# 2.1: Binary Relations - Mathematics

Definition

Let (S) be a non-empty set. Then any subset (R) of (S imes S) is said to be a relation over (S). In other words, a relation is a rule that is defined between two elements in (S). Intuitively, if (R) is a relation over (S), then the statement (a R b) is either true or false for all (a,bin S).

Example (PageIndex{1}):

Let (S={1,2,3}). Define (R) by (a R b) if and only if (a < b), for (a, b in S).

Then (1 R 2, 1 R 3, 2 R 3 ) and ( 2 ot R 1).

We can visualize the above binary relation as a graph, where the vertices are the elements of S, and there is an edge from (a) to (b) if and only if (a R b) , for (a,bin S).

The following are some examples of relation defined on (mathbb{Z}).

Example (PageIndex{2}):

1. Define (R) by (a R b) if and only if (a < b), for (a, b in mathbb{Z}).
2. Define (R) by (a R b) if and only if (a >b), for (a, b in mathbb{Z}).
3. Define (R) by (a R b) if and only if (a leq b), for (a, b in mathbb{Z}).
4. Define (R) by (a R b) if and only if (a geq b), for (a, b in mathbb{Z}).
5. Define (R) by (a R b) if and only if (a = b), for (a, b in mathbb{Z}).

Next, we will introduce the notion of "divides".

Definition

Let ( a) and (b) be integers. We say that (a) divides (b) is denoted (amid b), provided we have an integer (m) such that (b=am). In this case we can also say the following:

• (b) is divisible by (a)
• (a) is a factor of (b)
• (a) is a divisor of (b)
• (b) is a multiple of (a)

Example (PageIndex{3}):

(4 mid 12) and (12 otmid 4)

Theorem (PageIndex{1}): Divisibility inequality theorem

If (amid b), for (a, b in mathbb{Z_+}) then (a leq b),

Proof

Let (a, b in mathbb{Z_+}) such that (amid b), Since (amid b), there is a positive integer (m) such that (b=am). Since (m geq 1) and (a) is a positive integer, (b=am geq (a)(1)=a. )

Note that if (amid b), for (a, b in mathbb{Z_+}) then (a leq b), but the converse is not true. For example: (2 <3), but (2 otmid 3).

Example (PageIndex{4}):

According to our definition (0 mid 0).

Definition

An integer is even provided that it is divisible by (2).

## Properties of binary relation:

Definition

Let (S) be a set and (R) be a binary relation on (S). Then (R) is said to be reflexive if ( a R a, forall a in S.)

Example (PageIndex{5}): Visually

(forall a in S, a R a) holds.

Example (PageIndex{6}):

Define (R) by (a R b) if and only if (a < b), for (a, b in mathbb{Z}). Is (R) reflexive?

Counter Example:

Choose (a=2.)

Since ( 2 ot< 2), (R) is not reflexive.

Example (PageIndex{7}):

Define (R) by (a R b) if and only if (a mid b), for (a, b in mathbb{Z}). Is (R) reflexive?

Proof:

Let ( a in mathbb{Z}). Since (a=(1) (a)), (a mid a).

Thus (R) is reflexive. ( Box)

Definition

Let (S) be a set and (R) be a binary relation on (S). Then (R) is said to be symmetric if the following statement is true:

( forall a,b in S), if ( a R b ) then (b R a), in other words, ( forall a,b in S, a R b implies b R a.)

Example (PageIndex{8}): Visually

(forall a,b in S, a R b implies b R a.) holds!

Example (PageIndex{7}):

Define (R) by (a R b) if and only if (a < b), for (a, b in mathbb{Z}). Is (R) symmetric?

Counter Example:

(1<2) but (2 ot < 1).

Example (PageIndex{8}):

Define (R) by (a R b) if and only if (a mid b), for (a, b in mathbb{Z}). Is (R) symmetric?

Counter Example:

(2 mid 4) but (4 ot mid 2).

Definition

Let (S) be a set and (R) be a binary relation on (S). Then (R) is said to be antisymmetric if the following statement is true:

( forall a,b in S), if ( a R b ) and (b R a), then (a=b).

