Show its not a group for # where a#b=a+b-ab in the set of all real numbers

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SUMMARY

The discussion centers on the mathematical system S={set of all real numbers (ℝ), #} defined by the operation a#b=a+b-ab. It is established that this system does not form a group due to the failure of the associative property. The user attempts to demonstrate closure, identity, and inverses but encounters confusion regarding the associative nature of the operation. The conclusion is that the system cannot be classified as a group without further constraints on the set.

PREREQUISITES
  • Understanding of group theory, including closure, identity, and inverse properties.
  • Familiarity with set theory and operations on real numbers.
  • Knowledge of the associative property and its significance in group structures.
  • Basic algebraic manipulation skills to work with the defined operation a#b.
NEXT STEPS
  • Research the properties of group theory in-depth, focusing on associative operations.
  • Explore examples of non-group structures in set theory to understand common pitfalls.
  • Investigate the implications of defining operations on subsets of real numbers.
  • Learn about alternative algebraic structures, such as semigroups and monoids, that may apply to similar operations.
USEFUL FOR

Mathematics students, educators, and anyone studying abstract algebra or set theory who seeks to deepen their understanding of group properties and operations on real numbers.

brandy
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Homework Statement


In set theory, i have a two part question, the first is showing that the system S={set of all real numbers( \Re )}, #}
where a#b=a+b-ab
we have to show that it's not a group.
and then find what c is so that the system = { \Re \cap\overline{c}, # } is a group.

Homework Equations


the definition i have of a group is closed, associative, has an identity and each element has an inverse.

The Attempt at a Solution


i figured it has to be closed, as in a,b,a#b \in set of all real numbers
identity: i did a#e-ae=0 ->e(1-a)=0,e=0
inverse: a#b=e=a+b-ab, a+b-ab=0 ->a=-b/(1-b), then you have b-b/(1-b)+b^2/(1-b) which is equal to 0 so it works.
so far so good.
associative: i checked it and unless i made a mistake (a#b)#c=a#(b#c), it looks commutive so i tried (a#b)#(c#d) verses a#(b#c)#d but unless i made a mistake this wasn't equal. The system didnt look like it should be associative but by using the definition we were given, it was.
so I'm confused. can someone help point me in the right direction?
 
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a=-b/(1-b),

I can think of a value of b for which this doesn't work
 
*facepalm*
thanks!
 

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