(c*₂d) or (a*₂b)*₃(c*₁d), we will get the same result even for a non-associative operation, because we have the two operands for the last operation ready for it to operate on.

then why these two forms a*(b*c*d) & (a*b*c)*d are not included in the 5 form list making it total 7 forms of applying ASSOCIAITIVITY without changing the order of factors ---> thats bcoz these 2 extra FORMS does not include enough parentheses to make the order in which operations are to be performed clear . For a*(b*c*d) the order goes like this
1: first the order of operation goes to (b*c*d) part and then gets confused which one to execute (b*c) or (c*d). I HOPE MY REASON IS CORRECT.

Yes, I'd say you are correct, except that these are not "forms of applying associativity without changing the order of factors". Associativity doesn't come into this until you say that all those forms have the same value.

What do they say about their list? "A product of four elements may be written, without changing the order of the factors, in five possible ways". What they mean is that these are the ways you can place parentheses, just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation. Your additional forms a*(b*c*d) & (a*b*c)*d do not do so.

Out of 5 forms these 2 forms
(a*(b*c))*d and ( (a*b)*c)*d is expressing left - associativity implicitly but also a associative operation does not require left associativity or right.

I'm not sure you fully understand what left-associativity means. If we followed a left-associative rule ("conventionally evaluated from left to right", as Wikipedia says), then a*b*c*d would be evaluated the same as ((a*b)*c)*d, that is, a*₁b*₂c*₃d. It would not be evaluated the same as (a*(b*c))*d, where the second * would be done first.

this i was doing to meet the demands of defination of FPE.

You didn't answer my questions. Where do you get this "definition", and why? But if you are implying that you are no longer worrying about it, that's good.

 

[Saumyojit]

Whether we do (a*1b)*3(c*2d) or (a*2b)*3(c*1d), we will get the same result even for a non-associative operation, because we have the two operands for the last operation ready for it to operate on.

i got it that for divison (a non associative operation) if my expression is like this (3 operands)--> (a/b)/c or a/(b/c) my results will be different but for 4 operands -->(a*₁b)*3(c*₂d) or (a*₂b)*3(c*₁d), we will get the same result .

My point about independence seems to be one you are missing.

What independence?
My interpretation is That a expression of SAME ASSOCIATIVE operator is independent of any no of parentheses as result will be same anyway we evaluate order of OPERATIONS . Or u are talking in terms of NON ASSOCIATIVE ... see one thing i noticed in division if we have 4 operands keeping them like this (3/2)/(4/6) will give me 9/4 but keeping the same expression like this (3/2/4)/6--> first it will evaluate 3/2 in parenthis then result of 3/2 by 4 then the result of execution part in parenthis by 6 giving 1/16 final answer . In the parenthisis part , although i havent given another pair of parenthses still it should automatically happen that division will occur from 3/4 then /6 as DIVISON is LEFT ASSOCIATIVE if notation is not mentioned explicitly (MENTIONED IN WIKI). (3/2/4 ) this is the just parentheses part & LEFT associativity will happen even if i dont give brackets between any two operands .

Why do you say it means a*((b*c)*d)?

In post #19 for u inherently means

"literally means", I am talking about following a left-to-right rule, which technically is part of the order of operations we use for traditional addition and multiplication

And yes u were right

I have to guess that you mean something different by "inherently means".

FOR ME "inherently" meant I was not thinking from left to right perspective but I was saying from "rearranging parenthesis / associative property point of view " keeping in mind that i have to keep "b*c" inside 1 parenthesis. Thats why I gave these 2 arrangments
a*((b*c) *d) or (a(b*c))*d in post # 15

And why do you bring in a*(b*c*d), which is not fully parenthesized

I was thinking of rearranging parenthsis which will not change the result . a*(b*c*d) or (a*b*c)*d (* is ANY ASSOCIAITVE OPERATOR)not violating any rules of associative operation.

these are not "forms of applying associativity without changing the order of factors". Associativity doesn't come into this until you say that all those forms have the same value.

a*(b*c*d) or (a*b*c)*d (* is ANY ASSOCIAITVE OPERATOR) these 2 forms 's value will result to same answer like the mentioned 5 forms in wiki as * is a associative operator after all.

