Math Problem Statement

Let W = {1, 2, 3, 4} and R 1 = {(1, 2), (4, 3), (2, 2), (2, 1), (3, 1)} be the relation definition on W. Verify R={1} equivalence or not.

b) Let X = {1, 2, 3, 4} Y = {5, 7, 9, 11, 12} and f: X→Y defined by f(x) = 2x + 3 Represent f as

  1. Set of ordered pairs. il) an arrow diagram.

Solution

Part a) Verifying R1R_1 as an equivalence relation

To determine whether R1R_1 is an equivalence relation on the set W={1,2,3,4}W = \{1, 2, 3, 4\}, we need to check whether it satisfies the following three properties of an equivalence relation:

  1. Reflexivity: For every element xWx \in W, the pair (x,x)(x, x) should be in R1R_1.
  2. Symmetry: For every pair (a,b)R1(a, b) \in R_1, if (a,b)R1(a, b) \in R_1, then (b,a)(b, a) must also be in R1R_1.
  3. Transitivity: If (a,b)R1(a, b) \in R_1 and (b,c)R1(b, c) \in R_1, then (a,c)(a, c) must also be in R1R_1.

1. Reflexivity:

We check if (x,x)R1(x, x) \in R_1 for all xWx \in W.

  • (1,1)R1(1, 1) \notin R_1
  • (2,2)R1(2, 2) \in R_1 (this satisfies reflexivity for 2)
  • (3,3)R1(3, 3) \notin R_1
  • (4,4)R1(4, 4) \notin R_1

Since reflexivity is not satisfied for all elements of WW, R1R_1 is not an equivalence relation. Reflexivity alone is violated, so there's no need to check the other properties (symmetry and transitivity).

Part b) Function f:XYf: X \to Y defined by f(x)=2x+3f(x) = 2x + 3

Given:

  • X={1,2,3,4}X = \{1, 2, 3, 4\}
  • Y={5,7,9,11,12}Y = \{5, 7, 9, 11, 12\}
  • The function is defined as f(x)=2x+3f(x) = 2x + 3.

1. Set of ordered pairs representation of ff:

We evaluate f(x)f(x) for each xXx \in X:

  • f(1)=2(1)+3=5f(1) = 2(1) + 3 = 5
  • f(2)=2(2)+3=7f(2) = 2(2) + 3 = 7
  • f(3)=2(3)+3=9f(3) = 2(3) + 3 = 9
  • f(4)=2(4)+3=11f(4) = 2(4) + 3 = 11

Thus, the set of ordered pairs is: {(1,5),(2,7),(3,9),(4,11)}\{(1, 5), (2, 7), (3, 9), (4, 11)\}

2. Arrow diagram:

In the arrow diagram, we represent elements of XX as nodes on the left, elements of YY as nodes on the right, and draw arrows from xXx \in X to f(x)Yf(x) \in Y.

Here's the correspondence based on the function ff:

  • 151 \to 5
  • 272 \to 7
  • 393 \to 9
  • 4114 \to 11

This forms a simple mapping from XX to YY.


Do you want further details or have any questions?

5 related questions:

  1. How do we formally prove that a relation is symmetric?
  2. Can a relation still be transitive if it fails reflexivity or symmetry?
  3. How can we extend the function ff if more elements are added to XX or YY?
  4. What would happen if the function f(x)=2x+3f(x) = 2x + 3 mapped to values outside YY?
  5. What are some real-world applications of equivalence relations and functions?

Tip: When verifying if a relation is an equivalence relation, failing any one of reflexivity, symmetry, or transitivity is enough to disqualify it.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Set Theory
Relations
Functions
Equivalence Relations

Formulas

f(x) = 2x + 3

Theorems

Equivalence Relation Theorem

Suitable Grade Level

Grades 9-12