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$R \subset \Bbb R \times \Bbb R$

I have now encountered a couple of relations that have the following form:

$$R=\{(a,b)\in \Bbb R\times \Bbb R\,:\,a^2 = b^2\}$$

They seem to be always equivalence relations. Could it be that every relation that consists of $= $ is an equivalence relation?

Maik Klein
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2 Answers2

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It depends a bit on what you mean by "consists of $=$". But here is something which is true and probably what you generally encounter.

Let $A$ and $B$ be sets and $f: A\to B$ be a function. Then the set $\{(x,y)\in A\times A\mid f(x) = f(y)\}\subseteq A\times A$ is an equivalence relation on $A$.

In fact, this also works the other way around. If you have an equivalence relation on $A$ then there is a set $B$ and a function $f: A\to B$ such that the equivalence relation has the above form (take $B$ to be the set of equivalence classes of the relation).

Tobias Kildetoft
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First off, it is equivalence relation and not equivalent relation. That said, consider: $$R:=\{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a=2b\},$$ which is clearly not an equivalence relation.

b00n heT
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    Other easy examples can be constructed simply by negating the equality, e.g. $\lnot (a^2 = b^2)$. – Newb Jul 25 '16 at 09:56