Does the Operation in Set Theory Imply a Singleton Set?

set
Messages
58
Reaction score
0

Homework Statement


Let S be a set with an operation * which assigns an element a*b of S for any a,b in S. Let us assume that the following two rules hold:
1. If a, b are any objects in S, then a*b = a
2. If a, b are any objects in S, then a*b = b*a
(Herstein, Abstract Algebra, 2ed)


Homework Equations


Is it safe to assume that the symmetry, transitivity, and reflexibility hold?


The Attempt at a Solution


a = a*b = b*a = b
But I am not sure if this is sufficient as it is my first course (in fact, first problem!) in abstract algebra...
[Edit]
Or with the relation obtained from the axioms of S, shall I proceed with proof by contradiction?
 
Physics news on Phys.org
a = a*b = b*a = b
This is true for all a,b in S, and shows that S just has one single element. That is a strange problem.
The validity of those axioms will follow from that, but where is the point?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top