Relation

[thin spindle]

...I think it might be a typo and what it means is to show that R is a subset of A^2?

Because a set is never a subset of its own Cartesian square.

...wait, I don't get the hint.

[wicked geyser]

it's probably something else in place of "R^2", other than A^2

based on the hint

but yeah, very weird notation

[surreal oriole]

mm

so is the question not possible to solve with R is a subset of R^2?

[wicked geyser]

R being a subset of R^2 doesn't make sense

but in the hint it says you just need to prove that given an arbitrary pair (a, b) in R, then there is c in A with (a, c) and (c, b) in R

so just prove that instead

I suppose whoever made the question was thinking about the set
{ P in R^2 ; \exists x, y, z in A with ( (x, y), (y, z) ) = P}
which you can interpret as the set of 2-paths in the graph corresponding to R

so it's not completely arbitrary

[surreal oriole]

"(a, b) in R, then there is c in A with (a, c) and (c, b) in R"

isn't this the definition of transitivity

[wicked geyser]

no, transitivity is in the opposite direction

[surreal oriole]

(a, b), (b, c), then (a,c)?