For $x, y ∈$ $\mathbb{R}$, let $x△y = 2(x + y)$. Then $△$ is a binary operation on $\mathbb{R}$.
Show that there is no identity element for $△$ on $\mathbb{R}$.
I have tried $x△e = e△x=x$
I don't know what else to do.
For $x, y ∈$ $\mathbb{R}$, let $x△y = 2(x + y)$. Then $△$ is a binary operation on $\mathbb{R}$.
Show that there is no identity element for $△$ on $\mathbb{R}$.
I have tried $x△e = e△x=x$
I don't know what else to do.
Suppose we have an identity $e$. Then
$$0 \triangle e =0\implies2(0+e)=0\implies e=0.$$
But
$$2 \triangle e =2\implies2(2+e)=2\implies e=-1.$$
Thus $-1=e=0$. This is contradictory. So there is no such an identity $e$.
The operator is easily seen to be commutative, so we can just solve $$x\triangle e=2(x+e)=x\implies e=-x/2$$ But the identity cannot depend on $x$, so there is no identity in the first place.