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Let \(A=\\{1,2,3,4,5\\} .\) The identity relation on \(A\) is $$ I_{A}=\\{(1,1),(2,2),(3,3),(4,4),(5,5)\\}. $$ Determine an equivalence relation on \(A\) that is different from \(I_{A}\) or explain why this is not possible.

Short Answer

Expert verified
A different equivalence relation on A, which is not an identity relation, can be given by adding more pairs to the identity relation while maintaining the properties of reflexivity, symmetry, and transitivity. One example would be \(R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1)\}\).

Step by step solution

01

Understand the properties of equivalence relation

An equivalence relation must satisfy the following properties: 1. Reflexivity: For all elements x in A, (x,x) should belong to the relation, i.e., x ~ x. 2. Symmetry: For all elements x, y in A, if (x,y) belongs to the relation, then (y,x) must also belong to the relation, i.e., if x ~ y, then y ~ x. 3. Transitivity: For all elements x, y, z in A, if (x,y) and (y,z) belong to the relation, then (x,z) must also belong to the relation, i.e., if x ~ y and y ~ z, then x ~ z.
02

Add more pairs to the relation while keeping the properties

We can construct an equivalence relation by adding more pairs while still maintaining the three properties mentioned above. The identity relation on A, \(I_A\), is: \(I_A = \{(1,1), (2,2), (3,3), (4,4), (5,5)\}\) To create a different equivalence relation, we can add one or more non-identity pairs to \(I_A\). Let's add the pairs (1,2) and (2,1). Now, the new relation becomes: \(R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1)\}\)
03

Verify if the properties are maintained

Now, we need to check whether R satisfies all the properties of equivalence relation: 1. Reflexivity: All the elements on the diagonal (i.e., (1,1), (2,2), (3,3), (4,4), (5,5)) are present in the relation. Thus, reflexivity is satisfied. 2. Symmetry: For any non-diagonal pair (x,y) in the relation, its symmetric pair (y,x) is also in the relation, i.e., (1,2) and (2,1) are both present in R. So, symmetry is satisfied. 3. Transitivity: Consider the non-diagonal pairs present in the relation: a. (1,2) and (2,1): The transitive pair is (1,1), which is already in the relation. Since all three properties are satisfied, the relation R is an equivalence relation on A which is different from the identity relation \(I_A\): \(R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1)\}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reflexivity
Reflexivity is a fundamental property of an equivalence relation. It means that every element is related to itself. In simpler terms, for an equivalence relation \(R\) on a set \(A\), every element \(x\) in \(A\) must satisfy \((x, x) \in R\). This property ensures that each element is included in the relation in a self-referential way.

The identity relation \(I_A = \{(1,1), (2,2), (3,3), (4,4), (5,5)\}\) is an excellent example of reflexivity as it contains every element from the set \(A\) paired with itself. Any attempt to construct a new equivalence relation must ensure reflexivity is preserved. For the relation \(R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1)\}\), reflexivity is naturally maintained because all the self-pairs \((x,x)\) are still present in \(R\).

This property is crucial because it lays the foundation for symmetry and transitivity. Without reflexivity, the set wouldn't be equivalently connected within itself.
Symmetry
Symmetry in an equivalence relation means that if one element is related to another, then the second element is also related back to the first. Formally, a relation \(R\) on set \(A\) is symmetric if for all \(x, y \in A\), whenever \((x, y) \in R\), then \((y, x) \in R\) must also hold true. This two-way relationship allows the elements to freely swap positions without breaking the connection between them.

Looking at the relation \(R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1)\}\), symmetry is evident. The pair \((1,2)\) is mirrored by \((2,1)\), showing that the relation between 1 and 2 is mutual and reciprocal. It's crucial to verify that for every added pair, its symmetric counterpart is also included to maintain this essential property.

