Math Problem Statement

Consider S={1,2,3} and the following relation R on S: R={⟨1,2⟩,⟨2,3⟩,⟨3,1⟩,⟨1,1⟩,⟨2,2⟩,⟨3,3⟩}. Pick the right properties of R

:

Reflexive

Symmetric

Transitive

Total

Irreflexive: ∀x : ⟨x,x⟩∉R

Asymmetric: ∀x,y : ⟨x,y⟩∈R → ⟨y,x⟩∉R

Anti-symmetric

Intransitive: ∀x,y,z : (⟨x,y⟩∈R ∧ ⟨y,z⟩∈R) → ⟨x,z⟩∉R

Solution

Given the set S={1,2,3}S = \{1, 2, 3\} and the relation R={1,2,2,3,3,1,1,1,2,2,3,3}R = \{ \langle 1,2 \rangle, \langle 2,3 \rangle, \langle 3,1 \rangle, \langle 1,1 \rangle, \langle 2,2 \rangle, \langle 3,3 \rangle \}, let's analyze the properties of the relation RR in detail.

Reflexive:

  • A relation is reflexive if for every element xSx \in S, x,xR\langle x, x \rangle \in R.
  • Here, 1,1\langle 1,1 \rangle, 2,2\langle 2,2 \rangle, and 3,3\langle 3,3 \rangle are all in RR, so RR is reflexive.

Symmetric:

  • A relation is symmetric if for every pair x,yR\langle x, y \rangle \in R, we also have y,xR\langle y, x \rangle \in R.
  • In RR, we have 1,2\langle 1, 2 \rangle but not 2,1\langle 2, 1 \rangle, and 2,3\langle 2, 3 \rangle but not 3,2\langle 3, 2 \rangle, and similarly for 3,1\langle 3, 1 \rangle.
  • Therefore, RR is not symmetric.

Transitive:

  • A relation is transitive if whenever x,yR\langle x, y \rangle \in R and y,zR\langle y, z \rangle \in R, then x,zR\langle x, z \rangle \in R.
  • We can check:
    • 1,2R\langle 1, 2 \rangle \in R and (

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Math Problem Analysis

Mathematical Concepts

Set Theory
Relations
Properties of Relations

Formulas

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Theorems

Reflexivity
Symmetry
Transitivity

Suitable Grade Level

Grades 9-12