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Find a set \(A\) with \(n\) elements and a relation \(R\) on \(A\) such that \(R^{1}, R^{2}, \ldots, R^{t}\) are all distinct.

Short Answer

Expert verified
A cyclic relation on a set of \(n\) elements forms distinct powers up to \(R^n\).

Step by step solution

01

Understanding Relation Powers

Given a set \(A\) with \(n\) elements, we need a relation \(R\) on \(A\) such that all powers of \(R\) from \(R^1\) to \(R^t\) are distinct. \(R^k\) refers to the composition of the relation \(R\) with itself \(k-1\) times.
02

Choose a Simple Set

Choose a set \(A = \{1, 2, \, \ldots, n\}\) to make the creation of relations easier. We need a set on which we can define operations more intuitively.
03

Define the Relation \(R\)

Define a relation \(R\) on \(A\) such that it is a cyclic relation. For instance, let \(R = \{(1,2), (2,3), \, \ldots, (n-1,n), (n,1)\}\). This forms a cycle among the elements.
04

Calculate Relation Powers

Calculate the powers of the relation: - \(R^1 = R\) is simply the cycle relation.- \(R^2 = \{(1,3), (2,4), \, \ldots, (n-1,1), (n,2)\}\), each element pointing to the second next element. - Continue this process for \(R^3, R^4, \ldots, R^t\) to ensure each is distinct.
05

Verify Distinctness

Ensure that for \(t=n\), all \(R^k\) (for \(k=1\) to \(n\)) are distinct because each power shifts elements further around the cycle until \(R^n = I\) (the identity relation).
06

Achieve the Condition

We have constructed a set \(A\) and a relation \(R\) such that \(R^1, R^2, \ldots, R^n\) are all distinct because of the cyclic nature, fulfilling the problem's requirement.

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Key Concepts

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

Composition of Relations
In discrete mathematics, understanding the composition of relations is crucial when working with relations on sets. The composition of a relation, in simple terms, involves taking a relation and combining it with itself or another relation in a step-by-step manner. When you have a relation \( R \) on a set \( A \), the composition \( R^k \) is achieved by linking elements through multiple applications of \( R \).
This is similar to chaining functions, where the output of one relation becomes the input for the next.
To illustrate this with the original exercise, if \( R \) is a relation on set \( A = \{1, 2, \, \ldots, n\}\) defined as a cycle like \((1,2), (2,3), \ldots, (n-1,n), (n,1)\), then each subsequent power \( R^k \) means you continue pairing elements according to the structure of the previous relation \( R^{k-1} \).
This can lead to different sequences or pairings being formed with each increase in \( k \). It is a systematic shift around a cycle that helps to ensure distinct results for different powers.
Cyclic Relations
Cyclic relations form the foundation for creating dynamic and distinct compositions of relations. A cyclic relation on a set \( A \) creates a looping sequence amongst its elements.
Such a relation effectively rotates the entire set, so each element points to the next, and the final element points back to the start.
For example, consider \( A = \{1, 2, \, \ldots, n\} \) and the cyclic relation \( R = \{(1,2), (2,3), \, \ldots, (n-1,n), (n,1)\} \).
This simple yet powerful setup forms a cycle among the elements of set \( A \), and is what makes the different powers of the relation distinct when applied iteratively.
By definition, as each element moves to its next position, the entire set "cycles" through the elements, creating a clear pattern of distinct relations.
Identity Relation
The identity relation is an essential concept in discrete mathematics, acting as a baseline or reference point when examining relations on a set. It is essentially the relation where every element maps to itself.
For a set \( A \), the identity relation \( I \) is represented as \( \{(1,1), (2,2), \, \ldots, (n,n)\} \), which means every element is related only to itself.
In the context of the original exercise, the identity relation becomes important when dealing with cyclic relations and relation powers.
As we create successive powers \( R^1, R^2, ..., R^n \), eventually the sequence recycles precisely back to the corresponding elements, resulting in \( R^n \) being the identity relation. Here, each element has completed its full cycle and returns to map to itself, illustrating why all previous cycles must have been distinct.
Relation Powers
Understanding relation powers involves grasping how relations can be applied repetitively on a set. The power \( R^k \) of a relation \( R \) signifies the sequential composition of \( R \) with itself \( k-1 \) times. This concept is central to finding distinct relation compositions on a given set.
For the set \( A = \{1, 2, 3, ..., n\} \) with a cyclic relation \( R \), each power up to \( R^n \) exerts a unique transformation on \( A \).
- **\( R^1 \):** Simply \( R \) itself, cycling once.- **\( R^2 \):** Pairing each element with the second next, cycling twice.- **Up to \( R^n \):** Completing the cycle back to the identity relation.
It's this succession that ensures distinctness, as each uses the prior structure of the relation, gradually expanding the reach of each element around the cycle. In doing so, they provide a series of unique pairings, perfectly illustrating the power of relational compositions in discrete mathematics.

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Most popular questions from this chapter

Determine which of the following five relations defined on \(\mathbb{Z}\) are equivalence relations: (a) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a>0\) and \(b>0)\) or \((a<0\) and \(b<0)\\}\) (b) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a \geq 0\) and \(b>0\) ) or \((a<0\) and \(b \leq 0)\\}\) (c) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:|a-b| \leq 10\\}\) (d) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a \leq 0\) and \(b \geq 0)\) or \((a \leq 0\) and \(b \leq 0)\\}\) (e) \(\\{(a, b) \in \mathbb{Z} \times \mathbb{Z}:(a \geq 0\) and \(b \geq 0\) ) or \((a \leq 0\) and \(b \leq 0)\\}\)

Let \(R\) and \(S\) be equivalence relations on a set \(X\). (a) Show that \(R \cap S\) is an equivalence relation. (b) Show by example that \(R \cup S\) need not be an equivalence relation. (c) Show that \((R \cup S)^{*}\), the reflexive and transitive closure of \(R \cup S\), is the smallest equivalence relation containing both \(R\) and \(S\).

Identify the equivalence classes of \(\mathrm{N}\) for the following relations: (a) \(\equiv(\bmod 4)\) (b) \(\equiv(\bmod 6)\)

Let \(X=\\{0,1\\}\). Let \(B=\mathcal{P}(X \times X)\) be the set of all binary relations on \(X\). (a) List all the elements of \(B\). (b) Since elements of \(B\) are themselves relations, it makes sense to ask whether two of those relations are inverses of each other. Let $$\text { IslnerseOf }=\left\\{(R, S): R \in B \text { and } S \in B \text { and } R=S^{-1}\right\\}$$ List all elements of IslmerseOf. (c) Since IslmerseOf is a binary relation, it has an inverse. What is IsimerseOf \(^{-1}\) ? (d) What is Islmerse of o IslnverseOf?

Let \(X=\\{1,2,3,4,5,6\\},\) and define a relation \(R\) on \(X\) as $$R=\\{(1,2),(2,1),(2,3),(3,4),(4,5),(5,6)\\}$$ (a) Find the reflexive closure of \(R\). (b) Find the symmetric closure of \(R\). (c) Find the transitive closure of \(R\). (d) Find the reflexive and transitive closure of \(R\).

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