Proof of No Right Identity for Operation with Two Left Identities

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Homework Help Overview

The discussion revolves around the proof concerning operations with two left identities and the implications for the existence of a right identity. The subject area involves abstract algebra, particularly the properties of identity elements in algebraic structures.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of having two distinct left identities and the assumption of a right identity. They discuss evaluating expressions involving these identities to derive contradictions.

Discussion Status

The discussion has progressed with participants identifying potential contradictions arising from the assumption of a right identity. Some guidance has been provided regarding evaluating specific expressions, leading to further exploration of the implications.

Contextual Notes

Participants are working under the assumption that the identities are distinct and are attempting to derive a contradiction based on this premise. There is a focus on the logical structure of the proof without reaching a definitive conclusion.

Punkyc7
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If an operation has two left identities, show that it has no right identity.
_{}
pf/
Let e_{1} and e_{2} be left identities such that e_{1}≠e_{2}. Assume there exist a right identity and call it r.

Then we have that
e_{1}x=x
e_{2}x=x and
xr=x.


From here I want to try and show that there can not be a right identity but I don't see where to go.
 
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Punkyc7 said:
If an operation has two left identities, show that it has no right identity.
_{}
pf/
Let e_{1} and e_{2} be left identities such that e_{1}≠e_{2}. Assume there exist a right identity and call it r.

This is fine so far.

Try evaluating e1r. What two pieces of information can you conclude? Similarly...
 
wouldnt I get
e_{1}r=e_{1}=r

and

e_{2}r=e_{2}=r

So we get e_{2} and e_{1} are equal contradicting that they were distinct.
Is that right?
 
Correct.
 
thanks, I was trying to figure it out with the x's and I couldn't come to any contradiction
 

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