What do they say about their list? "A product of four elements may be written, without changing the order of the factors, in five possible ways". What they mean is that these are the ways you can place parentheses, just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation. Your additional forms a*(b*c*d) & (a*b*c)*d do not do so.

In the Generalized associative law part , they have clearly said a product of 4 elements which is a binary associative operator . WHy are u saying from the perspective of a non associative operation?

Then the 5 forms would not exist only as each form under non associative operation will give diffrent result in each case.
How and why did u came up with "to determine the value in the case of a non-associative operation"

If we followed a left-associative rule ("conventionally evaluated from left to right", as Wikipedia says), then a*b*c*d would be evaluated the same as ((a*b)*c)*d, that is, a*1b*2c*3d.

YEs a*b*c*d would be evaluated the same as ((a*b)*c)*d and it would not be evaluated as (a*(b*c))*d .
(a*(b*c))*d as u have told this is not following left assocaitive , but if i take this Expression a*b*(c*d) if it is operated under left-right rule should n't it be operated like this this (a*(b*c))*d but Hold on in the expression a*b*(c*d) it is given C*D under brackets which is telling the order of operation to start from C*D first and then after executing it , follow left to right which is -->(a*₂b)*3(c*₁d)

where in this case a*b*c*d there is no brackets from the beginning so this will translate to ((a*b)*c)*d under left associaitive.
RIGHT?

 

[Dr.Peterson]

This is too much material for me to work through carefully. I'll just make a couple comments.

First, did you at least try evaluating something like (6/3)/(8/2) to understand what I meant? You need to be willing to think, and then you might understand other people.

It doesn't matter whether I evaluate 6/3 = 2 or 8/2 = 4 first; I have to do both before doing the remaining division. Either way, I get (2)/(4) = 1/2. That is what I meant by "independent", and I expected you to figure that out by thinking about what I said: Neither division depends on the other. As I said, "here the two 1's are independent, and can be done in either order".

None of this has anything to do with "left associativity". I hope you understand that left associativity is an entirely different kind of thing than the associative property. The latter is a mathematical fact about certain operations, such as multiplication or addition, that is implied by their very definitions; the former is nothing more than an arbitrary convention we may choose to follow, and not a mathematical truth. Multiplication is associative; division may be treated as left-associative, or we could leave a/b/c undefined and require parentheses. As Wikipedia says, "This is simply a notational convention to avoid parentheses."

You keep mixing together statements about associative and non-associative operations. I blame Wikipedia for that; I wish they didn't put them together in one page. But you are responsible to read it carefully, and talk about one at a time. The point is that the distinction among (a*b)*(c*d), and so on, only matters when the operations is not associative; that's why much of what you say is either confused or confusing. Please stop talking about these things in the wrong contexts.

Hold on in the expression a*b*(c*d) it is given C*D under brackets which is telling the order of operation to start from C*D first and then after executing it , follow left to right which is -->(a*2b)*3(c*1d)

where in this case a*b*c*d there is no brackets from the beginning so this will translate to ((a*b)*c)*d under left associaitive.
RIGHT?

It's hard to follow you when you seem to be arguing with yourself, but I think you eventually came to the right conclusion here.

 

[Saumyojit]

PLEASE GO THROUGH IT ONCE EACH LINE.. my doubts will be cleared . I beg of u sir.

First, did you at least try evaluating something like (6/3)/(8/2) to understand what I meant? You need to be willing to think, and then you might understand other people.

i wrote it

for 4 operands -->(a*1b)*3(c*2d) or (a*2b)*3(c*1d), we will get the same result .

i got it ; anyway we do we will get the same result ...order does not matter in THIS case of non associative operation also.

here the two 1's are independent and can be done in either order

i got it.

division may be treated as left-associative, or we could leave a/b/c undefined and require parentheses. As Wikipedia says, "This is simply a notational convention to avoid parentheses."

