I'm sorry, but the more I study this the more confused I am getting. I posted the problem #2 ax=ay, ua=wa (reference link:http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215804"
In particular, the Proposition (Mappings on Finite and Infinite Sets). I think I understand the...
Other than problem 2 being over a finite set, what is the fundamental difference in these questions (i.e. I see problem 2 has left / right cancellation properties but not sure how it changes the answer for both these questions)
Problem 1:
If G is a set closed under an associative operation...
I understand that (N, +) meets conditions 3 and 4 yet has no inverse function and therefore is an answer to the problem. Could you explain how the right/left cancellation implies identity and inverse for finite sets but fails for infinite sets?
Thanks again.
Thank you.
Is the question getting at ranking of algebraic structures in the order of increasing complexity (i.e. magma, semigroup, monoid, group,...).
The original finite set with closure, associative, and a*b=a*c, and b*a=c*a is a group because the right and left cancellation...
Ok, the integers under multiplication (Z,x) is not a group because it fails the inverse. Therefore a finite set of this would also fail.
However, the problem asks for a finite set that is a group, but when extended to infinity fails the group definition. Is this not what the question is...
Ok, please bare with me. I'm really struggling with all this.
I understand that an infinite positive set fails because the inverse is negative (using addition) and is not in the set/group. But even a finite set would fail for the same reason. So I don't understand how this answers the...
I'm not picking up on what your trying to say...
Are your referring to mapping the natural integers to real numbers, in which case, its not onto, and therefore is not bijective?
Ok, now I'm really confused...
The set of all integers with an addition operation is a group that is closed, associative and
the inverse equal to -a and identity of 0.
Wouldn't any subgroup have the same properties?
Please help clarify.
I understand that an is the cardinality and that with an infinite set m=[FONT="Franklin Gothic Medium"][SIZE="4"]No (aleph-0), and a finite set has n=n and therefore they are not equal. Based on the above cardinality rules, the infinite set would fall under case 3 and is not bijective.
I am...
I understand for a Group there are 4 conditions that must be met.
1) Closure (if a,b are elements of G, then a*b must be elements of G)
2) Associativity (a*b)*c = a*(b*c)
3) Identify must exist e • a = a • e = a.
4) Inverse must exist a • b = b • a = e, where e is an identity element...
Let G be a finite nonempty set with an operation * such that:
1. G is closed under *.
2. * is associative.
3. Given with a*b=a*c, then b=c
4. Given with b*a=c*a, then b=c
Give an example to show that under the conditions above, G is no longer a group if G is an infinite set?