Page 205 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Ordering Relations 191
picking out pairs (x, y) in which y is, in a certain sense, at least as large as
x, it does not satisfy the definition of partial order. This example shows that
our description of partial orders as relations that indicate a sense in which
one object is at least as large as another should not be taken too seriously.
This was the motivation for the definition of partial order, but it is not the
definition itself.
3. Clearly every positive integer is divisible by itself, so D is reflexive. Also, as
we showed in Theorem 3.3.6, if x | y and y | z then x | z. Thus, if (x, y) ∈ D
and (y, z) ∈ D then (x, z) ∈ D,so D is transitive. Finally, suppose (x, y) ∈
D and (y, x) ∈ D. Then x | y and y | x, and because x and y are positive it
follows that x ≤ y and y ≤ x,so x = y. Thus, D is antisymmetric, so it is
a partial order. It is easy to find examples illustrating that D is not a total
order. For example, (3, 5) /∈ D and (5, 3) /∈ D.
Perhaps you were surprised to discover that D is a partial order. It doesn’t
seem to involve comparing the sizes of things, like the other partial orders
we’ve seen. But we have shown that it does share with these other relations
theimportantpropertiesofreflexivity,transitivity,andantisymmetry.Infact,
this is one of the reasons for formulating definitions such as Definition 4.4.2.
They help us to see similarities between things that, on the surface, might
not seem similar at all.
4. You should be able to check for yourself that G is a total order. Notice that
in this case it seems more reasonable to think of xGy as meaning that y is
as least as small as x rather than at least as large. The definition of partial
order, though motivated by thinking about orderings that go in one direction,
actually applies to orderings in either direction. In fact, this example might
−1
lead you to conjecture that if R is a partial order on A, then so is R .You
are asked to prove this conjecture in exercise 13.

Here’s another example of a partial order. Let A be the set of all words
in English, and let R ={(x, y) ∈ A × A | all the letters in the word x ap-
pear, consecutively and in the right order, in the word y}. For example, (can,
cannot), (tar, start), and (ball, ball) are all elements of R, but (can, anchor)
and (can, carnival) are not. You should be able to check that R is reflexive,
transitive, and antisymmetric, so R is a partial order. Now consider the set
B ={me, men, tame, mental}⊆ A. Clearly many ordered pairs of words in B
are in the relation R, but note in particular that the ordered pairs (me, me), (me,
men), (me, tame), and (me, mental) are all in R. If we think of xRy as meaning
that y is in some sense at least as large as x, then we could say that the word
me is the smallest element of B, in the sense that it is smaller than everything
else in the set.