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Binary Relations and Equivalence Relations, Study notes of Discrete Mathematics

Fayetteville State University (FSU)Discrete Mathematics

An introduction to binary relations, their properties, and the concept of equivalence relations. It covers reflexive, symmetric, and transitive properties, as well as partial and total orders. The document also includes examples and proofs for various relations and their properties.

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Pre 2010

Uploaded on 08/01/2009

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Ch. 3 - Relations
Sec. 3.1. Relations
• A binary relation R from a set X to a set Y
is a subset of the Cartesian product X Y.
• If (x, y) R, we write xRy and say that x is
related to y.
• If X=Y, R is a binary relation on X.
• The set {x X | (x, y) R for some y Y}
is the domain of R.
• The set {y Y | (x, y) R for some x X}
is the range of R.
• A function f from X to Y is a relation from X
to Y having the properties:
– The domain of f is equal to X; and
– For each x X, there is exactly one y Y
such that (x, y) f.
(ex) let X = {2, 3, 4} and Y = {3, 4, 5, 6, 7}. Defining
R: XY by (x, y) R if x divides y, we obtain R =
{(2,4), (2,6), (3,3), (3,6), (4,4)}. Rewriting R as a
table we have: X Y
2 4
2 6
3 3
3 6
4 4
The domain of R = {2,3,4} and the range = {3,4,6}.
(ex) let R be the relation on X = {1,2,3,4}
defined by (x, y) R if x y for x, y X.
Then R = {(1,1), (1,2), (1,3), (1,4), (2,2),
(2,3), (2,4), (3,3), (3,4), (4,4)}. The domain
of R = the range of R = X.
• We can picture a relation on a set with a
digraph.
• Steps:
(a) draw dots (or vertices) to represent the
elements of X.
(b) if (x, y) R, then draw an arrow (directed
edge) from x to y.
pf3
pf4
pf5

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Ch. 3 - Relations

Sec. 3.1. Relations

  • A binary relation R from a set X to a set Y is a subset of the Cartesian product X  Y.
  • If (x, y)  R, we write xRy and say that x is related to y.
  • If X=Y, R is a binary relation on X.
  • The set {x  X | (x, y)  R for some y  Y} is the domain of R.
  • The set {y  Y | (x, y)  R for some x  X} is the range of R.
  • A function f from X to Y is a relation from X to Y having the properties:
  • The domain of f is equal to X; and
  • For each x  X, there is exactly one y  Y such that (x, y)  f. (ex) let X = {2, 3, 4} and Y = {3, 4, 5, 6, 7}. Defining R: X Y by (x, y)  R if x divides y, we obtain R = {(2,4), (2,6), (3,3), (3,6), (4,4)}. Rewriting R as a table we have: X Y 2 4 2 6 3 3 3 6 4 4 The domain of R = {2,3,4} and the range = {3,4,6}. (ex) let R be the relation on X = {1,2,3,4} defined by (x, y)  R if x  y for x, y  X. Then R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)}. The domain of R = the range of R = X.
  • We can picture a relation on a set with a digraph.
  • Steps: (a) draw dots (or vertices ) to represent the elements of X. (b) if (x, y)  R, then draw an arrow ( directed edge ) from x to y.
  • A relation R on a set X is reflexive if (x, x)  R for every x  X. (ex) R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)} on X = {1, 2, 3, 4} is reflexive. (ex) R = {(1,2), (1,3), (1,4), (2,3), (2,4), (3,3), (3,4), (4,4)} on X = {1, 2, 3, 4} is not reflexive. - A relation R on a set X is symmetric if for every x, y  X, if (x, y)  R then (y, x)  R. (ex) R = {(a,b), (a,c), (a,d), (b,a), (b,c), (b,d), (c,a), (c,b), (c,d), (d,a), (d,b), (d,c)} on X = {a, b, c, d} is symmetric. (ex) R = {(a,a), (b,c), (c,b), (d,d)} on X = {a, b, c, d} is symmetric. (ex) R = {(a,a), (b,c), (d,d)} on X = {a, b, c, d} is not symmetric.
  • A relation R on a set X is antisymmetric if for every x, y  X, if (x, y)  R and x y, then (y, x)  R. (ex) R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)} on X = {1, 2, 3, 4} is antisymmetric but not symmetric. (ex) R = {(a,a), (b,c), (c,b), (d,d)} on X = {a, b, c, d} is symmetric but not antisymmetric. (ex) R = {(a,a), (b,b), (c,c), (d,d)} on X = {a, b, c, d} is symmetric and antisymmetric. (ex) R = {(a,a), (a,b), (b,c), (c,b), (d,d)} on X = {a, b, c, d} is not symmetric and not antisymmetric.
  • A relation R on a set X is transitive if for every x, y, z  X, if (x, y)  R and (y, z)  R, then (x, z)  R. (ex) R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)} on X = {1, 2, 3, 4} is transitive. (ex) R = {(a,a), (b,c), (c,b), (d,d)} on X = {a, b, c, d} is not transitive. (ex) R = {(a,a), (b,b), (b,c), (c,b), (c,c), (d,d)} on X = {a, b, c, d} is transitive.

