Math Problem Statement

Solve

Is the relation B on set A = {1, 2, 3, 4, 5} reflexive, symmetric, anti-symmetric, transitive, equivalence, or partial order? B is defined as B = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (5, 5)}

Solution

Let's analyze the relation BB on the set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\} to answer each of the six questions:

The relation BB is defined as: B={(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(4,5),(5,1),(5,3),(5,4),(5,5)}B = \{(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (5, 5)\}

(i) Is the relation reflexive?

A relation is reflexive if for every element xAx \in A, (x,x)(x, x) is in the relation BB.

For this set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}, we need to check if (1,1)(1, 1), (2,2)(2, 2), (3,3)(3, 3), (4,4)(4, 4), and (5,5)(5, 5) are present in BB.

  • (1,1)(1, 1) is missing.
  • (2,2)(2, 2) is present.
  • (3,3)(3, 3) is present.
  • (4,4)(4, 4) is present.
  • (5,5)(5, 5) is present.

Since (1,1)(1, 1) is missing, the relation is not reflexive.

(ii) Is the relation symmetric?

A relation is symmetric if whenever (x,y)B(x, y) \in B, (y,x)(y, x) is also in BB.

Checking pairs:

  • (1,2)(1, 2) is in BB, but (2,1)(2, 1) is also in BB.
  • (1,3)(1, 3) is in BB, but (3,1)(3, 1) is also in BB.
  • (1,4)(1, 4) is in BB, and (4,1)(4, 1) is in BB.
  • (1,5)(1, 5) and (5,1)(5, 1) are both in BB.

Similarly, checking other pairs such as (2,3)(2, 3), (3,2)(3, 2), etc., we find that for every (x,y)(x, y), (y,x)(y, x) is present in the relation.

Thus, the relation is symmetric.

(iii) Is the relation anti-symmetric?

A relation is anti-symmetric if whenever (x,y)B(x, y) \in B and (y,x)B(y, x) \in B, then x=yx = y.

We already found symmetric pairs such as (1,2)(1, 2) and (2,1)(2, 1), (1,3)(1, 3) and (3,1)(3, 1), etc., where xyx \neq y, meaning that the relation is not anti-symmetric.

Therefore, the relation is not anti-symmetric.

(iv) Is the relation transitive?

A relation is transitive if whenever (x,y)B(x, y) \in B and (y,z)B(y, z) \in B, then (x,z)B(x, z) \in B.

For example, take (1,2)(1, 2) and (2,3)(2, 3), we need to check if (1,3)(1, 3) is in BB, which it is. We can similarly check other combinations, but there are counterexamples where this property does not hold for all combinations.

Thus, the relation is not transitive.

(v) Is the relation an equivalence relation?

A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Since we found that the relation is not reflexive and not transitive, the relation is not an equivalence relation.

(vi) Is the relation a partial order?

A relation is a partial order if it is reflexive, anti-symmetric, and transitive.

Since the relation is neither reflexive nor anti-symmetric, the relation is not a partial order.


Would you like further details or clarifications on any point?

Related Questions:

  1. What are the conditions for a relation to be reflexive?
  2. Can a relation be symmetric but not anti-symmetric?
  3. How can we check for transitivity in a relation efficiently?
  4. Can a relation be both an equivalence relation and a partial order?
  5. What are real-world examples of symmetric and anti-symmetric relations?

Tip:

When checking for symmetry and anti-symmetry, always consider whether reversing the pair results in the same or a different outcome.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Relations
Properties of Relations

Formulas

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Theorems

Reflexivity
Symmetry
Anti-Symmetry
Transitivity
Equivalence Relation
Partial Order

Suitable Grade Level

Undergraduate