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Which of the following relations on the set of all people are reflexive? Symmetric? Antisymmetric? Transitive? Prove your assertions. (a) \(R(x, y)\) if y makes more money than \(x\). (b) \(R(x, y)\) if \(x\) and \(y\) are about the same height. (c) \(R(x, y)\) if \(x\) and \(y\) have an ancestor in common. (d) \(R(x, y)\) if \(x\) and \(y\) are the same sex. (e) \(R(x, y)\) if \(x\) and \(y\) both collect stamps. (f) \(R(x, y)\) if \(x\) and \(y\) like some of the same music.

Short Answer

Expert verified
(a) Antisymmetric; (b) Reflexive, Symmetric, Transitive; (c) Reflexive, Symmetric, Transitive; (d) Reflexive, Symmetric; (e) Reflexive, Symmetric; (f) Reflexive, Symmetric.

Step by step solution

01

Understanding Reflexivity

Reflexivity means every element is related to itself. For each relation, determine if every person is related to themselves. (a) A person cannot earn more money than themselves, so it's not reflexive. (b) A person's height is equal to themselves, making it reflexive. (c) Everyone shares an ancestor with themselves, hence reflexive. (d) Everyone is the same sex as themselves, thus reflexive. (e) If someone collects stamps, they collect them with themselves, so reflexive. (f) Everyone likes music with themselves, making it reflexive.
02

Checking for Symmetry

Symmetry means if \(x\) is related to \(y\), then \(y\) is related to \(x\). Check each relation for this property.(a) Symmetry doesn't hold if one earns more than the other.(b) If \(x\) is about the same height as \(y\), then \(y\) is about the same height as \(x\)—hence symmetric.(c) If \(x\) and \(y\) have a common ancestor, then \(y\) and \(x\) do too, making it symmetric.(d) If \(x\) is the same sex as \(y\), then \(y\) is the same sex as \(x\), so symmetric.(e) If both collect stamps, it’s symmetric with respect to each other.(f) If \(x\) and \(y\) like some music, \(y\) also likes it with \(x\), hence symmetric.
03

Evaluating Antisymmetry

Antisymmetry means if \(x\) is related to \(y\) and \(y\) to \(x\), then \(x\) must equal \(y\). Check each relation.(a) Antisymmetry holds; if \(x\) earns more than \(y\) and vice versa, they must be equal.(b) Antisymmetry doesn't apply as different people being same height doesn't imply equality.(c) Antisymmetry doesn't hold since shared ancestors don't equate equality.(d) Antisymmetric not applicable as the same sex doesn't imply \(x=y\).(e) and (f) both don't imply equality, hence not antisymmetric.
04

Determining Transitivity

Transitivity means if \(x\) is related to \(y\) and \(y\) to \(z\), then \(x\) must be related to \(z\). Analyze each relation.(a) Not transitive, as earning more than \(x\) doesn't extend beyond \(y\). (b) Transitive; if \(x\) and \(y\) & \(y\) and \(z\) are same height, \(x\) and \(z\) are similar too.(c) Transitive, if all share an ancestor, it retains consistency.(d) Not applicable; having common sex doesn't extend.(e) Not transitive just by collecting stamps.(f) Not transitive, no guaranteed extension across people from shared music.

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Key Concepts

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

Reflexivity
In discrete mathematics, a relation is said to be reflexive if every element is related to itself. Think about it as having a mirror image; each element reflects itself. Let's consider some scenarios to better understand:

* In relation (a), a person cannot have more money than themselves, so it's not reflexive.
* For relation (b), a person's height is obviously equal to themselves, making it reflexive.
* In relation (c), everyone shares an ancestor with themselves, hence the relation is reflexive.
* Relation (d) shows that everyone is the same sex as themselves, thus it's reflexive.
* Relation (e) implies that if someone collects stamps, they collect them with themselves, so it is reflexive.
* Lastly, in relation (f), everyone enjoys music with themselves, making it reflexive.

Reflexivity essentially confirms that a 'self-relationship' exists, an interesting property especially when considering sets and logical equivalence in mathematics.
Symmetry
A relation is symmetric if whenever an element \( x \) is related to an element \( y \), \( y \) is related back to \( x \). Imagine a two-way street where both parties can reciprocate the relationship equally. Here's how symmetry applies to different scenarios:

* In relation (a), the relation is not symmetric, since one person earning more doesn't imply the other returns the favor financially.
* On the other hand, relation (b) is symmetric; if \( x \) is about the same height as \( y \), then \( y \) is the same height relative to \( x \).
* Relation (c) is symmetric as well. If \( x \) and \( y \) share an ancestor, then \( y \) shares that ancestor with \( x \) too.
* Relation (d) maintains symmetry; being the same sex is mutual between \( x \) and \( y \).
* In relation (e), if both collect stamps, the bond is evidently symmetric.
* Similarly, relation (f) sees symmetry as liking the same music inherently includes both parties reciprocating that enjoyment.

