Define a relation $R$ on $\mathbb{R}$ as follows: $(x,y) \in R \iff (x-y)(x^{2} + y^{2} - 1) = 0$.

Question

Question:

Define a relation $R$ on $\mathbb{R}$ as follows: $(x,y) \in R \iff (x-y)(x^{2} + y^{2} - 1) = 0$. Is this an equivalence relation?

My attempt:

Reflexive ?

$ \forall x \in \mathbb{R}, (x-x)(x^{2} + x^2 - 1) = 0$. So $(x,x) \in R$. Hence the relation is reflexive.

Symmetric?

If $ \ (x,y) \in R \implies (x-y)(x^2 + y^2 -1) = 0 \implies -(y-x)(x^2 + y^2 -1) = 0 \implies (y-x)(y^2 + x^2 -1) = 0 \implies (y,x) \in R$.

Hence, symmetric.

Transitive?

I don't think its transitive since $(-1,0) \in R, (0,1) \in R$ but $(-1,1) \notin R$.

Is my approach and proof correct?

Answer 1

Not that there is nothing wrong with doing what you did for the question, but all you need to say is:

Because $(-1, 0) \in R$, and $(0, 1) \in R$ but $(-1, 1) \notin R$, $R$ is not transitive. Hence $R$ is not an equivalence relation.

Otherwise, everything is correct. But keep in mind there are tradeoffs between comprehensiveness and conciseness.