In wiki "However, mathematicians agree on a particular order of evaluation for several common non-associative operations " if we leave a/b/c undefined then division has to be treated as (a/b)/c as mathematicians agreed that division will be a left associative operation if no parantheses is given.

MAIN DOUBT: I asked u in post#22 why my extra 2 forms will not be included in 5 form list . U replied to me in post #23

Associativity doesn't come into this until you say that all those forms have the same value

"Associaitivity doesn't come into this until you say that all those forms have the same value" -->SEE a*(b*c*d) or (a*b*c)*d

(* is A ASSOCIAITVE OPERATOR) these 2 forms will result to same answer after execution like the other mentioned 5 forms in wiki as * is a associative operator after all.
SO what are u implying by saying "until you say that all those forms have the same value" ?


THEN U SAY on the same line

What they mean is that these are the ways you can place parentheses, just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation. Your additional forms a*(b*c*d) & (a*b*c)*d do not do so

In the Generalized associative law part , they have clearly said they are talking about a * binary associative operator not a "not associative operator". THEN Why are u bringing the case of "NON ASSOCIAITVE" in this particular context/LINE--> { just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation.} ?

OKAY i am kind of interpeting ur implication of this ... when u say " a*(b*c*d) & (a*b*c)*d do not do so" i am taking * as a non associaitve operator which can be DIVISION ; then this expression --> a*(b*c*d) will be a/ (b/c/d) so the order of EVALUATION will be first it will go to the parenthisis part , THEN division will occur in b/c ; then /d ; then a shall be divided by b/c/d as mathematicians agreed that DIVISON is LEFT ASSOCIATIVE BY DEFAULT if there is ambiguity .
I mean by "AMBIGUITY " in this particular eg of a/(b/c/d) is that inside the b/c/d part ; the question has not mentioned after order goes to b/c/d part which division operator to execute which naturally raises a question q1

q1: Can it happen that inside a subexpression i.e (b/c/d) of the whole expression i.e a/ (b/c/d) Left asso can be applied inside of subexpression or Left asso can be applied to only a whole expression ??
if YES , then division will occur first in b/c ; (b/c) dividede by /d ; then a shall be divided by b/c/d which i have written in the above para.



So if ur answer is yes to q1; then " a*(b*c*d) & (a*b*c)*d " these two will work even * is non associaitve .

----- END OF MAIN DOUBT-------

 

[Dr.Peterson]

In wiki "However, mathematicians agree on a particular order of evaluation for several common non-associative operations " if we leave a/b/c undefined then division has to be treated as (a/b)/c as mathematicians agreed that division will be a left associative operation if no parantheses is given.

The point is that this is just an agreement, not something we are forced to. We COULD choose not to make any agreement, so that a/b/c would be undefined, and parentheses would be required. In fact, parentheses are still strongly recommended, as it is easy to misread such an expression even if you have been taught the "rule". Do you not understand this??

"Associaitivity doesn't come into this until you say that all those forms have the same value" -->SEE a*(b*c*d) or (a*b*c)*d

(* is A ASSOCIAITVE OPERATOR) these 2 forms will result to same answer after execution like the other mentioned 5 forms in wiki as * is a associative operator after all.
SO what are u implying by saying "until you say that all those forms have the same value" ?

I'm really not sure why you are so obsessive about this list of five forms. They simply list five ways to place parentheses, which, in the absence of knowledge about the operation (including whether it is associative, or whether additional rules of interpretation are to be applied), are potentially different. You want to add cases which are ambiguous in the absence of such knowledge, and therefore do not belong in that list.

q1: Can it happen that inside a subexpression i.e (b/c/d) of the whole expression i.e a/ (b/c/d) Left asso can be applied inside of subexpression or Left asso can be applied to only a whole expression ??
if YES , then division will occur first in b/c ; (b/c) dividede by /d ; then a shall be divided by b/c/d which i have written in the above para.