Symmetry supports balanced exchanges within a relation, creating an aura of equality and mutual connection among elements of the set \(A\). Without symmetry, equivalence relations would lack one of their defining characteristics, akin to having a lopsided connection that isn't fair or mutual.
Transitivity
Transitivity is the property that allows an equivalence relation to extend beyond direct pairs. If an element is related to a second, and the second is related to a third, transitivity dictates that the first element is also related to the third. Formally, for a relation \(R\) on set \(A\), it must hold that for all \(x, y, z \in A\), if \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\) must be part of \(R\) as well.

In the constructed relation \(R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,1)\}\), transitivity can be a bit less obvious as there are fewer chains to verify. For example, look at the pairs \((1,2)\) and \((2,1)\). Although they circle back to themselves via reflexivity, this doesn't give rise to new pairs that need to be checked. This simplicity in design ensures that there are no transitive incompletions, confirming that the existing relation is closed under transitivity.

Transitivity adds a depth of connection, allowing an equivalence relation to relate not just pairs of elements but extendable sequences. This property stitches the fabric of the set into a cohesive and comprehensive equivalence relation.

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Most popular questions from this chapter

For \((a, b),(c, d) \in \mathbb{R} \times \mathbb{R},\) define \((a, b) \sim(c, d)\) if and only if \(a^{2}+b^{2}=\) \(c^{2}+d^{2} .\) In Exercise (15) of Section 7.2, we proved that \(\sim\) is an equivalence relation on \(\mathbb{R} \times \mathbb{R}\). (a) Determine the equivalence class of (0,0) . (b) Use set builder notation (and do not use the symbol \(\sim\) ) to describe the equivalence class of (2,3) and then give a geometric description of this equivalence class. (c) Give a geometric description of a typical equivalence class for this equivalence relation. (d) Let \(\mathbb{R}^{*}=\\{x \in \mathbb{R} \mid x \geq 0\\}\). Prove that there is a one-to-one correspondence (bijection) between \(\mathbb{R}^{*}\) and the set of all equivalence classes for this equivalence relation.

Let \(U\) be a finite, nonempty set and let \(\mathcal{P}(U)\) be the power set of \(U\). That is, \(\mathcal{P}(U)\) is the set of all subsets of \(U\). Define the relation \(\sim\) on \(\mathcal{P}(U)\) as follows: For \(A, B \in \mathcal{P}(U), A \sim B\) if and only if \(A \cap B=\emptyset .\) That is, the ordered pair \((A, B)\) is in the relation \(\sim\) if and only if \(A\) and \(B\) are disjoint. Is the relation \(\sim\) an equivalence relation on \(\mathcal{P}(U) ?\) If not, is it reflexive, symmetric, or transitive? Justify all conclusions.

Repeat Exercise (6) using the function \(f: \mathbb{R} \rightarrow \mathbb{R}\) that is defined by \(f(x)=\) \(x^{2}-3 x-7\) for each \(x \in \mathbb{R}\).

Use mathematical induction to prove that if \(n\) is a nonnegative integer, then \(10^{n} \equiv 1(\) mod 3\()\). Hence, for congruence classes modulo \(3,\) if \(n\) is a nonnegative integer, then \(\left[10^{n}\right]=[1]\)

Let \(n \in \mathbb{N}\) and let \(s(n)\) denote the sum of the digits of \(n\). So if we write $$ n=\left(a_{k} \times 10^{k}\right)+\left(a_{k-1} \times 10^{k-1}\right)+\cdots+\left(a_{1} \times 10^{1}\right)+\left(a_{0} \times 10^{0}\right) $$ then \(s(n)=a_{k}+a_{k-1}+\cdots+a_{1}+a_{0} .\) Use the result in Exercise (7) to help prove each of the following: (a) \([n]=[s(n)],\) using congruence classes modulo 3. (b) \(n \equiv s(n)(\bmod 3)\) (c) \(3 \mid n\) if and only if \(3 \mid s(n)\).

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