Math Problem Statement

Prove that (P, ≤) is a partially ordered set (POSET), where P = {1, 2, 3, 4, 6, 12} and the relation “≤” is defined as follows: a ≤ b if a divides b (i.e., b = ka) for some integer k.

Solution

To prove that (P,)(P, \leq) is a partially ordered set (POSET), we must show that the relation \leq, defined as "a divides b", satisfies the three properties of a partial order: reflexivity, antisymmetry, and transitivity. Let P={1,2,3,4,6,12}P = \{1, 2, 3, 4, 6, 12\}.


1. Reflexivity:

A relation \leq is reflexive if for all aPa \in P, aaa \leq a. In our case, aaa \leq a means aa divides aa, which is true for any integer since any integer divides itself.

Verification: For a{1,2,3,4,6,12}a \in \{1, 2, 3, 4, 6, 12\}, we have a=kaa = k \cdot a where k=1k = 1. Thus, aaa \leq a holds for all aPa \in P.

\therefore Reflexivity is satisfied.


2. Antisymmetry:

A relation \leq is antisymmetric if for all a,bPa, b \in P, aba \leq b and bab \leq a imply a=ba = b. In our case, aba \leq b means b=kab = ka for some integer kk, and bab \leq a means a=mba = m \cdot b for some integer mm. If both hold, aa and bb must be the same.

Verification: Suppose aba \leq b and bab \leq a. Then: b=kaanda=mb.b = k \cdot a \quad \text{and} \quad a = m \cdot b. Substituting b=kab = k \cdot a into a=mba = m \cdot b, we get: a=m(ka)=(mk)a.a = m \cdot (k \cdot a) = (m \cdot k) \cdot a. For this to hold, mk=1m \cdot k = 1, which implies m=k=1m = k = 1 (since mm and kk are integers). Thus, a=ba = b.

\therefore Antisymmetry is satisfied.


3. Transitivity:

A relation \leq is transitive if for all a,b,cPa, b, c \in P, aba \leq b and bcb \leq c imply aca \leq c. In our case, aba \leq b means b=kab = k \cdot a, and bcb \leq c means c=mbc = m \cdot b. We must show that c=nac = n \cdot a for some integer nn.

Verification: Suppose aba \leq b and bcb \leq c. Then: b=kaandc=mb.b = k \cdot a \quad \text{and} \quad c = m \cdot b. Substituting b=kab = k \cdot a into c=mbc = m \cdot b, we get: c=m(ka)=(mk)a.c = m \cdot (k \cdot a) = (m \cdot k) \cdot a. Let n=mkn = m \cdot k, which is an integer. Thus, c=nac = n \cdot a, meaning aca \leq c.

\therefore Transitivity is satisfied.


Conclusion:

Since the relation \leq on PP satisfies reflexivity, antisymmetry, and transitivity, (P,)(P, \leq) is a partially ordered set (POSET).


Would you like to explore specific examples or visualizations (e.g., a Hasse diagram) to better understand the structure?
Here are five related questions to deepen your understanding:

  1. What is the difference between a partial order and a total order?
  2. Can you draw the Hasse diagram for the given POSET?
  3. How would adding a new element to PP (e.g., 8) affect the POSET structure?
  4. What is the greatest lower bound (GLB) of two elements in this POSET?
  5. Can a partially ordered set have cycles? Why or why not?

Tip: A Hasse diagram is a useful way to visualize a POSET by representing elements and their direct relationships (covering relations) without

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Math Problem Analysis

Mathematical Concepts

Partially Ordered Sets (POSET)
Relations
Number Theory

Formulas

a ≤ b if b = ka, where k is an integer

Theorems

Properties of Partial Orders: Reflexivity, Antisymmetry, Transitivity

Suitable Grade Level

Grades 9-12