In other words, ( forall a,b in S), ( a R b wedge b R a implies a=b.)

Example (PageIndex{9}): VISUALLY

( forall a,b in S), ( a R b wedge b R a implies a=b ) holds!

Example (PageIndex{10}):

Define (R) by (a R b) if and only if (a < b), for (a, b in mathbb{Z}). Is (R) antisymmetric?

Example (PageIndex{11}):

Define (R) by (a R b) if and only if (a mid b), for (a, b in mathbb{Z_+}). Is (R) antisymmetric?

Definition

Let (S) be a set and (R) be a binary relation on (S). Then (R) is said to be transitive if the following statement is true

( forall a,b,c in S,) if ( a R b ) and (b R c), then (a R c).

In other words, ( forall a,b,c in S), ( a R b wedge b R c implies a R c).

Example (PageIndex{12}): VISUALLY

( forall a,b,c in S), ( a R b wedge b R c implies a R c) holds!

Example (PageIndex{13}):

Define (R) by (a R b) if and only if (a < b), for (a, b in mathbb{Z}). Is (R) transitive?

Example (PageIndex{14}):

Define (R) by (a R b) if and only if (a mid b), for (a, b in mathbb{Z_+}). Is (R) transitive?

Summary:

In this section we learned about binary relation and the following properties:

Reflexive

Symmetric

Antisymmetric

Transitive

## Binary Relations (Types and Properties)

In this article, I discuss binary relations. I first define the composition of two relations and then prove several basic results. After that, I define the inverse of two relations. Then the complement, image, and preimage of binary relations are also covered.

Let $X$ be a set and let $X imes X=<(a,b): a,b in X>.$ A (binary) relation $R$ is a subset of $X imes X$. If $(a,b)in R$, then we say $a$ is related to $b$ by $R$. It is possible to have both $(a,b)in R$ and $(a,b’)in R$ where $b’ eq b$ that is any element in $X$ could be related to any number of other elements of $X$. It is also possible to have some element that is not related to any element in $X$ at all.

Definition. Let $R$ and $S$ be relations on $X$. The composition of $R$ and $S$ is the relation $Scirc R =<(a,c)in X imes X : exists , bin X, (a,b)in R land (b,c)in S>.$

Compositions of binary relations can be visualized here.

Theorem. If $R$, $S$ and $T$ are relations on $X$, then $Rcirc (Scirc T)=(Rcirc S)circ T$.

Proof. The proof follows from the following statements. egin (x,y)in & Rcirc (Scirc T) & Longleftrightarrow exists zin X, (x,z)in Scirc T land (z,y)in R & Longleftrightarrow exists zin X, [ exists win X, (x,w)in T land (w,z)in S ] land (z,y)in R & Longleftrightarrow exists w, zin X, (x,w)in T land (w,z)in S land (z,y)in R & Longleftrightarrow exists win X, [exists zin X, (w,z)in S land (z,y)in R] land (x,w)in T & Longleftrightarrow exists win X, (x,w)in T land (w,y)in Rcirc S & Longleftrightarrow (x,y)in (Rcirc S) circ T end

Theorem. If $R$, $S$ and $T$ are relations on $X$, then $Rcirc (Scup T)=(Rcirc S)cup (Rcirc T)$.

Proof. The proof follows from the following statements. egin (x,y) & in Rcirc (Scup T) & Longleftrightarrow exists zin X, (x,z)in S cup T land (z,y)in R & Longleftrightarrow exists zin X, [(x,z)in S lor (x,z)in T ] land (z,y)in R & Longleftrightarrow exists zin X, [(x,z)in S land (z,y)in R] lor [(x,z)in T land (z,y)in R] & Longleftrightarrow (x,y)in Rcirc S lor (x,y)in Rcirc T & Longleftrightarrow (x,y)in (Rcirc S)cup (R circ T) end

Theorem. If $R$, $S$ and $T$ are relations on $X$, then $(Scup T)circ R=(Scirc R)cup (Tcirc R)$.

