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#1 abanjo

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Posted 4 September 2016 - 07:03 PM

Hi,

an easy one I guess but its been a while since I last stepped into a class room so I need some help.

 

I need a function that fulfill  f(f(a,b),c) = f(a, f(b,c)) e.g. simple multiplication or XOR, but in theory how does this family of functions are called?

 

Thanks!


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#2 ajb

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Posted 4 September 2016 - 07:21 PM

This is the associative property for a binary operation on a set. For example, standard multiplication of real numbers satisfies this property.
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#3 abanjo

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Posted 4 September 2016 - 07:25 PM

Thanks  :-)

exactly what I was looking for


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#4 blue89

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Posted 4 September 2016 - 08:45 PM

~ IMPLICIT FUNCTION ~

 

 for any E set  ,E ⊂ R3   ;

 

F : E → R 

 

G ⊂ R2   ,    f function defined    such that 

 

f: G → R ,  x3 = f(x1,x2)  

 

if f function provides  every   (x1,x2) ϵ G  ,   F ( x1,x2,f(x1,x2) ) ≡ 0 equivalency , then f function  will be a solution of  F(x1,x2,x3) = 0  function.

 

 

any type of f , g solutions which would provide x1= h(x2,x3) or x= f(x1,x3) equalities  , will be implicit function for implicit defined function F.

 

 

example 1

 

x12+ x22 - x3 =0 equation ,defines only one implicit function on R2 as x3= x12+x22

 

 

example 2

 

x12+ x22+y2-= 0 , this equation defines 2 implicit functions  on G = { (x1,x2) ϵ R2 : x12 + x2≤ 1 }

 

f  , h :  G → R , y = f (x1, x2) = √ (1 - x12 - x22   and  y = h (  (x1, x2) = - ( √ (1 - x12 - x22


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#5 ajb

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Posted 4 September 2016 - 09:01 PM

I don't think we are dealing with implicit functions here - the example given in the opening post suggest that we are dealing with binary operations that are associative.

An examples of something that does not satisfy the associative property are Lie algebras.
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#6 blue89

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Posted 4 September 2016 - 09:14 PM

mmm , really there was already a strange appearance in writing ;

f ( f(a,b) ,c) ,because this is not original notation for implicit functions , it should have been F(a,b,f(a,b))as I did.

mm furthermore ,presumably I have been a bit myopic 

I could not realise the second writing of f letter well.

 

What was implied with "multiplication"

 

  this one 

 

(fog)(x) = f(g(x)) ?


Edited by blue89, 4 September 2016 - 09:14 PM.

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#7 ajb

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Posted 4 September 2016 - 09:20 PM

I think the word 'function' in the opening post is not correct - rather we have a binary operation.

It is true that composition of functions is associative...
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#8 blue89

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Posted 4 September 2016 - 09:38 PM

already, I could not understand anything clearly. 

Now ,I think that it should be compulsory to write ORIGINAL AND FULL TEXT of QUESTION.

 

:-) :-)

 

but it would be good to say 

 

there is nothing which is DIFFICULT in mathematics.

 

all existences are things what require studying hard and more carefully :-)


I have to leave ....the time is 00.36 ,in turkey

Unless I sleep ,it will be worse for my eyes. 

good nights.

 

sincerely

blue


Edited by blue89, 4 September 2016 - 09:33 PM.

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#9 ajb

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Posted 4 September 2016 - 09:40 PM

He or she has asked for the generic name of binary operations that satisfy f(f(a,b),c) = f(a, f(b,c)). Such binary operations are known as associative binary operations. They can now simply 'google' this for more examples.
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#10 blue89

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Posted 5 September 2016 - 06:14 AM

is it the same defiition with homomorphism / isomorphism (1-1 ,onto) ?


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#11 ajb

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Posted 5 September 2016 - 10:22 AM

is it the same defiition with homomorphism / isomorphism (1-1 ,onto) ?


As long as the domain and range are compatible one can make sense of associativity - so partial binary operations can be associative. For example, when dealing with groupoids where we have a partial multiplication.
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#12 blue89

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Posted 5 September 2016 - 11:32 AM

As long as the domain and range are compatible one can make sense of associativity - so partial binary operations can be associative. For example, when dealing with groupoids where we have a partial multiplication.

