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

Specify the truth or falsity of (a∗b)∗c=a∗(b∗c) if∃ a non-empty set S;a,b,c∈S;∗ is a binary operation on S such that ∃ i∈S such that x∗i=x for any x∈S;x∈S  ⟹  ∃ y∈S such that x∗y=i;p,q,r∈S  ⟹  p=q if p∗r=q∗r or r∗p=r∗q.\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.
If that is what the problem is specifying, I personally am lost. Is S\mathbb S finite so I can play with Cayley tables. Does ∗* have closure? If we are allowed to use the definition for ∗* 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.

6,7∈Z≥0 and 6−7∉Z≥0.(18−6)−5=7≠17=18−(6−5).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).
(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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