Determining Associativitiy

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Zelda22

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(Z,*) is invertible and has cancellation property, does it imply that is associative?

I know that if it is invertible and is associative, then it is a group and has cancellation property.

I'm confused. Can I say it is associative, only known that is invertible and has cancellation?

Please help, an example will be highly appreciated. Thanks
 

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You have to be kidding me.

Why not just test associativity? Let a, b and c be in Z. Does (a+b)+c = a+(b+c)?????
 
Zelda22 said:
Yes, I know is associative, I tested it. But question part d is asking if I can tell just solely based on the answers of a,b and c.
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Exactly, determine if (a+b)+c = a+(b+c) using just a-c.
You claim that you know that is associativity. Did you conclude that using a-c? If yes, then you're done.
 
I think the question of part d is referring to my answers to parts a, part b, and part c.

part a- is has a neutral element
part b- it is invertible
part c- it has cancellation property
 
Zelda22 said:
I think the question of part d is referring to my answers to parts a, part b, and part c.

part a- is has a neutral element
part b- it is invertible
part c- it has cancellation property
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The question is can you convert a*(b*c) to (a*b)*c using just those properties? As Steven G says it's almost certainly easier by just using the definition. I don't see how to do it from a - c.

-Dan
 
topsquark said:
The question is can you convert a*(b*c) to (a*b)*c using just those properties? As Steven G says it's almost certainly easier by just using the definition. I don't see how to do it from a - c.

-Dan
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Me neither. I think I should answer No to part d.
 
Zelda22 said:
I think the question of part d is referring to my answers to parts a, part b, and part c.

part a- is has a neutral element
part b- it is invertible
part c- it has cancellation property
Click to expand...
So use them to check for associativity.
 
It is not possible, solely on the basis of the answers to parts (a) - (c), to determine whether or not (Z,∗) is associative because there exist both associative and non-associative operational systems that have a neutral element, are invertible, and have the cancellation property. However, it so happens that (Z,∗) actually is associative, although this cannot be determined from the answers to parts (a) - (c)
 
Zelda22 said:
It is not possible, solely on the basis of the answers to parts (a) - (c), to determine whether or not (Z,∗) is associative because there exist both associative and non-associative operational systems that have a neutral element, are invertible, and have the cancellation property. However, it so happens that (Z,∗) actually is associative, although this cannot be determined from the answers to parts (a) - (c)
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So give a counter example.
 
I think the question has been very badly posed.

Yes, (Z,*) is associative.

But the real question is whether a structure that has two specific properties can be proved to have a third specific property. The mention of (Z, *) just confuses the issue.
 
@Steven G

Yes, I am really perplexed. The only ways to show that the answer is “no“ is to prove that the addition of associativity to the other properties leads to a contradiction or to give a counter-example. But (Z, *) shows that associativity can be joined to the other elements without contradiction so I do not see that the answer of “no” is justified without a counter-example.

As I said, it is not a well worded question, and I cannot find it in my heart to blame the student for a bad question.
 
(a*b) = a+b-50 is given

[(a*b) * c] = [(a+b-50) * c] = (a+b-50) + c -50 = a+b+c-100

[a * (b*c)] = [a*(b+c-50)] = a+(b+c-50) -50 = a+b+c-100

Am I not allowed to use the definition of * along with a-c? If yes, then assoc holds with a-c.
 
@Steven G

But that is the flaw in this question. What does “solely” mean? Given the “correct” answer, it appears to mean that we are being asked

[math]\text {Specify the truth or falsity of } (a * b) * c = a * (b * c) \text { if}\\ \exists \text { a non-empty set } \mathbb S;\\ a,b,c \in \mathbb S; \\ * \text { is a binary operation on } \mathbb S \text { such that }\\ \exists \ i \in \mathbb S \text { such that } x * i = x \text { for any } x \in \mathbb S;\\ x \in \mathbb S \implies \exists \ y \in \mathbb S \text { such that } x * y = i;\\ p, q, r \in \mathbb S \implies p = q \text { if } p * r = q * r \text { or } r * p = r * q. [/math]
If that is what the problem is specifying, I personally am lost. Is [imath]\mathbb S[/imath] finite so I can play with Cayley tables. Does [imath]*[/imath] have closure? If we are allowed to use the definition for [imath]*[/imath] given on the problem, the problem has an answer different from the so-called “correct“ answer.
 
How can you even talk about having associativity if * is not (well) defined?
 
Steven G said:
How can you even talk about having associativity if * is not (well) defined?
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@Steven G I like you way too much to squabble over a poorly worded problem. Associativity is not a necessary property for an operation. Using the non-negative whole numbers as our set and arithmetic subtraction as our binary operation, we have an algebraic structure that has neither closure nor associativity.

[math]6,7 \in \mathbb Z^{\ge 0} \text { and } 6 - 7 \not \in \mathbb Z^{\ge 0}.\\ (18 - 6) - 5 = 7 \ne 17 = 18 - (6 - 5).[/math]
(I know we avoid this as a problem in elementary algebra by adopting the notational solution of PEMDAS, but PEMDAS is not a principle of abstract algebra.)

In abstract algebra, properties like closure and associativity must be definitional or theorems. I believe octonians and sedenions are not associative.

Can we please just agree that this is a badly worded problem?
 
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