So if ur answer is yes to q1; then " a*(b*c*d) & (a*b*c)*d " these two will work even * is non associaitve .

I'll assume that you ask this separately, and not as part of the discussion of the list of five forms.

If we treat an operation as left-associative, then that applies everywhere, including within parentheses. It would be silly to do otherwise!

But when you're talking about an unspecified operation "*", you can't assume anything about it. So a*(b*c*d) and (a*b*c)*d would be ambiguous.

By the way, could you please stop using your wild mixes of font sizes, cases, and bolding? That only makes what you write even more confusing, and when I'm dizzy, I don't have a good attitude.

 

[JeffM]

Dr. Peterson has been infinitely patient. I’d like to summarize what I think are the main points on which no further questions need to be asked. Please do not ask questions that concern these points or that refer back to previous posts in what is now a long thread of long posts.

(1) Associativity is an attribute of binary operations that applies when that operation is present in an expression more than once.

(2) As de Saussure has explained, a word is defined by contrasting what it is versus what it is not.
Associative means that the order of performing the operation does not matter. Example: addition of real numbers.
Non-associative means that the order of performing the operation does matter. Example: division of real numbers.
That’s it. No need for mystification of a very simple idea.

(3) It is a convention of mathematical communication that grouping symbols generally, especially parentheses, are used to specify order of operations. A convention of communication is not a mathematical fact; it is a human agreement subject to change.

(4) Whenever that convention is in place, parentheses are not needed in an expression consisting solely of repetitions of an associative operation although they may be used.

(5) When dealing with the basic arithmetic operations, additional conventions of mathematical communication further specify order of operations In order to reduce the number of needed grouping symbols without introducing ambiguity.

(6) In abstract algebra, specifying whether or not an operation is associative is necessary because the operations are defined only in terms of such general attributes. The additional conventions of communication used for dealing with the basic arithmetic operations such as PEMDAS do not apply In abstract algebra because we are deliberately trying to explore the consequences of operations that are not necessarily arithmetic.

(7) No one here is responsible for a wiki article that everyone here agrees will be found opaque by anyone who has not been introduced to abstract algebra. You keep fussing about differences in the exact wording of different posts and the wiki article while completely losing sight of the very simple points of conceptual agreement among the helpers’ posts and the wiki article.

 

[Saumyojit]

Please do not ask questions that concern these points or that refer back to previous posts

I dont know whether i will able to maintain it or not . Sorry from the beginning . PLease forgive me and my questions. But as long as i am having doubts which are legit to me I will ask .

Whenever that convention is in place, parentheses are not needed in an expression consisting solely of repetitions of an associative operation although they may be used.

@JeffM I got ur point in a row of same associative operator parenthses are not needed although we may use. BUT what do u mean by that convention is in place?

@Dr.Peterson

which, in the absence of knowledge about the operation

In the five form they have clearly said that * is any binary associaitve operator and they have told * is actually multiplication in those 5 forms. They have listed product in different arrangments . So * is multiplication ...no ambiguity from now on.
Every form is being written using multiplication operator and so i dont have the absence of knowledge about the operation rather i know that multiplication is being done.

you're talking about an unspecified operation

So a*(b*c*d) and (a*b*c)*d are unambiguous as * denotes multiplication. There is no unspecified operation but the operation is MULTIPLICATION.