Proof. The proof follows from the following statements. egin (x,y) & in (Scup T)circ R & Longleftrightarrow exists zin X, (x,z)in R land (z,y)in Scup T & Longleftrightarrow exists zin X, (x,z)in R land [(z,y)in Slor (z,y)in T] & Longleftrightarrow exists zin X, [(x,z)in R land (z,y)in S] lor [(x,z)in R land (z,y)in T] & Longleftrightarrow (x,y)in (Scirc R) lor (x,y)in (Tcirc R) & Longleftrightarrow (x,y)in (Scirc R)cup (Tcirc R) end

Theorem. If $R$, $S$ and $T$ are relations on $X$, then $Rcirc (Scap T) subseteq (Rcirc S)cap (Rcirc T)$.

Proof. The proof follows from the following statements. egin qquad quad & (x,y) in Rcirc (Scap T) & qquad Longleftrightarrow exists zin X, (x,z)in Scap T land (z,y)in R & qquad Longleftrightarrow exists zin X, [(x,z)in S land (x,z)in T] land (z,y)in R & qquad Longleftrightarrow exists zin X, [(x,z)in S land (z,y)in R] land (x,z)in T & qquad Longleftrightarrow exists zin X, [(x,z)in S land (z,y)in R] land [(x,z)in T land (z,y)in R] & qquad Longrightarrow [exists zin X, [(x,z)in S land (z,y)in R] land [ exists win X, (x,w)in T land (w,y)in R] & qquad Longleftrightarrow (x,y)in Rcirc S land (x,y)in Rcirc T & qquad Longleftrightarrow (x,y)in (Rcirc S) cap (Rcirc T) end

Theorem. If $R$, $S$ and $T$ are relations on $X$, then $Rsubseteq S implies Rcirc T subseteq Scirc T$.

Proof. The proof follows from the following statements. egin & (x,y)in Rcirc T Longleftrightarrow exists zin X, (x,z)in T land (z,y)in R & qquad Longrightarrow exists zin X, (x,z)in T land (z,y)in S Longleftrightarrow (x,y)in Scirc T end

Theorem. If $R$, $S$ and $T$ are relations on $X$, then $Rsubseteq S implies Tcirc R subseteq Tcirc S$.

Proof. The proof follows from the following statements. egin & (x,y)in Tcirc R Longleftrightarrow exists zin X, (x,z)in R land (z,y)in T & qquad Longrightarrow exists zin X, (x,z)in S land (z,y)in T Longleftrightarrow (x,y)in Tcirc S end

Definition. Let $R$ and $S$ be relations on $X$. The inverse of $R$ is the relation $R^<-1>=<(b,a)in X imes X : (a,b)in R>.$

Theorem. If $R$ and $S$ are relations on $X$, then $(R^<-1>)^<-1>=R$.

Proof. $(x,y)in (R^<-1>)^ <-1>Longleftrightarrow (y,x)in R^ <-1>Longleftrightarrow (x,y)in R$

Theorem.If $R$ and $S$ are relations on $X$, then $(Rcup S)^<-1>=R^<-1>cup S^<-1>$.

Proof. egin & (x,y)in (Rcup S)^ <-1>Longleftrightarrow (y,x)in Rcup S Longleftrightarrow (y,x)in R lor (y,x)in S & qquad Longleftrightarrow (x,y)in R^ <-1>lor (x,y)in S^ <-1>Longleftrightarrow (x,y)in R^<-1>cup S^ <-1>end

Theorem. If $R$ and $S$ are relations on $X$, then $(Rcap S)^<-1>=R^<-1>cap S^<-1>$.

Proof. egin & (x,y)in (Rcap S)^ <-1>Longleftrightarrow (y,x)in Rcap S Longleftrightarrow (y,x)in R land (y,x)in S & qquad Longleftrightarrow (x,y)in R^ <-1>land (x,y)in S^ <-1>Longleftrightarrow (x,y)in R^<-1>cap S^ <-1>end

Theorem. If $R$ and $S$ are relations on $X$, then $(Rcirc S)^<-1>=S^<-1>circ R^<-1>$.