Orders on Sets

  • Relations can be used to order the elements of a set.
  • A relation R on a set X is called a partial order if R is reflexive, antisymmetric and transitive.
  • If R is a partial order on a set X, we sometimes write x= y to indicate that (x, y)  R.
  • Let R be a partial order on set X. For x, y  X, if either xRy or yRx, then x and y are said to be comparable.
  • If x, y  X and (xRy) and (yRx), then x and y are incomparable. (ex) given R based on divides, 2 and 4 are comparable because (2,4)  R, i.e., 2 divides 4. (ex) given R based on divides, 2 and 3 are incomparable because (2,3)  R and (3,2)  R, i.e., 2 doesn’t divide 3 and 3 doesn’t divide 2. - Let R be a partial order on X. If every pair of elements in X is comparable, then R is a total order. (ex)  on the positive integers is a total order because for all positive integers x and y, either x  y or y  x. (ex) the divides relation on the positive integers is not a total order because for some positive integers x and y, x doesn’t divide y and y doesn’t divide x. (ex) Prove R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)} on X = {1, 2, 3, 4} is a total order. Proof: (1) R is a partial order (shown earlier). (2) 1R1, 1R2, 1R3, 1R4, 2R2, 2R3, 2R4, 3R3, 3R4 and 4R4.  for every x,y  X, either xRy or yRx.
  • Since R is a partial order and for every x,y  X, x and y are comparable, R is a total order. - Let R be a relation from X to Y. The inverse of R, R-^1 , is the relation from Y to X defined by R-^1 = {(y, x) | (x, y)  R}. (ex)  is the inverse relation of . (ex) R-^1 = {(1,1), (2,1), (2,2), (3,1), (3,2), (3,3), (4,1), (4,2), (4,3), (4,4)} is the inverse of R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)}
  • Let R 1 be a relation from X to Y and R 2 be a relation from Y to Z. The composition of R 1 and R 2 , denoted R 2 ° R 1 , is the relation from X to Z defined by: R 2 ° R 1 = {(x, z) | (x, y)  R 1 and (y, z)  R 2 for some y  Y}.

Composition of Relations

(ex) The composition of the relations R 1 = {(1,2), (1,6), (2,4), (3,4), (3,6), (3,8)} and R 2 = {(2,u), (4,s), (4,t), (6,t), (8,u)} is R 2 ° R 1 = {(1,u), (1,t), (2,s), (2,t), (3,s), (3,t), (3,u)}

Sec. 3.2 – Equivalence Relations

  • A partition of a set X is a collection S of nonempty subsets of X such that every element of X belongs to exactly one member of S.
  • Thm. 3.2.1 – Let S be a partition of a set X. Define xRy to mean that for some set s  S, both x and y belong to s, i.e., x and y belong to the same member of the partition S. Then R is reflexive, symmetric and transitive. Proof of Thm 3.2.1: (1) Let x  X. By the definition of a partition, x belongs to some s  S. Thus xRx and so R is reflexive. (2) Suppose xRy, then both x and y belong to some s  S. Therefore, yRx and so R is symmetric. (3) Suppose xRy and yRz. Then both x and y belong to some s  S, and both y and z belong to some t  S. But since y only belongs to one set in S, s = t.  both x and z belong to s and so xRz.  R is transitive.
  • A relation that is reflexive, symmetric

and transitive on a set X is an

equivalence relation on X.

(ex) Prove R = {(1,1), (1,3), (1,5), (2,2), (2,4), (3,1), (3,3), (3,5), (4,2), (4,4), (5,1), (5,3), (5,5)} on the set X = {1, 2, 3, 4, 5} is an equivalence relation on X. Proof: (1) 1R1, 2R2, 3R3, 4R4, 5R5.  R is reflexive. (2) 1R3 and 3R1, 1R5 and 5R1, 2R4 and 4R2, 3R5 and 5R3.  R is symmetric. (3) 1R3 and 3R5 and 1R5.  R is transitive.  R is an equivalence relation on X.

  • Thm 3.2.8 – Let R be an equivalence relation on a set X. For each a  X, let [a] = {x  X | xRa}, then S = {[a] | a  X} is a partition of X. The sets [a] are called the equivalence classes of X given by the relation R. (ex) There are two equivalence classes on X = {1, 2, 3, 4, 5} defined by R = {(1,1), (1,3), (1,5), (2,2), (2,4), (3,1), (3,3), (3,5), (4,2), (4,4), (5,1), (5,3), (5,5)}, namely, [1] = [3] = [5] = {1, 3, 5} and [2] = [4] = {2, 4}. These two classes partition X.