Denoting Like Inequalities

Question

I wish to assert that the equivalence of ordered pairs (a,b) and (c, d) is contingent upon the common direction of their inequality. a could be less or greater than b but (c, d) could be considered eligible for equality with (a, b) only if the inequality between c and d is of the same kind.

For example, (5, 1) and (4, 1) both have the ">" inequality whereas (5, 1) and (1, 5) have ">" and "<," respectively. What is the traditional/proper way in which to say or notate that the inequalities match as with [(5, 1) and (4, 1)] and not [(5, 1) and (1, 5)]?

Answer 1

One way you can do it is say $$(a,b) \equiv (c,d) \quad if \quad (a-b)(c-d)>0$$

Notice ordered pairs $(a,b)$ with the attribute ">" will have $a-b>0$. Similarly, those with "<" will have $a-b<0$. Thus if 2 order pairs $(a,b)$ and $(c,d)$ have the same direction (both ">" or both "<"), then $(a-b)(c-d)>0$

Also notice that this definition excludes ordered pairs of the form $(a,a)$. To include these pairs, we would alter our equivalence to be $(a,b) \equiv (c,d) \quad if \quad (a-b)(c-d)>0 \,\, or \,\, (a-b)=(c-d)$

Answer 2

For two ordered pairs $(a,b)$ and $(c,d)$ s.t. $a,b,c,d$ are elements of a set $S$ with a relation $R$, you can define an "equality" relation $T$, by $(a,b)T(c,d)$ iff $aRb$ and $cRd$. So $(5,1)T(4,1)$ and $(5,1)\not T(1,5)$.