Proof. egin & (x,y)in (Rcirc S)^ <-1>Longleftrightarrow (y,x)in Rcirc S & qquad Longleftrightarrow exists zin X, (y,z)in S land (z,x)in R & qquad Longleftrightarrow exists zin X, (z,y)in S^ <-1>land (x,z)in R^ <-1> & qquad Longleftrightarrow exists zin X, (x,z)in R^ <-1>land (z,y)in S^ <-1> & qquad Longleftrightarrow (x,y)in S^ <-1>circ R^ <-1>end

Theorem. If $R$ and $S$ are relations on $X$, then $Rsubseteq S implies R^<-1>subseteq S^<-1>$.

Proof. If $Rsubseteq S$, then $R^<-1>subseteq S^<-1>$. egin (x,y)in & R^ <-1>Longleftrightarrow (y,x)in R Longrightarrow (y,x)in S Longleftrightarrow (x,y) in S^ <-1>end

Theorem. If $R$ and $S$ are relations on $X$, then $(R^c)^<-1>=(R^<-1>)^c$.

Proof. egin & (x,y)in (R^c)^ <-1>Longleftrightarrow (y,x)in R^c Longleftrightarrow (y,x)in X imes X land (y,x) otin R & qquad Longleftrightarrow (x,y)in X imes X land (x,y) otin R^ <-1>Longleftrightarrow (x,y)in (R^<-1>)^c end

Theorem. If $R$ and $S$ are relations on $X$, then $(Rsetminus S)^<-1>=R^<-1>setminus S^<-1>$.

Proof. egin & (x,y)in (Rsetminus S)^ <-1>Longleftrightarrow (y,x)in Rsetminus S Longleftrightarrow (y,x)in R land (y,x) otin S & qquad Longleftrightarrow (x,y)in R^ <-1>land (y,x) otin S Longleftrightarrow (x,y)in R^ <-1>land (x,y) otin S^ <-1> & qquad Longleftrightarrow (x,y)in R^<-1>setminus S^ <-1>end

Definition. Let $R$ and $S$ be relations on $X$. The image of $Asubseteq X$ under $R$ is the set $R(A)=.$

Theorem. If $R$ and $S$ are relations on $X$ and $A, Bsubseteq X$, then $Asubseteq B implies R(A)subseteq R(B)$.

Proof. If $Asubseteq B$, then $R(A)subseteq R(B)$. egin qquad yin R(A) Longleftrightarrow exists xin A, (x,y)in R implies exists xin B, (x,y)in R Longleftrightarrow yin R(B) end

Theorem. If $R$ and $S$ are relations on $X$ and $A, Bsubseteq X$, then $R(Acup B)=R(A)cup R(B)$.

Proof. egin qquad & yin R(Acup B) Longleftrightarrow exists xin X, xin Acup B land (x,y)in R & qquad Longleftrightarrow exists xin X, (xin A lor xin B) land (x,y)in R & qquad Longleftrightarrow exists xin A, (x,y)in R lor exists xin B, (x,y)in R Longleftrightarrow yin R(A) cup R(B)end

Theorem. If $R$ and $S$ are relations on $X$ and $A, Bsubseteq X$, then $R(Acap B)subseteq R(A)cap R(B)$.

Proof. egin qquad & yin R(Acap B) Longleftrightarrow exists xin X, xin Acap B land (x,y)in R & qquad Longleftrightarrow exists xin X, (xin A land xin B) land (x,y)in R & qquad Longrightarrow exists xin A, (x,y)in R land exists xin B, (x,y)in R Longleftrightarrow yin R(A) cap R(B) end

Theorem. If $R$ and $S$ are relations on $X$ and $A, Bsubseteq X$, then $R(A)setminus R(B)subseteq R(Asetminus B)$.

Proof. egin yin R(A)setminus R(B) & Longleftrightarrow yin R(A)land y otin R(B) & Longleftrightarrow exists xin A, (x,y)in R land forall zin B, (z,y) otin R & Longleftrightarrow exists xin Asetminus B, (x,y)in R Longleftrightarrow yin R(Asetminus B) end

Theorem. If $R$ and $S$ are relations on $X$ and $R(x)=S(x)$ for all $xin X$, then $R=S$.

Proof. Assume $R(x)=S(x)$ for all $xin X$, then $(x,y)in R Longleftrightarrow yin R(x) Longleftrightarrow yin S(x) Longleftrightarrow (x,y)in S$ completes the proof.

