Articles

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.

We will follow the template below to answer the question about reflexive.

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!

We will follow the template below to answer the question about symmetric.

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!

We will follow the template below to answer the question about anti-symmetric.

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!

We will follow the template below to answer the question about transitive.

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 )

You've made it really long.

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 )


You've made it really long.

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?

Elite Member
Elite Member

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

This is one of over 2,400 courses on OCW. Explore materials for this course in the pages linked along the left.

MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum.

No enrollment or registration. Freely browse and use OCW materials at your own pace. There's no signup, and no start or end dates.

Knowledge is your reward. Use OCW to guide your own life-long learning, or to teach others. We don't offer credit or certification for using OCW.

Made for sharing. Download files for later. Send to friends and colleagues. Modify, remix, and reuse (just remember to cite OCW as the source.)

About MIT OpenCourseWare

MIT OpenCourseWare is an online publication of materials from over 2,500 MIT courses, freely sharing knowledge with learners and educators around the world. Learn more »

© 2001&ndash2018
Massachusetts Institute of Technology

Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use.


Welcome!

This is one of over 2,400 courses on OCW. Explore materials for this course in the pages linked along the left.

MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum.

No enrollment or registration. Freely browse and use OCW materials at your own pace. There's no signup, and no start or end dates.

Knowledge is your reward. Use OCW to guide your own life-long learning, or to teach others. We don't offer credit or certification for using OCW.

Made for sharing. Download files for later. Send to friends and colleagues. Modify, remix, and reuse (just remember to cite OCW as the source.)


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).


Q: compare and contrast serial processing operating system and simple batch processing operating system.

A: Answer: One of the main functionality of the Os is resource management. The resource may be hardwire.

Q: Three part question: 1) How does indexing work in terms of database management systems? 2) How does .

A: I have provided solution in step 2.

Q: Provide a reason to explain that requirements engineering and design are interlaced activities (rele.

A: Requirements engineering (RE) is the process of defining, documenting, and maintaining requirements .

Q: Write a program named CreateCustomerFile.java that allows you to create a file of customers (Custome.

A: Note: As you have asked multiple questions, as per our policy, we will solve the first question for .

Q: Write a program to find out the grade of a student based on the marks obtained in three subjects. Th.

A: As the programming language is not mentioned so, I am using C language to write a code. If you want .

Q: In MC68000, name addressing mode that are not allowed for destination, along with the reasons

A: When people say that they have built their own computer, what they really mean is they really mean i.

Q: Write in c programming language. Make class triangle having three sides as class members. Use parame.

A: C language is not an object oriented language so there is no concept of classes in C but C++ does ha.

Q: write a program c++ If money is left in a particular bank for more than 5 years, the bank pays inter.

A: Given: write a program c++ If money is left in a particular bank for more than 5 years, the bank pay.

Q: Write a function get_total_price(class_list) that takes in a list of class callnumbers and outputs t.

A: import catalog as catl#get total price methoddef get_total_price(class_list): #iterate through th.


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.


Watch the video: Binary Relations (December 2021).