Prove R is a partial order: $((a,b),(c,d))∈R → [a\leq c$ and $a+b \leq c+d]$

Question

$((a,b),(c,d))∈R → [a\leq c$ and $a+b \leq c+d]$

I feel like I have the transitive part down, but with the antisymmetric and reflexive parts I need a lot more detail...

Antisymmetric: Suppose $(a,b)R(c,d)$ and $(c,d)R(a,b)$ then $c\le a$ and $c+d\le a+b$

Reflexive: Let $a=c$ and $b=d$ the $c\le c$ and $c+d\le c+d$

How do I prove $R$ is antisymmetric and reflexive?

Answer 1

Since $a\leq a$ and $a+b\leq a+b$ then $(a,b)\,R\, (a,b)$ so relation is reflexsive.

Say $(a,b)\,R\, (c,d)$ and $(c,d)\,R\,(a,b)$

• From we get $(a,b)R (c,d)$ we get $a\leq c$ and $a+b\leq c+d$ and

• from we get $(c,d)R (a,b)$ we get $c\leq a$ and $c+d\leq a+b$ so we get $a=c$ and $a+b =c+d$ so $b=d$ and thus $(a,b)= (c,d)$.