Definition. Let $R$ and $S$ be relations on $X$. The preimage of $Bsubseteq X$ under $R$ is the set $R^<-1>(B)=.$

Theorem. Let $R$ be a relation on $X$ with $A, Bsubseteq X$. Then $Asubseteq B implies R^<-1>(A)subseteq R^<1->(B)$.

Proof. egin xin R^<-1>(A) & Longleftrightarrow exists yin A, (x,y)in R & implies exists yin B, (x,y)in R Longleftrightarrow xin R^<-1>(B) end

Theorem. Let $R$ be a relation on $X$ with $A, Bsubseteq X$. Then $R^<-1>(Acup B)=R^<-1>(A)cup R^<-1>(B)$.

Proof. egin & xin R^<-1>(Acup B) Longleftrightarrow exists y in Acup B, (x,y)in R & qquad Longleftrightarrow exists yin A, (x,y)in R lor exists yin B, (x,y)in R & qquad Longleftrightarrow xin R^<-1>(A)lor R^<-1>(B) Longleftrightarrow xin R^<-1>(A)cup R^<-1>(B) end

Theorem. Let $R$ be a relation on $X$ with $A, Bsubseteq X$. Then $R^<-1>(Acap B)subseteq R^<-1>(A)cap R^<-1>(B)$.

Proof. egin & xin R^<-1>(Acap B) Longleftrightarrow exists yin A cap B, (x,y)in R & qquad Longleftrightarrow exists yin X, yin A land yin B land (x,y)in R & qquad Longrightarrow xin R^<-1>(A) land xin R^<-1>(B) Longleftrightarrow xin R^<-1>(A) cap xin R^<-1>(B)end

Theorem. Let $R$ be a relation on $X$ with $A, Bsubseteq X$. Then $R^<-1>(A)setminus R^<-1>(B)subseteq R^<-1>(Asetminus B)$.

Proof. egin & xin R^<-1>(A)setminus R^<-1>(B)Longleftrightarrow xin R^<-1>(A) land eg(xin R^<-1>(B)) & qquad Longleftrightarrow xin R^<-1>(A)land [forall yin B, (x,y) otin R] & qquad Longleftrightarrow exists yin A, (x,y)in R land [forall yin B, (x,y) otin R] & qquad Longrightarrow exists yin Asetminus B, (x,y)in R Longleftrightarrow xin R^<-1>(Asetminus B)end

Theorem. Let $R$ and $R_i$ be relations on $X$ for $iin I$ where $I$ is an indexed set. Then $Rcirc left(igcup_ R_i ight)=igcup_(Rcirc R_i)$.

Proof. egin (x,y)in Rcirc left(igcup_ R_i ight) & Longleftrightarrow exists zin X, (x,z)in igcup_ R_i land (z,y)in R & Longleftrightarrow exists zin X, exists iin I, (x,z)in R_i land (z,y)in R & Longleftrightarrow exists iin I, (x,y)in Rcirc R_i & Longleftrightarrow (x,y) in igcup_(Rcirc R_i) end

Theorem. Let $R$ and $R_i$ be relations on $X$ for $iin I$ where $I$ is an indexed set. Then $left(igcup_ R_i ight)circ R=igcup_(R_icirc R)$.

Proof. egin (x,y)in left(igcup_ R_i ight)circ R & Longleftrightarrow exists zin X, (x,z)in R land (z,y)in igcup_ R_i & Longleftrightarrow exists zin X, exists iin I, (x,z)in R land (z,y)in R_i & Longleftrightarrow (x,y)in igcup_(R_icirc R) end

Theorem. Let $R$ be a relation on $X$. Then $(R^n)^<-1>=(R^<-1>)^n$ for all $ngeq 1$.

