Answer:
R is reflexive
R is not symmetric
R is transitive
Step-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].
