Functions and Relations 2 A relation R is defined on ordered pairs of integers as follows :
$(x,y) R(u,v)$ if $x<u$ and $y>v.$ 
Then R is 


*

*Neither a Partial Order nor an Equivalence relation

*A Partial Order but not a Total Order

*A Total Order 

*An Equivalence relation
 A: It's not even reflexive, i.e. we don't have that $(x,y)R(x,y)$. So it has no chance of being a partial order or an equivalence relation.
A: The answer depends on details of definition in your course! Some definitions of partial order ask that for $R$ to be a partial order, we can never have simultaneously  $xRy$ and $yRx$. Call this sort of definition Definition $1$. 
Other definitions specify that $xRy$ and $yRx$ if and only if $x=y$. Call this sort of definition Definition $2$. 
Under Definition $2$, the correct answer is $1$. For, as pointed out already by Alex P., we have that $(1,1)R(1,1)$ does not hold. That simultaneously kills the possibility of being a partial order under Definition $2$, and also the possibility of being an equivalence relation.
The situation is different under Definition $1$, for then $R$ is a partial order, and $(2)$ becomes the correct answer. 
The relation is $R$ is then a partial order. To do this, you need to verify that the conditions for partial order hold. We need to check that we cannot have simultaneously $(a,b)R(c,d)$ and $(c,d)R(a,b)$. This is because if $(a,b)R(c,d)$ holds, then $a\lt c$, which rules out  $(c,d)R(a,b)$.  We also have to check transitivity, that is, to show that if $(a,b)R(c,d)$ and $(c,d)R(e,f)$ then
$(a,b)R(e,f)$. This is because if $a\lt c$ and $c\lt e$, then $a\lt e$. Also, if $b\gt d$ and $d\gt f$ then $b\gt f$. 
Note that even under Definition $1$, the relation $R$ is not a total order. For consider the pairs $(1,2)$ and $(3,4)$. It is easy to check that $(1,2)R(3,4)$ fails, since $3$ is not greater than $4$. Also, $(3,4)R(1,2)$ fails, since $3$ is not less than $1$. so $(1,2)$ and $(3,4)$ are incomparable. 