 

ok. thanks although I have forgotten to give some particular instructions ,I see you have not missing such details (like red coloured) 

 

could you gice an example clearly please. 

 

For instance if   < G,* > and < H,ß > be each group  :

 

F : G → H and  if F provide the requirements below this will be homomorphism or isomorphism 

 

for every a , b ϵ G 

 

F ( a*b) = F (a) ß F(b)        ⟹  F is homomorphism

 

When F provides 1-1 and onto characters at the same time , then F is isomorphism.

of course there will be very many variety of groups (I know up to nilpotent groups but there exists many other types too) 

Could you give an example for binary operation?

(I could not find the best terminology ,so definition (!) ) :) 


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#13 ajb

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Posted 5 September 2016 - 12:17 PM

Could you give an example for binary operation?


Standard multiplication and addition of real or complex numbers; Group multiplication; vector addition; matrix multiplication... all these are associative.

Non-associative examples include multiplication of octonians and Lie brackets.

Associativity means that you can make sense of a*b*c without having to put in the parenthesis -- that is a*(b*c) = (a*b)*c
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#14 blue89

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Posted 5 September 2016 - 01:11 PM

Standard multiplication and addition of real or complex numbers; Group multiplication; vector addition; matrix multiplication... all these are associative.

Non-associative examples include multiplication of octonians and Lie brackets.

Associativity means that you can make sense of a*b*c without having to put in the parenthesis -- that is a*(b*c) = (a*b)*c

 

isn't this a bit basic? or ..what its importance?

 

for instance if there exists any discussion about such sets ,I don't know the reason why we learnt thems. look ,

 

M ≠ Ø  < M , o > provides qualifications below.

 

1) every a,b  ϵ M aob ϵ M (closed) 

 

2) every a,b ϵ M (aob)oc = ao(boc) (associative)

 

3) e unit element , every a ϵ M eoa = aoe = a

 

-----

although there exist many many many subjects & titles at algebra , I do not remember we used such sets commonly.

I see you are using "Lie Algebra" in your papers

what is the usage of this definition : this is "Monoid",but has it importance? ) 

this set does not provide the last requirement to be group!

 

last requirement :

4)every a,b  ϵ M  one x must exist  such that aox=b    ˅  

,4* A) e ϵ M any a ϵ M , aoe = eoa   ˄  B) for any a ϵ M , a*  ϵ M such that aoa* = e 

4 and 4* requirements are equivalent.

 

and was the binary oparetaion there "o"?

correctance: e is not unit element ,it should be available only. but not element of M as I know.  


Edited by blue89, 5 September 2016 - 01:55 PM.

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#15 ajb

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Posted 5 September 2016 - 05:17 PM

isn't this a bit basic? or ..what its importance?


basic yes, but the point is that we do have binary operations that are not associative.
 

M ≠ Ø  < M , o > provides qualifications below.
 
1) every a,b  ϵ M aob ϵ M (closed) 
 
2) every a,b ϵ M (aob)oc = ao(boc) (associative)
 
3) e unit element , every a ϵ M eoa = aoe = a


This is the definition of a monoid.
 

although there exist many many many subjects & titles at algebra , I do not remember we used such sets commonly.


Monoids are not uncommon, but for sure they are less well known that say a group. An example of a monoid are the real numbers with standard multiplication.

I see you are using "Lie Algebra" in your papers


Lie algebras are fundamental in mathematical physics and differential geometry. They are also of great interest from a pure algebra point of view. I would be very suprised if you have not encountered these in your studies - at least just the basics and some simple examples.

what is the usage of this definition : this is "Monoid",but has it importance? ) 
this set does not provide the last requirement to be group!


Yes, you have defined a monoid. Typically a monoid is a 'group' for which not all the elements have an inverse. So, for the case of the real numbers and muliplication, 0 has no inverse.
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#16 blue89

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Posted 5 September 2016 - 08:33 PM

I thought that this text was not suitable to be shown under this thread, it has been folded & sent to ajb by private message

 

 

...

 

blue


Edited by blue89, 5 September 2016 - 08:40 PM.

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