Proof. By induction. The basis step is obvious: $(R^<1>)^<-1>=(R^<-1>)^1$. In fact, $(R^2)^<-1>=(Rcirc R)^<-1>=R^<-1>circ R^<-1>=(R^<-1>)^2$. The induction step is $(R^n)^<-1>=(R^<-1>)^nimplies (R^)^<-1>=(R^<-1>)^.$ The result now follows from the argument: egin (x,y)in (R^)^ <-1>& Longleftrightarrow (y,x)in R^ & Longleftrightarrow exists zin X, (y,z)in R land (z,x)in R^n & Longleftrightarrow exists zin X, (z,y)in R^ <-1>land (x,z)in (R^n)^<-1> & Longleftrightarrow exists zin X, (x,z)in (R^n)^ <-1>land (z,y)in R^<-1> & Longleftrightarrow exists zin X, (x,z)in (R^<-1>)^n land (z,y)in R^ <-1> & Longleftrightarrow (x,y)in (R^<-1>)^ end

Theorem. Let $R$ be a relation on $X$. Then $left( igcup_ R^n ight)^ <-1>= igcup_ (R^<-1>)^$.

Proof. egin (x,y)in & left( igcup_ R^n ight)^ <-1>Longleftrightarrow (y,x)in igcup_ R^n & Longleftrightarrow exists ngeq 1, (y,x)in R^n =R^circ R & Longleftrightarrow exists ngeq 1, exists zin X, (y,z)in R land (z,x)in R^ & Longleftrightarrow exists ngeq 1, exists zin X, (z,y)in R^ <-1>land (x,z)in (R^)^<-1> & Longleftrightarrow exists ngeq 1, exists zin X, (x,z)in (R^)^ <-1>land (z,y)in R^ <-1> & Longleftrightarrow exists ngeq 1, exists zin X, (x,z)in (R^<-1>)^ land (z,y)in R^ <-1> & Longleftrightarrow exists ngeq 1, (x,y)in (R^<-1>)^n Longleftrightarrow (x,y)in igcup_(R^<-1>)^n end

Theorem. Let $R$ be a relation on $X$. Then $R^n cup S^nsubseteq (Rcup S)^n$ for all $ngeq 1$.

Proof. The basis step is obvious. The induction step is: $R^n cup S^nsubseteq (Rcup S)^n implies R^ cup S^subseteq (Rcup S)^$ The result holds by egin (Rcup S)^ & =(Rcup S)^ncirc (Rcup S) & supseteq (R^ncup S^n) circ (R cup S) & = [(R^ncup S^n)circ R] cup (R^ncup S^n) circ S & = R^ cup (S^n circ R) cup (R^ncirc S) cup S^ & supseteq R^cup S^. end

Theorem. Let $R$ be a relation on $X$. Then $(x,y)in R^n$ if and only if there exists $x_1, x_2, x_3, ldots, x_in X$ such that $(x,x_1)in R, (x_1,x_2)in R , ldots, (x_,y)in R$.

Proof. Bases case, $i=1$ is obvious. We assume the claim is true for $j$. Then egin& (x,y)in R^ Longleftrightarrow (x,y)in R^jcirc R & Longleftrightarrow exists x_1in X, (x,x_1)in R land (x_1,y)in R^j & Longleftrightarrow exists x_1in X, (x,x_1)in R land exists x_2, ldots, x_in X, (x_2, x_3), ldots, (x_,y)in R & Longleftrightarrow exists x_1in X, x_2, ldots, x_in X, (x,x_1), (x_2, x_3), ldots, (x_,y)in R end as needed to complete induction.

(try using the X 2 tag just above the Reply box )

Hint: how many pairs are there in A? And how many are there in T t ?

(try using the X 2 tag just above the Reply box )

Hint: how many pairs are there in A? And how many are there in T t ?

i dont totally understand the question, but looking at your equivalence realtion, along with Tiny Tim's comments:

first, there's more than 4 pairs that can be made from A, & more than 8 in T^t

Let T = <(0,2), (1,0), (2,3), (3,1)>
then the way i read it, following through T gives: 0

1, so everything is equivalant in the transitive closure (ie contains all binary pairs).

uhh? you found 12 in T t .

And A has 4 elements, so how many different pairs of elements are there (counting, for example, <0,3>and <3,0>as the same pair).

(if you can't calculate it, then just list them )

uhh? you found 12 in T t .

And A has 4 elements, so how many different pairs of elements are there (counting, for example, <0,3>and <3,0>as the same pair).