If we treat an operation as left-associative, then that applies everywhere, including within parentheses

U agreed it can be applied within parentheses . NOW u had previously said

Yes, I'd say you are correct

I hope u said YES to my argument about this confusion in below

1: first the order of operation goes to (b*c*d) part and then gets confused which one to execute (b*c) or (c*d). I HOPE MY REASON IS CORRECT

If u say that left associative operation applies everywhere, including within parentheses then u should not have said YES as a reply Coz in

1. a*(b*c*d) expression where * is MULTIPLICATION (Multiplication is left asso ; as u have told in post 19 although does not matter)
2. ORDER OF OPERATION: first it will go to the parenthisis part
3. (b*c*d)
4. , THEN multiplication will start from b*c ; then
5. (b*c)*d
6. then
7. (b*c*d) with * a

So now is my two forms elgible ?

 

[JeffM]

I dont know whether i will able to maintain it or not . Sorry from the beginning . PLease forgive me and my questions. But as long as i am having doubts which are legit to me I will ask .

@JeffM I got ur point in a row of same associative operator parenthses are not needed although we may use. BUT what do u mean by that convention is in place?

That is a legitimate question that I am happy to answer. I meant the convention that parentheses are grouping symbols specifying order of operations. There is nothing in the world that precludes the mathematical community from deciding that parentheses mean something else, either generally or in some special field. Conventions are arbitrary but communal decisions like language itself.

So my meaning was that, so long as parentheses are merely grouping symbols specifying order of operations, they can be omitted with respect to multiple repetitions of an associative operation. If we change the meaning of parentheses, then that omission cannot be assumed.

I was trying to get you to see that conventions of communication are alterable in principle and that any conclusion based on a convention is contingent on the unaltered continuance of the convention. It was a subtle point. It does not materially affect the argument.

You need to get used to the fact that people like me whose careers were at least partially based on writing very exactly put in as explicit qualifications things that common sense may view as obvious even if implicit. When you are writing something that may lead to litigation, you tend to leave nothing implicit. It is sometimes a flaw in the context of pure explication.

 

[Saumyojit]

@Dr.Peterson I hope that sir replies to my recent post .
Hats of to ur patience sir for staying with my post ...I also wrote this last comment at 2 am from India .
I am also quite frustrated with myself why I can't get ur implications . I am having self doubt that I am extremely dumb

 

[Saumyojit]

@Dr.Peterson i am waiting for u reply
please help

 

[Dr.Peterson]

In the five form they have clearly said that * is any binary associaitve operator and they have told * is actually multiplication in those 5 forms. They have listed product in different arrangments . So * is multiplication ...no ambiguity from now on.
Every form is being written using multiplication operator and so i dont have the absence of knowledge about the operation rather i know that multiplication is being done.
So a*(b*c*d) and (a*b*c)*d are unambiguous as * denotes multiplication. There is no unspecified operation but the operation is MULTIPLICATION.

U agreed it can be applied within parentheses . NOW u had previously said

I hope u said YES to my argument about this confusion in below

If u say that left associative operation applies everywhere, including within parentheses then u should not have said YES as a reply Coz in

1. a*(b*c*d) expression where * is MULTIPLICATION (Multiplication is left asso ; as u have told in post 19 although does not matter)
2. ORDER OF OPERATION: first it will go to the parenthisis part
3. (b*c*d)
4. , THEN multiplication will start from b*c ; then
5. (b*c)*d
6. then
7. (b*c*d) with * a

So now is my two forms elgible ?

Again, I can't take the time to track back through the thread to figure out what response to what you are asking about. Let me just quote the part from Wikipedia that I think we are discussing:

If a binary operation is associative, repeated application of the operation produces the same result regardless of how valid pairs of parentheses are inserted in the expression. This is called the generalized associative law. For instance, a product of four elements may be written, without changing the order of the factors, in five possible ways:​

If the product operation is associative, the generalized associative law says that all these formulas will yield the same result. So unless the formula with omitted parentheses already has a different meaning (see below), the parentheses can be considered unnecessary and "the" product can be written unambiguously as​

The list itself does not assume that the operation is associative; and the operation is not assumed to be multiplication in the usual sense, but some operation that they are writing as a "product". (This is why they say "if" the operation is associative.) They also do not assume it is treated as left-associative (that's dealt with later, and is utterly unrelated to associativity, as I have said).

