Questions tagged [relations]

For questions concerning partial orders, equivalence relations, properties of relations (transitive, symmetric, etc), a composition of relations, or anything else concerning a relation on a set.

A relation $R$ on a set $X \times Y$ (sometimes also called a relation between $X$ and $Y$) is any subset of $X\times Y$. So a relation is any set of ordered pairs $(x,y)$ such that $x\in X$ and $y \in Y$. Often we write $x\mathrel R y$ instead of $(x,y)\in R$. Sometimes we say a relation is defined on a single set $X$, but this just means that we're letting the relation $R$ be a subset of $X \times X$.

Here are some examples of relations:

  • For a very abstract example of a relation, take $X = \{1,2,3\}$ and $Y = \{a,b,c,d\}$, and let $R = \{(1,b),(3,c),(3,d),(2,c),(2,d)\}$. This $R$ is a relation on $X \times Y$. Fun fact, there are $2^{12}$ distinct relations you could put on $X \times Y$.

  • A function $f\colon X\to Y$ can be defined as a relation on $X \times Y$. The pairs in $X \times Y$ that are members of the relation are the input-output pairs $(x, f(x))$.

  • A partial order is a relation that mimics the notion of one element being greater/less than the other. An example of a partial order is the relation $\leq$ on $\mathbf{Z} \times \mathbf{Z}$ where for two integers $n$ and $m$ we say $n \leq m$ if $n$ is less than or equal to $m$.

  • Given a set $X$, let $\mathcal{P}(X)$ denote the set of all subsets of $X$. We can define a partial order $\subseteq$ on $\mathcal{P}(X)$ by saying that for two susbsets $A$ and $B$ of $X$, $A \subseteq B$ if $A$ is a subset of $B$.

  • An equivalence relation is a relation that mimics the notion of two things being "equal". For example, on the set $\mathbf{Z}$ of integers let's define the relation $\equiv_{37}$ on $\mathbf{Z}\times \mathbf{Z}$ by saying $a\equiv_{37} b$ if both $a$ and $b$ give the same remainder when divided by $37$. If $a \equiv_{37} b$ we say that $a$ and $b$ are congruent modulo $37$.

  • Let $T$ be the set of all triangles in the plane. An example of an equivalence relation on $T$ is the relation of two triangles being congruent.

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If $R$ is transitive and $S$ is reflexive, Prove $(R\ ; S\ ;R)^2 \subseteq (R\ ;S)^3$

$ R $ and $S$ are two relations. Given $R$ is transitive and $S$ is reflexive How can I Prove $(R\ ; S\ ;R)^2 \subseteq (R\ ;S)^3$
TechJ
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For Relation $R$ Prove $R^*=I \cup R^+$

Given R is a relation, I need to prove 2 things. 1) $R^+=R \ ;R^* $ 2) $R^*=I \cup R^+$ For (1) I proceeded as $ \Rightarrow R^+$ $ \Rightarrow \bigcup\limits_{n=1}^\infty R^n \qquad \qquad \qquad $ {definition of $R^+$} $…
TechJ
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Is this relation anti symmetric

Was asked to Work out transitive closure of: R = {(1, 1),(1, 3),(2, 2),(2, 1),(3, 3),(4, 4),(4, 3),(4, 2)} I did using Warshall's, getting: R*={(1,1)(1,3)(2,1)(2,2)(2,3)(3,3)(4,1)(4,2)(4,3)(4,4)} Is R* antisymmetric? I understand antisymmetric…
aqibjr1
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Set notation help

I'm working on this questions: Let $A= \{-2,-1,0,1,2\}$. Let $R$ be the relation defined by $xRy$ if and only if $y=-x$. Let $S$ be the relation defined by $xSy$ if and only if $y = |x|$ Whilst I've figured out most of the question, what I don't…
Mel
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all subsets of R are reflexive?

Prove or refute: If R is reflexive than all subsets of R are reflexive. This my solution: Let {1,2,3) in R --> We know that R is reflexive -> R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),*(3,3)}. We also know that R subset means: R is a…
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Operator: 'Greater Than Or Equal To' is not 'equasive' or 'inequasive' So what is it?

Lets say we use the following symbols for operators: < (less than) <= (less than or equal to) > (greater than) >= (greater than or equal to) = (equal to) /= (not equal to) Now, extrapolating from wikipedia... This type of operator…
JW01
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Alegebra- Function or NOT a function

if you have a relation with the domains: 0,1,2,3,4 and a range of: 3,1,2,4,2 does this mean it is not a function because there are two outputs of the number 2? Or can it only not be function if there are two of the same inputs with a different…
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Discrete Math: Relations

Why is it necessary for a relation to be a subset of the Cartesian product of two sets. Why couldn't we say that a relation is a relationship between any two elements of one or more sets.
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Is this always true about symmetric relations?

Statement : If $R_1$ and $R_2$ are both symmetric, then $R_1 \cap R_2$ is symmetric. Is this always true? As far as I could understand this with an example, suppose I have a set $A = \{1,2,3\}$ And $R_1$ and $R_2$ be symmetric relations where,…
Siddharth Thevaril
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Relation and closure

I have difficulties understanding what is exactly the relations $\leftrightarrow$ (symmetric closure) (and friends: $\stackrel{+}{\leftrightarrow}$ (transtive symmetric closure), $\stackrel{*}{\leftrightarrow}$ (reflexive transtive symmetric…
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Trivial Question : Union of Relation

$ A,B $ are sets. Let Relation $ R \subseteq A \times A. \; $ Relation $ S \subseteq B \times B $. Can we have $ R \cup S $ ? where the underlying sets are different and if so, what is the underlying set of the union ?
lapin
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Can I make the following assumption about symmetric relations?

If $R$ is a symmetric relation then: $$(x,y) \not \in R \rightarrow (y,x) \not \in R$$
Julia
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How can an asymmetric relation be antisymmetric?

Let R be asymmetric. So we have: Assumption: (x,y)∈R⟹(y,x)∉R We need to show R is antisymmetric, i.e. (x,y)∈R∧(y,x)∈R⟹x=y Since 2 is a conditional, we can assume Assumption:(x,y)∈R∧(y,x)∈R And try to show x=y. But if we simplify LHS of 3…
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antisymmetric operators

Is the symbol ">" antisymmetric? For example, if I say $(x>y) \wedge(y>x) \rightarrow \exists(x,y)|x=y$ is vacuously true since the premise cannot be true. This means that $x>y$ is antisymmetric? I do know that $\geq$ is antisymmetric because…
Jabernet
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Define a relation $M$ on $\mathbb{Z} \times \mathbb{Z}$...

Update #2 (7.21.15): Here is a screenshot of the corrected question, in case anyone was interested. No need to look at the first update or original post to anyone viewing this for the first time. Mia Update #1 (07.19.15): Sorry about the…