Discrete Math - Hasse Diagrams

Question

This is one of many questions of similar type I have to do for an assignment and im troubled with what to do. The question is as follows:

Consider a relation R defined on the set A = {−7, −6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7}. Determine for the following if the relations are reflexive, symmetric, anti-symmetric, transitive, partial orders, equivalence relations. (a) R = {(a, b)|a ≤ b}, if this is a partial order, draw the Hasse diagram.

I understand the definitions of what symmetric, anti-symmetric, reflexive and transitive are but im troubled with understanding a partial order. Also how can one draw a hasse diagram? Is there any logic/rules to apply?

thanks!

Answer 1

To draw a Hasse diagram of a finite poset follow the algorithm below, but first notice that on any finite poset $P$, given $x,y\in P$ such that $x\neq y$, then $x$ and $y$ aren't comparable or there is a chain of covers from one to the other.

Now for the algorithm: Let $x,y,z\in P$

1. To each $p\in P$ associate a point $(a_p,b_p)$ on the euclidean plane in such a way that $x<y$, then $b_x<b_y$, if $x\neq y$.
2. If $x$ is covered by $y$, then draw a line from $(a_x,b_x)$ to $(a_y,b_y)$.
3. Ensure that, if $x\neq z\neq y$, $(a_z, b_z)$ is not in the line that goes from $(a_x,b_x)$ to $(a_y,b_y)$.

Answer 2

A partial order is any relation that is reflexive, antisymmetric, and transitive.

An equivalence relation is any relation that is reflexive, symmetric, and transitive.

See if you can show that your relation $R$ is a partial order on set $A$ and why it is, but it is not an equivalence relation and why it is not.

A Hasse diagram shows visually, for a partially ordered set, how elements "compare": in this case, you have a partially ordered set, with a total order, and can be thought of as a chain.