(if you can't calculate it, then just list them )

Sorry, meant 12 elements in T t

I still don't follow. Why would <0,3>and <3,0>be the same?

OK By defintion:
The transitive closure of T is the binary relation T t on A that satisfies the following three properties:
1. T t is transitive.
2. T is a subset of T t .
3. If S is any other transitive relation that contains T, then T t is a subset of S.

## Contents

A binary relation R over sets X and Y is a subset of X × Y . [1] [8] The set X is called the domain [1] or set of departure of R, and the set Y the codomain or set of destination of R. In order to specify the choices of the sets X and Y, some authors define a binary relation or correspondence as an ordered triple (X, Y, G) , where G is a subset of X × Y called the graph of the binary relation. The statement ( x , y ) ∈ R reads "x is R-related to y" and is denoted by xRy. [4] [5] [6] [note 1] The domain of definition or active domain [1] of R is the set of all x such that xRy for at least one y. The codomain of definition, active codomain, [1] image or range of R is the set of all y such that xRy for at least one x. The field of R is the union of its domain of definition and its codomain of definition. [10] [11] [12]

## Binary Relations

(2,3))
There is one that does not belong which is it. Are there others left out?

#### Devil2euz

##### New member

(2,3))
There is one that does not belong which is it. Are there others left out?

#### HallsofIvy

##### Elite Member

The ordered pairs from A= <1, 2, 3, 4>are <(1, 1), (1, 2). (1, 3), (1, 4). (2, 1). (2, 2). (2, 3), (2, 4). (3, 1), (3, 2), (3, 3), (3, 4). (4. 1), (4, 2), (4, 3), (4, 4)>. As pka said there are 16 of those- 4 that start with 1, 4 that start with 2, 4 that start with 3. and 4 that star with 4.

The "relation" "3a+ 5b is a prime number" consists of all those pairs, (a. b), that satisfy. The only way to do this is to actually calculate 3a+ 5b for each pair.

(1,1): 3(1)+ 5(1)= 8. That is not a prime number so (1, 1) is not in the relation.
(1,2) 3(1)+ 5(2)= 13. That is a prime number so (1, 2) is in the relation.
(1,3): 3(1)+5(3)= 18. That is not a prime number so (1, 3) is not in the relation.

## 1.7 Binary Relations

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Massachusetts Institute of Technology

## Welcome!

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## Arbitrary Relations

From the definition of a binary relation, we can easily generalize it to that of an arbitrary relation. Since a binary relation involves two sets, an arbitrary relation involves an arbitrary collection of sets. More specifically, a relation R is a subset of some Cartesian product ( http://planetmath.org/GeneralizedCartesianProduct ) of a collection of sets. In symbol, this is

where each A i is a set, indexed by some set I .

From this more general definition, we see that a binary relation is just a relation where I has two elements. In addition, an n -ary relation is a relation where the cardinality of I is n ( n finite). In symbol, we have

It is not hard to see that any n -ary relation where n > 1 can be viewed as a binary relation in n - 1 different ways, for

 R ⊆ A 1 × A 2 × ⋯ × A n = ∏ i = 1 j A i × ∏ i = j + 1 n A i ,

where j ranges from 1 through n - 1 .

A common name for a 3 -ary relation is a ternary relation. It is also possible to have a 1 -ary relation, or commonly known as a unary relation, which is nothing but a subset of some set.

Following from the first remark from the previous section , relations of higher arity can be inductively defined: for n > 1 , an ( n + 1 ) -ary relation is a binary relation whose domain is an n -ary relation. In this setting, a “unary relation” and relations whose arity is of “arbitrary” cardinality are not defined.

A relation can also be viewed as a function (which itself is a relation). Let R ⊆ A := ∏ i ∈ I A i . As a subset of A , R can be identified with the characteristic function

where χ R ⁢ ( x ) = 1 iff x ∈ R and χ R ⁢ ( x ) = 0 otherwise. Therefore, an n -ary relation is equivalent to an ( n + 1 ) -ary characteristic function. From this, one may say that a 0 -ary, or a nullary relation is a unary characteristic function. In other words, a nullary relation is just a an element in < 0 , 1 >(or truth/falsity).

