Associativity of Group Operation

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SUMMARY

The discussion focuses on the associativity of a custom group operation defined as x*y = x+y+1. Participants clarify how to demonstrate that this operation is associative by evaluating x*(y*z) and showing it equals (x*y)*z. The key transformation involves recognizing that the operation does not represent standard multiplication but a defined operation, allowing for the simplification of expressions. The final result confirms that x*(y*z) = x+y+z+2, validating the associativity of the operation.

PREREQUISITES
  • Understanding of group theory concepts, particularly associativity.
  • Familiarity with custom operations and their definitions.
  • Basic algebraic manipulation skills.
  • Knowledge of notation used in abstract algebra.
NEXT STEPS
  • Study the properties of group operations in abstract algebra.
  • Learn about custom binary operations and their implications.
  • Explore examples of non-standard operations and their associative properties.
  • Investigate the role of identity and inverse elements in group theory.
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Students of abstract algebra, mathematicians exploring group theory, and anyone interested in understanding custom operations and their properties.

port31
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Homework Statement


Im looking at this example and trying to figure out how they showed it was associative.
They start out with x*y=x+y+1
then they add in z to show it is associative.
x*(y*z)=x*(y+z+1)=x+(y+z+1)+1=x+y+z+2
I don't know how they go from
this x*(y+z+1)=x+(y+z+1)+1
and end up with the equation on the right.
If i solved for x from x*y=x+y+1 and plugged it in I wouldn't get what they got.
Maybe there is some trick that I am missing.
 
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port31 said:

Homework Statement


Im looking at this example and trying to figure out how they showed it was associative.
They start out with x*y=x+y+1
then they add in z to show it is associative.
x*(y*z)=x*(y+z+1)=x+(y+z+1)+1=x+y+z+2
I don't know how they go from this x*(y+z+1)=x+(y+z+1)+1
and end up with the equation on the right.
They're simply applying the rule for *. If the general rule is x'*y'=x'+y'+1, and you have x*(y+z+1) then you can match up x' with x and y' with (y+z+1):
x*(y+z+1) = x'*y' = x'+y'+1 = x+(y+z+1)+1
If i solved for x from x*y=x+y+1
What do you mean you 'solved it'? The * doesn't stand for normal multiplication here. x*y=x+y+1 is defining an operation '*'.
 
oh ok i see now, thanks for your help.
 

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