Relations

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# Relations - PowerPoint PPT Presentation

Relations. A binary relation R from a set A to a set B is a subset of A  B. Domain of R = { a  A | (a,b)  R for some b  B} Range of R = {b  B | (a,b)  R for some a  A} Student Course Adam CS271 Adam CS301 Bill Phy101 Bill CS271

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Relations
• A binary relation R from a set A to a set B is a subset of
• AB.
• Domain of R = { a  A | (a,b)  R for some b B}
• Range of R = {b  B | (a,b)  R for some a  A}
• Student Course
• Bill Phy101
• Bill CS271
• Beth Bio101
• Beth CS271

Beth CS301

Bill Phy101

Bio101

Relation on a set

A relation R on a set A is a relation from A to A.

Example: A = { 1, 2, 3, ,4}

R = { (x,y) | x,y  A, x is no greater than y and xy is a square}

R= { (1,1), (1,4), ( 2,2), (4,4)}

Also write xRy if (x,y) R.

Properties of a relation
• A relation R on a set A is
• reflextive: (x,x) R for all x A.
• symmetric: for all x,y  A, if (x,y) R then (y,x)  R.
• antisymmetric: all x,y  A, if (x,y) R and x y, then (y,x) R.
• transitive: for all x,y,z A, if (x,y) R and (y,z)  R, then (x,z)  R.
Examples

N = the set of natural numbers.

(1) R= { (x,y) | x,y N, x > y }

antisymm, trans.

(2) R={ (x,y) | x,y N, x is no less than y}

reflexive, antisymm, trans.

(3) R = { (x,y) | x,y N, xy is even}

reflexive, symm, trans.

Transitivity: for all x,y,z N, suppose

(x,y) R and (y,z) R.

Then

A relation that is symmetric and antisymmetric?

R = { (x,x) | xN}

Every relation on N that is both symmetric and antisymmetric must be a subset of R.

Enough to show that such a relation cannot contain (x,y) for xy.

Otherwise, R being symmetric implies (y,x) also in R, but since xy, antisymmerty is violated.

Partial order

A relation on a set which is reflexive, antisymmetric and transitive.

Total order: a partial order with respect to which every pair of elements is comparable.

Eg. Divide | : partial

: total

Given a set of partially ordered tasks.

To schedule the tasks without violating the partial order.

Eg. (1,2), (2,4), (3,5), (3,1)

3,1,2,5,4

Equivalence

A relation R on a set A is an equivalence relation if R is

reflexive, symmetric and transitive.

A partition on a set A is a collection  of nonempty subsets of A such that every element of A belongs to exactly one member of .

Example: ={O,E} where E is the set of even integers and O the set odd integers is a partition of the set of integers.

Example:  = { { 1,2,3}, {a,b,c}} is a partition of {1,a,2,b,3,c}.

Theorem:

Let  be a partition of a set A. Define a relation R on A such that for every pair x,yA, xRy if and only if x,y belong to the same member of . Then R is an equivalence relation.Proof:

For all xA, let S(x) be the member of  which contains x.

Reflexive: xRx for all xA, since S(x)=S(x).

Symmetric: xRy means S(x)=S(y). So S(y)=S(x), so yRx.

Transitive: if xRy and yRz, then S(x)=S(y) and S(y)=S(z), so S(x)=S(z), so xRz.

Theorem

Let R be an equivalence relation on a set A.

Define [a] = { xA | xRa} (the equivalence class of a). Then  = {[a] | aA} is a partition of A.

Proof

Since R is reflexive aRa for all aA, hence

[a] is nonempty for all aA.

Next show that every xA belong s to exactly one member of . That is, if x [a] and x[b], then [a]=[b].

Claim: aRb (and similarly bRa).

Pf: So suppose x [a] and x[b]. Then xRa and xRb.

Since R is symmetric, xRa implies aRx.

Now aRx and xRb, and since R is transitive, it follows that aRb.

Claim: [a]  [b]

Pf: For all y [a] we have yRa. Since aRb, and R is transitive, it follows that yRb, hence y[b]. Therefore [a]  [b].

Similarly, [b]  [a].

Therefore [a] = [b].

Example

R is the relation on the set of integers so that for

integers x and y, xRy iff x-y is even.

Reflexive: For every integer x, xRx since x-x = 0.

Symmetric: For integers x and y, if xRy, then x-y is even,

since x-y= -(y-x), y-x is also even, hence yRx.

Transitive: For integers x,y and z, if xRy and yRz, then

x-y and y-z are even, hence x-z = (x-y)+(y-z) is also even hence xRz.

Therefore R is an equivalence.

How many equivalence classes are there?

[0] = the set of even integers

[1] = the set of odd integers

Example:

R is the relation on the set of integers so that for

integers x and y, xRy iff 5 | x-y.

Reflexive: 5 | x-x for all integers x.

Symmetric: 5 | x-y implies 5 | y-x = -(x-y) for all integers x,y.

Transitive: for all integers x,y,z, if 5 | x-y and 5 | y-z

then 5 | (x-y)+(y-z) = x-z.

equivalence classes: [0], [1], [2], [3], [4]