Symmetry is easy to spot by considering mutual characteristics where roles can be reverted without changing the direction of the relation.
Antisymmetry
Antisymmetry in relations means that if \( x \) is related to \( y \) and \( y \) is related to \( x \), then it implies that \( x \) must equal \( y \). To visualize this, imagine it like a one-way street for overlapping traits that demand equivalence.

* Checking relation (a), let's say if \( x \) earns more than \( y \), and vice versa, they must essentially earn the same amount, hence antisymmetric.
* However, in relation (b), different people being of the same height doesn't directly mean they're the same individual, so it's not antisymmetric.
* For relation (c), sharing an ancestor doesn't imply the individuals themselves are the same, failing to meet antisymmetric conditions.
* Relation (d) indicates the same sex but doesn't entail equality in identity, thus not antisymmetric.
* Relations (e) and (f) neither suggest identity, so they are also not antisymmetric.

In antisymmetry, keep an eye out where a loop-back to self must preserve identity for the property to hold.
Transitivity
Transitivity is where a relation stipulates that if \( x \) relates to \( y \), and \( y \) relates to \( z \), then \( x \) must relate to \( z \) as well. It forms a connected chain of relationships. Let's delve into some examples for clarity:

* In relation (a), earning more than \( x \) does not guarantee that this extends to another person \( z \), so it is not transitive.
* Relation (b) follows transitivity; if \( x \) is about the same height as \( y \), and \( y \) is about the same height as \( z \), then \( x \) and \( z \) are of similar height too.
* As for relation (c), sharing a common ancestor establishes a continuous relationship, exemplifying transitivity.
* In relation (d), the property of having the same sex doesn’t naturally extend beyond two people; thus, it's not transitive.
* Similarly, relation (e) does not construct a transitive relationship merely based on stamp collection.
* Lastly, relation (f) doesn’t satisfy transitivity since liking the same music does not guarantee an extended liking across a group.

Transitivity builds a bridge from start to end in a sequence of relations, fundamentally stringing together shared characteristics across a set.

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Most popular questions from this chapter

There is an old, fallacious proof that if a relation is both symmetric and transitive, it is reflexive. We give this "proof" below. What is the error? Suppose \(R\) is a symmetric and transitive relation on a set \(X\). Pick an \(x \in X\). We need to show \(x R x\). So, take any \(y\) where \(x R y .\) By symmetry, it follows that \(y R x .\) By transitivity, it follows that \(x R x\).

For a relation \(R\) on a set \(X\), let \(R^{*}\) denote the reflexive and transitive closure of \(R\). (a) For any relation \(R\) on a set \(X\), define a relation \(\sim\) on \(X\) as follows: \(x \sim y\) if and only if \(x R^{*} y\) and \(y R^{*} x .\) Prove that \(\sim\) is an equivalence relation. (b) Let \(x_{1} \sim x_{2}\) and \(y_{1} \sim y_{2} .\) Show that \(x_{1} R^{*} y_{1}\) if and only if \(x_{2} R^{*} y_{2}\).

For the set of all people, prove that the relation "weighs less than" is not a linear order.

Find the reflexive, symmetric, and transitive closures of the following relations: \((a)=\) on \(\mathbb{N}\) (b) \(<\) on \(\mathbb{N}\) (c) \(\leq\) on \(\mathbb{N}\) (d) \(R\) on \(\mathbb{N}\) where \(R(x, y)\) if and only if \(y=x+1\) (e) \(R\) on \(\mathbb{R}\) where \(R(x, y)\) if and only if \(y=x+1\) (f) \(R\) on \(\mathbb{R}\) where \(R(x, y)\) if and only if \(|x-y|<0.0005\).

(a) For \(x, y \in \mathbb{N},\) define \(\left.x\right|_{p r} \| y\) if, for some \(z \in \mathbb{N}, z \neq 0, z \neq 1, z \cdot x=y .\) We say \(x\) is a proper divisor of \(y .\) Is \(\left.\right|_{p r} \mathbb{N}\) a linear ordering on \(\mathbb{N} ?\) (b) In the real numbers \(\mathbb{R}\), define \(x \operatorname{lpr} \mathrm{R} y\) if, for some \(z \in \mathbb{R}, z \neq 0, z \neq 1, z \cdot x=\) \(y .\) Is \(\left.\right|_{p r} \mathbb{R}\) a linear ordering on \(\mathbb{R} ?\)

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