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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.

mku
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    Assume it has an identity $e$. Compute $e \triangle 0 =0$ to get a value for $e$. Now do it for $e \triangle 1$ and get a contradiction. – Randall Feb 20 '23 at 00:31

2 Answers2

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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 $e$.

HeroZhang001
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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.

Parcly Taxel
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