The list claims merely to be the ways in which "valid pairs of parentheses can be inserted"; it is about how the expression is written. It doesn't say explicitly in what sense they claim these are all the possibilities, so we have to work that out from what they show. That is, if you think they missed something, then you are misunderstanding what they mean! This is an important principle of communication, and may underlie much of your difficulty, though I've talked about something like this before: You need to interpret what someone else says on their own terms, and assume they are the experts about what they mean. There is no sense telling them that they don't mean what they think they mean. They simply haven't bothered to explain the list fully, because that isn't central to what they are saying.

The fact that you think they are wrong should lead you first to think deeply about whether you are misunderstanding their intent. Only when it is clear that there is no way to reconcile what they say, should you consider that they got the list wrong. They make mistakes (Wikipedia is constantly being edited, and I've seen plenty of errors), but look first for errors in your own thinking before saying they are wrong. A little humility is important, especially when you are reading about something you've never heard of before. But also, the fact that the idea of abstract operations is new to you means that your lack of understanding is normal, not a sign of stupidity. It is only your failure to let it go and move on that suggests foolishness.

 

[Saumyojit]

lead you first to think deeply about whether you are misunderstanding their intent

Seriously what u told just now such type of analysis did not came into my mind .

not assumed to be multiplication in the usual sense, but some operation that they are writing as a "product

Seriously things go complex ..why they are writing "product " of four elements and then contradiciting themselves by saying " if product is associative" ; anyone with a basic knowledge will get the first impression that product means multiplication which is Associative and will overlook this line or words which has a hidden meaning

if the product operation is associative

I assume from this line only u had analysed what they were trying to say and came to a conclusion i.e some operation that they are writing as a "product"

Associativity doesn't come into this until you say that all those forms have the same value

Still did not got the meaning of this line .

What they mean is that these are the ways you can place parentheses, just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation. Your additional forms a*(b*c*d) & (a*b*c)*d do not do so.

Why my additional forms dont work . Perhaps u gave ur reason i.e mentioned below

you're talking about an unspecified operation "*", you can't assume anything about it. So a*(b*c*d) and (a*b*c)*d would be ambiguous.

If my additional forms are ambiguous when the operation is itself unspecified , I dont understand how those 5 forms are legit or valid or unambiguous when they are using " Product' operation which is also not the real product operation that we genrally know of ?

If u are saying those 5 forms are valid in the sense that they are there only to show " when valid pairs of parentheses are inserted in the expression AND repeated application of the operation produces the same result " then my additional forms are also using the same operation that those 5 forms are using and also shows " If valid pairs of parentheses are inserted in the expression repeated application of the operation produces the same result" STILL THE REASON IS UNCLEAR.

I understood ur last post acc to me .
@Dr.Peterson

 

[JeffM]

Seriously what u told just now such type of analysis did not came into my mind .

Seriously things go complex ..why they are writing "product " of four elements and then contradiciting themselves by saying " if product is associative" ; anyone with a basic knowledge will get the first impression that product means multiplication which is Associative and will overlook this line or words which has a hidden meaning I assume from this line only u had analysed what they were trying to say and came to a conclusion i.e some operation that they are writing as a "product"

We have tried to tell you from almost the beginning of this thread that the context of the wiki article is abstract algebra.

In abstract algebra, we use letters to stand for mathematical objects, which may not be numbers. We use + and * to stand for operations, which may not be the familiar arithmetic operations of addition and multiplication. We use 0 and 1 for identity elements, which may not be numbers. You are trying to read this article as though it is about arithmetic and the elementary algebra associated with the system of real numbers. That is not what the article is about AT ALL.

I hope you will take my word for it that I was a good math student. I took a course in abstract algebra without ever having taken linear algebra, which was a mistake because linear algebra does not quite work like elementary algebra. Consequently, I found abstract algebra to be a very difficult course even after years of studying math.