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## 2.1: Binary Relations - Mathematics

Definition (Union): The union of sets A and B , denoted by A B , is the set defined as

Example 1: If A = < 1, 2, 3 >and B = < 4, 5 >, then A B = < 1, 2, 3, 4, 5 >.

Example 2: If A = < 1, 2, 3 >and B = < 1, 2, 4, 5 >, then A B = < 1, 2, 3, 4, 5 >.

Note that elements are not repeated in a set.

Definition (Intersection): The intersection of sets A and B , denoted by A B , is the set defined as

Example 3: If A = < 1, 2, 3 >and B = < 1, 2, 4, 5 >, then A B = < 1, 2 >.

Example 4: If A = < 1, 2, 3 >and B = < 4, 5 >, then A B = .

Definition (Difference): The difference of sets A from B , denoted by A - B (or A B ), is the set defined as

Example 5: If A = < 1, 2, 3 >and B = < 1, 2, 4, 5 >, then A - B = < 3 >.

Example 6: If A = < 1, 2, 3 >and B = < 4, 5 >, then A - B = < 1, 2, 3 >.

Note that in general A - B B - A

Definition (Complement): For a set A , the difference U - A , where U is the universe, is called the complement of A and it is denoted by .
Thus is the set of everything that is not in A .

--> Definition (ordered pair): An ordered pair is a pair of objects with an order associated with them.
If objects are represented by x and y , then we write the ordered pair as ( x, y ) .

Two ordered pairs ( a, b ) and ( c, d ) are equal if and only if a = c and b = d . For example the ordered pair ( 1, 2 ) is not equal to the ordered pair ( 2, 1 ) .

Definition (Cartesian product): The set of all ordered pairs ( a, b ) , where a is an element of A and b is an element of B , is called the Cartesian product of A and B and is denoted by A B .
Example 1: Let A = <1, 2, 3>and B = . Then
A B = < ( 1, a ) , ( 1, b ) , ( 2, a ) , ( 2, b ) , ( 3, a ) , ( 3, b ) >.

Example 2: For the same A and B as in Example 1,
B A = < ( a, 1 ) , ( a, 2 ) , ( a, 3 ) , ( b, 1 ) , ( b, 2 ) , ( b, 3 ) >.

As you can see in these examples, in general, A B B A unless A = , B = or A = B .
Note that A = A = because there is no element in to form ordered pairs with elements of A .

The concept of Cartesian product can be extended to that of more than two sets. First we are going to define the concept of ordered n-tuple .

Definition (ordered n-tuple): An ordered n -tuple is a set of n objects with an order associated with them . If n objects are represented by x 1 , x 2 , . x n , then we write the ordered n -tuple as ( x 1 , x 2 , . x n ) .

Definition (Cartesian product): Let A 1 , . A n be n sets. Then the set of all ordered n -tuples ( x 1 , . x n ) , where x i A i for all i , 1 i n , is called the Cartesian product of A 1 , . A n , and is denoted by A 1 . A n .

Definition (equality of n -tuples): Two ordered n -tuples ( x 1 , . x n ) and ( y 1 , . y n ) are equal if and only if x i = y i for all i , 1 i n .
For example the ordered 3 -tuple ( 1, 2, 3 ) is not equal to the ordered n -tuple ( 2, 3, 1 ) .

Definition (binary relation):
A binary relation from a set A to a set B is a set of ordered pairs ( a, b ) where a is an element of A and b is an element of B .
When an ordered pair ( a, b ) is in a relation R , we write a R b , or ( a, b ) R . It means that element a is related to element b in relation R .
When A = B , we call a relation from A to B a (binary) relation on A .

Examples:
If A = < 1, 2, 3 >and B = < 4, 5 >, then < (1, 4), (2, 5), (3, 5) >, for example, is a binary relation from A to B .
However, < (1, 1), (1, 4), (3, 5) >is not a binary relation from A to B because 1 is not in B .
The parent-child relation is a binary relation on the set of people. (John, John Jr.) , for example, is an element of the parent-child relation if John is the father of John Jr.