# Thread: How would I write a proof for these partial orders?

1. ## How would I write a proof for these partial orders?

Let R 1 and R 2 be relations on N defined by

xR 1 y if and only if y=a+x for some a∈N 0 .

xR 2 y if and only if y=xa for some a∈N

For all x;y∈N.

Also N 0 denotes all integers x≥0 , while N denotes all integers x≥1 .

There are two different things I want to prove with this.
1.
I want to write a proof to show that R 1 is a partial order on N .

2.
I want to write a proof to show that R 2 is a partial order on N .

I have others that I want to try and do but for now if someone could model how to write a proof for these that would be great as I could reference to it.

2. ## Re: How would I write a proof for these partial orders?

Originally Posted by mj2323
Let R 1 and R 2 be relations on N defined by

xR 1 y if and only if y=a+x for some a∈N 0 .

xR 2 y if and only if y=xa for some a∈N

For all x;y∈N.

Also N 0 denotes all integers x≥0 , while N denotes all integers x≥1 .

There are two different things I want to prove with this.
1.
I want to write a proof to show that R 1 is a partial order on N .

2.
I want to write a proof to show that R 2 is a partial order on N .

I have others that I want to try and do but for now if someone could model how to write a proof for these that would be great as I could reference to it.
I see that A relation R on S that is reflexive, anti-symmetric, and transitive is said to be a partial order on S.

So you need to show each of these for your relation.

reflexive means (x R x)

antisymmetric means if (x R y) and (y R x) then x = y

transitive means if (x R y) and (y R z) then (x R z)

Ok so now show each of these for your relations.

R1: (x R1 y) iff y=a+x for a a non-negative integer.

does (x R1 x)? x = a+x --> a = 0, and 0 is a non-negative integer so yes R1 is reflexive

suppose (x R1 y), then y = a+x, if (y R1 x) then x = a+y, so x = a + a + x, a = 0, and thus x = y. So R1 is anti-symmetric

suppose (x R1 y), and (y R1 z). then y = a1+x, z = a2 + y, so z = (a1+a2) + x, (a1+a2) is in N0, and so (x R1 z), and R1 is transitive.

Now you do R2.