You are making yourself feel stupid because you are trying to interpret, in a foreign language, a somewhat ineptly written article on an advanced mathematical topic, one far away from the everyday world and one dependent on a great deal of mathematical knowledge that you do not have yet.

In one sense, you can think of abstract algebra as a game, where we make up the rules and see what results. It has a practical benefit. When we need to make up a new mathematics for some special but practical purpose, we do not need to develop the basic rules for it if we can show that it is essentially the same as some abstract system that has already been explored. This is the reason that abstract algebra FORMALLY has as little contact with anything experienced in the everyday world as possible. But it grew out of trying to make sense of things that did arise in the everyday world and seemed to differ from arithmetic in ways that were not initially understood. Here (from wiki) is a brief history in italics.

As in other parts of mathematics, concrete problems and examples have played important roles in the development of abstract algebra. Through the end of the nineteenth century, many – perhaps most – of these problems were in some way related to the theory of algebraic equations. Major themes include:

• Solving of systems of linear equations, which led to linear algebra
• Attempts to find formulas for solutions of general polynomial equations of higher degree that resulted in discovery of groups as abstract manifestations of symmetry
• Arithmetical investigations of quadratic and higher degree forms and diophantine equations, that directly produced the notions of a ring and ideal.

Numerous textbooks in abstract algebra start with axiomatic definitions of various algebraic structures and then proceed to establish their properties. This creates a false impression that in algebra axioms had come first and then served as a motivation and as a basis of further study. The true order of historical development was almost exactly the opposite. For example, the hypercomplex numbers of the nineteenth century had kinematic and physical motivations but challenged comprehension. Most theories that are now recognized as parts of algebra started as collections of disparate facts from various branches of mathematics, acquired a common theme that served as a core around which various results were grouped, and finally became unified on a basis of a common set of concepts.

Associativity is one of those common themes. It applies to the arithmetic operations of addition and multiplication on real numbers, but not on the arithmetic operations of subtraction and addition. Simple. You are being distracted from understanding the mathematics that every educated person should know by worrying about a concept that has only a foundational role in branches of advanced mathematics that virtually no one who is not a mathematician ever studies. (I was a very odd history student and an even odder banker, but I went to Columbia, which at that time believed no one who graduated should be ignorant of certain basic knowledge. I was required to learn a lot of math because I found organic chemistry boring.)

 

[Dr.Peterson]

I strongly recommend that you simply drop this topic. You are simply not ready to understand it, and nothing I say can change that.

And stop reading Wikipedia or other random sources as if you could learn things from them for the first time. That is like hopping from one planet to another and expecting to understand each culture without a guide. (If you read Wikipedia as I do, skipping over things that I don't have the background for, then you could survive. But often in mathematical topics, you don't even realize that something is unfamiliar because the words seem superficially familiar, as in this case.)

Go through one textbook or video series that takes you through the mathematics you want to learn in a systematic way, ensuring that you are prepared to understand each new topic that they present to you. And even there, if you find yourself failing to understand something, stop and back up to find where they introduced the idea you don't understand. Don't just plow through thinking you should be able to understand everything without knowing the prerequisites.

 

[JeffM]

One of the great mathematicians was Newton. Personally, I suspect Newton was clinically insane, but there is no doubt that he was a major mathematician and physicist. He said that he read Descartes until he was lost and then went back to start from the very beginning, over and over again. He did not jump from one source to another (easy to do then because there were so few sources), nor did he try to go past what he was sure he understood. In the end, Newton’s methodical and limited approach allowed him to surpass by far Descartes and even Leibniz, who was possibly as intelligent and definitely far saner but not as intensely focused as Newton.

If you want to understand a lot superficially, it is OK to flirt with ideas. Flirting, however, is not a relationship even though it’s a frequent way to start one. If you want to learn something, you must start at the beginning and proceed step by step. Jumping from arithmetic to abstract algebra is like learning about relationships by starting with divorce.

