# Discrete Mathematics,binary relation

Both $$R_1$$ and $$R_2$$ are equivalence relations on the set $$A$$.($$A$$ is finite)

Define the binary relation $$R$$ on $$A\times A$$as follows.

$$R = \{((a_1, a_2),(b_1, b_2)) | (a_1, b_1) \in R_1,(a_2, b_2) \in R_2\}$$

Prove that $$R$$ is an equivalence relation.

I know a binary relation including reflexivity, Symmetry Transitivity，and assuming $$A=\{a_1,a_2,\ldots,a_n\}$$ then $$A \times A=\{(a_1,a_1),(a_1,a_2), \ldots,(a_n,a_n)\}$$

But I have no idea how to prove it.

One way to unravel the definition is to use infix notation for relation $$R.$$ This is a relation on $$A \times A,$$ so it holds or not between two pairs $$(a,b)$$ and $$(c,d)$$ in the product $$A \times A.$$ in infix notation then: $$(a,b)R(c,d) \leftrightarrow aR_1 c\ \rm{and}\ bR_2 d. \tag{1}$$
Reflexive property: This holds provided for all $$(x,y) \in A \times A$$ we have $$(x,y)R(x,y)$$ or using $$(1),$$ $$xR_1 x \ \rm{and} \ yR_2 y,$$ which follows from reflexivity of $$R_1$$ and $$R_2$$ individually, since they are assumed equivalence relations.
Symmetric property: This holds provided for all $$w,x,y,z$$ in $$A$$ we have if $$(w,x) R (y,z)$$ then also $$(y,z)R (w,x),$$ and again unraveling by using $$(1)$$ we see this holds because $$R_1$$ and $$R_2$$ are each individually symmetric since they are asumed to be equivalence relations.
Transitive property: Here we take any six elements $$u,v,w,x,y,z \in A$$ and then assume that both $$(u,v)R(w,x)$$ and $$(w,x)R(y,z)$$ hold. We then wznt to show that then $$(u,v)R(y,z)$$ holds also. Again a careful application of the re-write in $$(1)$$ finishes things, this time since $$R_1$$ and $$R_2$$ are each individually transitive.
Added note: The above argument did not assume $$A$$ is finite. But of course it would thus cover that case. In my opinion, the notation would have been more cumbersome if we had to work with a specific listing of the elements of $$A$$ in the finite case.
• Thanks any way, but according to my statement ,$R$ should be $$(a,c)R(b,d) \leftrightarrow aR_1 b\ \rm{and}\ cR_2 d$$ , instand of $$(a,b)R(c,d) \leftrightarrow aR_1 b\ \rm{and}\ cR_2 d$$. But I don't think this hurt the provement.Thanks again! Mar 29, 2021 at 5:15
• @ChenJason You're right. In my setup I was defining $(a,b)R(c,d)$ and should have put $aR_1c \ \rm{and} \ bR_2d.$ If you interchange $b$ and $c$ in your correction it matches mine. Mar 29, 2021 at 10:34