Hello Snooks02Suppose X is the set {1, 2, 3, 4} and R is the relation on X: xRy if and only if x > y.

Then R = {(2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)}

Suppose now that we define a subset Y as {1, 2, 3}.

Then Y x Y = {(1, 1), (2, 1), (3, 1), (1, 2), (2, 2), (3, 2), (1, 3), (2, 3), (3, 3)}

So the restriction of R to Y is those elements (ordered pairs) of R that are also elements of Y x Y; in other words {(2, 1), (3, 1), (3, 2)}.

So, in the question you are given, instead of all values of x and y (positive and negative) that satisfy you need only those that are positive. On a Cartesian (x-y) plane, instead of the whole circle centre O, radius 1, we just get the quarter of this circle that lies in the first quadrant. OK?

Grandad