- #1
Punkyc7
- 420
- 0
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.
[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.