Determining Partial Order Relations in Discrete Mathematics, Lecture notes of Discrete Mathematics

Solutions to examples of determining whether given relations are partial order relations in the context of discrete mathematics. The examples involve defining relations on sets of integers and words in the english language, and analyzing their reflexivity, antisymmetry, and transitivity properties.

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Rosen, Discrete Mathematics and Its Applications, 6th edition
Extra Examples
Section 8.6—Partial Orderings
Page references correspond to locations of Extra Examples icons in the textbook.
p.566, icon at Example 1
#1. Let A={(x, y)|x, y integers}. Define a relation Ron Aby the rule
(a, b)R(c, d)acor bd.
Determine whether Ris a partial order relation on A.
Solution:
Ris reflexive: (a, b)R(a, b) for all elements (a, b) because aaor bbis always true.
Ris not antisymmetric: For example, (1,4)R(3,2) because 1 3, and (3,2)R(1,4) because 2 4. But
(1,4) =(3,2).
Ris not transitive: For example, (1,4)R(3,2) because 1 3, and (3,2)R(0,3) because 2 3. But
(1,4)R/ (0,3) because 1 ≤ 0and4≤ 3.
Therefore, Ris not a partial order relation because Ris neither antisymmetric nor transitive.
p.566, icon at Example 1
#2. Let A={(x, y)|x, y integers}. Define a relation Ron Aby the rule
(a, b)R(c, d)a=cor b=d.
Determine whether Ris a partial order relation on A.
Solution:
Ris reflexive: (a, b)R(a, b) for all elements (a, b) because a=aand b=bare always true.
Ris not antisymmetric: For example, (1,2)R(1,3) and (1,3)R(1,2) because 1 = 1, but (1,2) =(1,3).
Ris not transitive: For example, (1,2)R(1,3) because 1 = 1, and (1,3)R(4,3) because 3 = 3. But (1,2) =
(4,3) because 1 =4and2=3.
Therefore, Ris not a partial order relation because Ris neither antisymmetric nor transitive.
p.567, icon at Example 4
#1. Let Rbe the relation on the set of words in the English language where xRy if xprecedes (that is,
comes before) yin the dictionary. Show that Ris not a partial ordering.
Solution:
Note that Ris antisymmetric because if xprecedes yin the dictionary, where xand yare English words,
then ydoes not precede x.AlsonotethatRis transitive, for if xprecedes yin the dictionary and yprecedes
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Rosen, Discrete Mathematics and Its Applications, 6th edition

Extra Examples

Section 8.6—Partial Orderings — Page references correspond to locations of Extra Examples icons in the textbook.

p.566, icon at Example 1

#1. Let A = {(x, y) | x, y integers}. Define a relation R on A by the rule

(a, b)R(c, d) ↔ a ≤ c or b ≤ d.

Determine whether R is a partial order relation on A.

Solution: R is reflexive: (a, b)R(a, b) for all elements (a, b) because a ≤ a or b ≤ b is always true.

R is not antisymmetric: For example, (1, 4)R(3, 2) because 1 ≤ 3, and (3, 2)R(1, 4) because 2 ≤ 4. But (1, 4) = (3, 2).

R is not transitive: For example, (1, 4)R(3, 2) because 1 ≤ 3, and (3, 2)R(0, 3) because 2 ≤ 3. But (1, 4)R/ (0, 3) because 1 ≤ 0 and 4 ≤ 3.

Therefore, R is not a partial order relation because R is neither antisymmetric nor transitive.

p.566, icon at Example 1

#2. Let A = {(x, y) | x, y integers}. Define a relation R on A by the rule (a, b)R(c, d) ↔ a = c or b = d. Determine whether R is a partial order relation on A.

Solution: R is reflexive: (a, b)R(a, b) for all elements (a, b) because a = a and b = b are always true.

R is not antisymmetric: For example, (1, 2)R(1, 3) and (1, 3)R(1, 2) because 1 = 1, but (1, 2) = (1, 3).

R is not transitive: For example, (1, 2)R(1, 3) because 1 = 1, and (1, 3)R(4, 3) because 3 = 3. But (1, 2) = (4, 3) because 1 = 4 and 2 = 3.

Therefore, R is not a partial order relation because R is neither antisymmetric nor transitive.

p.567, icon at Example 4

#1. Let R be the relation on the set of words in the English language where xRy if x precedes (that is, comes before) y in the dictionary. Show that R is not a partial ordering.

Solution: Note that R is antisymmetric because if x precedes y in the dictionary, where x and y are English words, then y does not precede x. Also note that R is transitive, for if x precedes y in the dictionary and y precedes

1

z in the dictionary, where x, y, and z are English words, then x precedes z in the dictionary. However, R is not reflexive because no word precedes itself in the dictionary. This means that R is not a partial ordering.

p.574, icon at Example 20

#1. Referring to this Hasse diagram of a partially ordered set, find the following:

(a) all upper bounds of {d, e}. (b) the least upper bound of {d, e}. (c) all lower bounds of {a, e, g}. (d) the greatest lower bound of {a, e, g}. (e) greatest lower bound of {b, c, f }. (f) least upper bound of {h, i, j}. (g) greatest lower bound of {g, h}. (h) least upper bound of {f, i}.

Solution: (a) There are no upper bounds of {d, e}. (b) Because there are no upper bounds of {d, e}, there is no least upper bound of {d, e}. (c) The only lower bound of {a, e, g} is i. (d) The glb of {a, e, g} is the only lower bound of {a, e, g}, namely i. (e) Both h and i are lower bounds of {b, c, f }. But there is no greatest lower bound. (f) Both a and c are upper bounds of {h, i, j}. The element c is the least upper bound. (g) There is no lower bound of {g, h}. Hence there is no greatest lower bound. (h) The elements a, c, and f are upper bounds of {f, i}. The element f is the least upper bound.