S is the set of ordered pairs of integers and (x1, x2) R(y1, y2) means that x1= y1and x2≤ y2Demonstrate whether R exhibits the reflexive property or not.Demonstrate whether R exhibits the symmetric property or not.Demonstrate whether R exhibits the transitive property or not.

Answer:R is reflexiveR is not symmetricR is transitiveStep-by-step explanation:R is reflexive. To show this, we have to verify that for any pair of integers [tex](x_1,x_2)[/tex] [tex](x_1,x_2)R(x_1,x_2)[/tex]. But this is obvious because [tex]x_1=x_1[/tex] and [tex]x_2\leq x_2[/tex]. R is not symmetric. To show it, we need to find two pairs [tex](x_1,x_2)[/tex] and [tex](y_1,y_2)[/tex] such that [tex](x_1,x_2)R(y_1,y_2)[/tex] but [tex](y_1,y_2) \not \mathrel{R} (x_1,x_2)[/tex] For example (1,1) and (1,2). [tex](1,1)R(1,2)[/tex] for 1=1 and [tex]1\leq 2[/tex] but  [tex](1,2) \not \mathrel{R} (1,1)[/tex] because [tex]2\not \leq 1[/tex] Finally, R is transitive. If we take 3 pairs of integers [tex](x_1,x_2), (y_1,y_2)[/tex] and [tex](z_1,z_2)[/tex] Such that [tex](x_1,x_2)R(y_1,y_2)[/tex] and [tex](y_1,y_2)R(z_1,z_2)[/tex] then [tex]x_1=y_1[/tex] and [tex]x_2\leq y_2[/tex] [tex]y_1=z_1[/tex] and [tex]y_2\leq z_2[/tex] But then, [tex]x_1=z_1[/tex] and [tex]x_2\leq z_2[/tex] So  [tex](x_1,x_2)R(z_1,z_2)[/tex].