Is the following relation, an Equivalence relation?

Question

Relation R is defined as:

$(a,b)R(c,d) \iff (a-c)(b-d)=0$ where $a,b,c,d$ are Real numbers.

Is this relation an Equivalence relation?

I think it is not. It is a symmetry relation and a reflexive relation but I think it is not transitive. but I read in a book that said it is transitive too. which one is true?

Answer 1

Indeed it is not transitive since

$$(1,2)R(3,2)\quad \text{and}\quad (3,2)R(3,4)\quad \text{but}\quad(1,2)\not R(3,4)$$

Moreover, if this relation is defined in any non-trivial commutative ring then

$$(1,0)R(1,1)\quad \text{and}\quad (1,1)R(0,1)\quad \text{but}\quad(1,0)\not R(0,1)$$

Answer 2

You are right, this relation is not transitive. You have

$(1,0)R(1,2)$ and $(1,2)R(3,2)$, but $(1,0)R(3,2)$ doesn't hold.

Answer 3

We have

$$(1,2)R(1,3)$$ and $$(1,3)R(2,3)$$

but ...