Proof of No Right Identity for Operation with Two Left Identities

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

Then we have that
e[itex]_{1}[/itex]x=x
e[itex]_{2}[/itex]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.
[itex]_{}[/itex]
pf/
Let e[itex]_{1}[/itex] and e[itex]_{2}[/itex] be left identities such that e[itex]_{1}[/itex]≠e[itex]_{2}[/itex]. 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[itex]_{1}[/itex]r=e[itex]_{1}[/itex]=r

and

e[itex]_{2}[/itex]r=e[itex]_{2}[/itex]=r

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