Key ideas from section 6.3: reflexive, antisymmetric,
transitive; partial ordering, poset, Hasse diagram;
symmetric, equivalence relation
In the divisibility poset D6, the relation “divides” (“|”)
consists of the ordered pairs (1,1), (1,2), (1,3), (1,6), (2,2),
(2,6), (3,3), (3,6), and (6,6).
More generally, for any positive integer n, the set Dn of
divisors of n becomes a poset with the relation “|”.
Definition: If x and z are elements of a poset L, we say z
covers x iff x < z and there does not exist any y such that x
< y and y < z.
Example: In D6, 6 covers 2 and 3 but not 1.
Theorem: In a finite poset L, x < y iff and only if there
exists a sequence w1, w2, …, wk of elements of L with w1 =
x and wk = y, where w1 is covered by w2, which is covered
by w3, …, which is covered by wk–1.
So we can reconstruct the partial order on L if we know
which elements of L cover which other elements of L.
In a Hasse diagram, we draw an edge joining x and y iff y
covers x, and we draw y above x.
Group work: determine the poset D4 as a set of ordered
pairs and as a directed graph, and draw its Hasse diagram.
Let L be a poset and a,c be elements of L. We say that c
covers a iff there exists no b in L such that a ≤ b ≤ c.
The Hasse diagram consists of all edges {a,c} where c
covers a; c is drawn above a.
[Draw the Hasse diagram for D6. Also draw the Hasse
diagram for Pow({1,2}) shown in Figure 6.3.6 and note
that they are the same picture: only the labels have
changed.]
In Discrete Structures II we’ll learn to say that these
partially ordered sets are isomorphic.
Draw the Hasse diagram of D6 and compare with the Hasse
diagram in Figure 6.3.6.
Why is the divisibility relation a partial ordering on the
positive integers?
Recall that a|b iff there exists an integer x such that ax=b.
Let’s restrict to the case where a and b are positive integers,
so that the integer x (when it exists) must be positive too.
“|” is reflexive: ax=a always has the solution x=1.
“|” is antisymmetric: If ax=b and by=a for positive integers
x and y, then a=by=axy, so xy=1, implying x=y=1; this
implies a=b.
“|” is transitive: If ax=b and by=c for positive integers x
and y, then xy is a positive integer satisfying a(xy) = (ax)y
= by = c, so a divides c.
Questions on section 6.3?
Group work: 6.3.1, 6.3.3, 6.3.5
Group work: Prove that congruence modulo m is transitive
Other questions on section 6.3?
Key ideas from section 6.4: adjacency matrix of a relation;
Boolean arithmetic (write “⋅ ” and “+” for “and” and “or”)
[Draw the Boolean matrix for D6.]
Questions on section 6.4?
Let r be the relation “divides” from the set {1,2} to the set
{3,4}, and let s be the relation “divides” from the set {3,4}
to the set {5,6}, so that rs is the composition of the two
relations.
Then the Boolean matrix for the relation rs is equal to the
product of the Boolean matrix for the relation r and the
Boolean matrix for the relation s.
In fact, for ANY relation r from a set A to a set B, and
ANY relation s from a set B to a set C, the Boolean matrix
for rs is equal to the product of the (rectangular) Boolean
matrix for r and the (rectangular) Boolean matrix for s, as
long as we use a modified version of arithmetic in which
1+1=1 rather than 2 (the usual value) or 0 (the mod-2
value).
This seems less bizarre if we think about 0 and 1 as
representing “false” and “true”; then multiplication
corresponds to “and”, addition corresponds to “or”, and our
bizarre formula “1+1=1” corresponds to the assertion “true
∨ true = true”.
Group work: 6.4.1, 6.4.5
Other questions on section 6.4?
Return HW and exams; collect section summaries and HW