The point I am trying to make is that Newton was systematic. He did not bounce from one topic, one teacher, to another, but that single mindedness ended up making him the single most important figure of the 17th century.

 

[Saumyojit]

THANKING U FOR UR INFINTE PATIENCE .
BUT I AM curious to know ....
once u make me understand these last bits of doubt the post is over I hope .


@Dr.Peterson @JeffM
Summary of where I Stand

Associativity is one of those common themes. It applies to the arithmetic operations of addition and multiplication on real numbers, but not on the arithmetic operations of subtraction and addition.

I know this from the beiginning it is very basic.

What they mean is that these are the ways you can place parentheses, just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation.

why did u bring up the case of " a noon asso operation" . Its a honest request ... i think u can end the doubt here only just by giving a explanation .

Associativity doesn't come into this until you say that all those forms have the same value.

. please make me understand or give an eg explaining this line .

What i have understood is that if all the forms give the same result using a asso operator * , then associativty is happening .

THIS is what u are trying to say? If yes , then why did u say it .. i know its very obvious and basic of asso operation .


This is not a doubt -->: What i have understood is : If * is associative operator then writing any combinations of forms is legal as the RESULT is same ..those 5 mentioned + mine 2 + creating extra combo for example like (a)*(b)*c*d, to (((((a*)))*((b*(c))*d))) are all ok as long as * is ASSO.

The five cases all include enough parentheses to make the order in which operations are to be performed clear, without additional rules such as "left-associativity"

I think now i understand what u are trying to refer. U are saying these five forms are sufficent enough to give us a view of THE combinations by rearranging parentheses where NO ADDTIONAL RULES OF CONVENTION ( LEFT ASSO) is needed.

ALTHOUGH my extra two forms are not invalid (If * is asso) but they dont fit into these 5 forms as my forms will be needing Left asso in order of operation which will break the pattern somewhat.

But can u make understand why my two forms are invalid ? in the case of NON asso operation u have mentioned this in earlier post.
I know its valid when * is asso as result will come same.

FPE" is important? I didn't. The only place it would matter is with a non-associative operation

I UNDERSTAND fpe does not matter if * is asso but why and how does it matter if operation is non associaitive.

IF operation is non associative then using that operation and creating new kinds of expression by reaaranging parenthesis will surely not give the same result . What was ur context . Please reply to this question i am curious to know .

 

[Dr.Peterson]

The reason we stopped responding to this thread is that, as we indicated, you don't have the background understanding needed even to ask proper questions about it, much less to fully understand any answers we give. I have humored you more than was good for you. Please believe me: Answers to these questions will not help you understand anything that matters at your level of knowledge.

Just to give one example, your last question here is about "FPE" as if it were a formal term I could explain at length. I had happened to use the phrase "fully parenthesized expressions" in passing, and you apparently found a definition in some other source that I don't think you ever identified for us. (See #21: I just used the term as an attempt to describe what you were asking about.) If it matters at all, it is only as part of some different subject. It will not help you if I explain what I was thinking when I made the wild guesses I made about it.

Please drop the subject. It is an unhelpful obsession, and wastes our time as well as yours.

 

[Saumyojit]

What they mean is that these are the ways you can place parentheses, just enough to determine the order of evaluation sufficiently to determine the value in the case of a non-associative operation. Your additional forms a*(b*c*d) & (a*b*c)*d do not do so.

So * can be any arbitrary operation .
And it is named " product"

The list claims merely to be the ways in which "valid pairs of parentheses can be inserted

The 5 ways that they have given keeping in mind that these are Just enough to show any type of arbitrary operation whether it will be Associative or non associative are just "Fully parenthesized" which is extremely important if operation is non associative!


The list is just fully parenthesized so that any non associative operation can be evaluated without any problem as the order in Which
operation will be performed via " parentheses" has been clearly described Which subexpression to evaluate in order

that's why my two forms are not validate in this list.