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Chapter 7: Problem 21

Define the relation \(\mathscr{\text { on the set }} \mathbf{Z}\) by \(a \Re b\) if \(a-b\) is a nonnegative even integer. Verify that \(\mathscr{\text { defines a partial order for }} \mathbf{Z}\). Is this partial order a total order?

Short Answer

Expert verified
The relation \(\Re\), defined as 'a is related to b if the difference between a and b is a nonnegative even integer', is a partial order but not a total order.

Step by step solution

01

Prove Reflexivity

For \(a \Re a\), we get \(a - a = 0\), which is a nonnegative even number. Thus, the relation \(\Re\) is reflexive.
02

Prove Antisymmetry

Assuming both \(a \Re b\) and \(b \Re a\) are true, we get two conditions: \(a - b\) is a nonnegative even integer and \(b - a\) is a nonnegative even integer. Because the difference between two integers is even if and only if the two integers are the same, we have \(a = b\). Thus, \(\Re\) is antisymmetric.
03

Prove Transitivity

Assuming that \(a \Re b\) and \(b \Re c\), we get \(a - b\) is a nonnegative even integer and \(b - c\) is a nonnegative even integer. Adding these gives \(a - c\) = (a - b) + (b - c), which is the sum of two even numbers, and so is even. Thus, \(\Re\) is transitive.
04

Check for Total Order

For \(\Re\) to be a total order, for all \(a\) and \(b\) where \(a \neq b\), \(a \Re b\) or \(b \Re a\) needs to be true. Let's assume \(a = 1\) and \(b = 2\). Neither \(1 \Re 2\) nor \(2 \Re 1\) is true, since neither 1 - 2 nor 2 - 1 are nonnegative even numbers. Therefore, \(\Re\) is not a total order.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reflexivity
In mathematics, a relation is reflexive if every element is related to itself. Think of it like looking in a mirror; you always see yourself. For a relation \( \Re \) on a set \( \mathbf{Z} \), the test for reflexivity is to check whether, for every integer \( a \), \( a \Re a \) holds true.

In the given exercise, we consider \( a \Re a \) which means \( a - a = 0 \). The result zero is not only nonnegative but also an even number. Hence, the relation \( \Re \) is reflexive.

Reflexivity is a fundamental requirement for a partial order because it ensures every element maintains some relationship with itself.
Antisymmetry
Antisymmetry in a relation means that if two elements are related in both directions, then they must be identical. For our relation \( \Re \), if \( a \Re b \) and \( b \Re a \) both hold, it implies \( a = b \). Let's break it down:
  • Condition \( a \Re b \): the difference \( a-b \) is a nonnegative even number.
  • Condition \( b \Re a \): the difference \( b-a \) is also a nonnegative even number.
If both conditions are true, it means both \( a-b \) and \( b-a \) are even. The only integer solution that satisfies this is when \( a = b \).

Therefore, \( \Re \) satisfies antisymmetry. Antisymmetry is crucial in partial orders as it prevents two distinct elements from being related symmetrically both ways.
Transitivity
Transitivity connects the relationship between three elements through a chain-like rule. For a relation \( \Re \) to be transitive, whenever \( a \Re b \) and \( b \Re c \) hold true, \( a \Re c \) must also be valid. It goes like this:
  • Given \( a \Re b \), with \( a-b \) being a nonnegative even number.
  • Given \( b \Re c \), with \( b-c \) also a nonnegative even number.
The sum, \( (a-b) + (b-c) = a-c \), results in a nonnegative even number. Hence, \( a \Re c \) holds true.

Transitivity helps to build a "chain reaction" in relations, ensuring the consistency of order across multiple elements. This property is necessary for reflexive and antisymmetric relations to define a partial order.

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Most popular questions from this chapter

Let \(n \in \mathbf{Z}^{+}\)with \(n>1\), and let \(A\) be the set of positive integer divisors of \(n\). Define the relation \(\Re\) on \(A\) by \(x \Re y\) if \(x\) (exactly) divides \(y\). Determine how many ordered pairs are in the relation 9 when \(n\) is (a) 10 ; (b) 20 ; (c) 40 ; (d) 200 ; (e) 210 ; and, (f) 13860 .

a) For \(q u=\\{1,2,3\\}\), let \(A=\mathscr{( q u )}\). Define the relation \(\mathscr{R}\) on \(A\) by \(B \mathscr{C}\) if \(B \subseteq C\). How many ordered pairs are there in the relation \(\mathscr{R} ?\) b) Answer part (a) for \(q=\\{1,2,3,4\\}\). c) Generalize the results of parts (a) and (b).

Let \((A, \mathscr{})\) be a totally ordered poset. If for all \(\emptyset \neq B \subseteq A\) the (totally ordered) poset \((B,(B \times B) \cap 9 R)\) has a least element, then \((A, \Re)\) is called well ordered. (We saw this idea in Section 4.1, where we used the well-ordering of \(\left(\mathbf{Z}^{+}, \leq\right)\)to establish the principle of mathematical induction.) For each of the following (totally ordered) posets, determine whether the poset is well ordered. a) \((\mathrm{N}, \leq)\) b) \((\mathbf{Z}, \leq)\) c) \((Q, S)\) d) \(\left(\mathbf{Q}^{+}, \mathbf{S}\right)\) e) \((P, \leq)\), where \(P\) is the set of all primes f) \((A, \leq)\), where \(A\) is a nonempty subset of \(\mathbf{Z}^{+}\) g) \((A, \leq)\), where \(\boldsymbol{0} \neq A \subset \mathbf{Z}\), and \(A\) is finite

For sets \(A, B\), and \(C\) with relations \(\mathscr{R}_{1} \subseteq A \times B\) and \(\mathscr{R}_{2} \subseteq B \times C\), prove or disprove that \(\left(\Re{R}_{1} \circ \mathscr{R}_{2}\right)^{c}=\mathscr{R}_{2}^{c} \odot \mathscr{R}_{1}^{c}\).

For \(A=\\{(-4,-20),(-3,-9),(-2,-4),(-1,-11),(-1,-3),(1,2),(1,5),(2,10)\), \((2,14),(3,6),(4,8),(4,12)\\}\), define the relation \(\Re\) on \(A\) by \((a, b) \Re R(c, d)\) if \(a d=b c\). a) Verify that \(\mathscr{\text { is an equivalence relation on }} A\). b) Find the equivalence classes \([(2,14)],[(-3,-9)]\), and \([(4,8)]\). c) How many cells are there in the partition of \(A\) induced by \(\mathscr{\text { ? }}\)
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